gfale  Bicentennial  Duplications 


THE  MECHANICS  OF  ENGINEERING 


THE 


MECHANICS    OF    ENGINEERING 


VOLUME  I 

KINEMATICS,  STATICS,  KINETICS,  STATICS  OF 
RIGID  BODIES  AND  OF  ELASTIC  SOLIDS 


BY 

A.   JAY    DuBOIS,    C.E.,    PH.D. 

PROFESSOR  OF  CIVIL  ENGINEERING  IN  THE  SHEFFIELD  SCIENTIFIC  SCHOOL 
OF   YALE   UNIVERSITY 


FIJtST     EDITION 
FIRST    THOUSAND 


NEW   YORK 

JOHN  WILEY   &   SONS 

LONDON:    CHAPMAN  &    HALL,   LIMITED 

I9°5 


Copyright,  1902, 

BY 
A.  JAY  DuBOIS. 


Librar 


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PREFACE. 


THIS  work  is  presented  as  the  first  volume  of  a  series  dealing  with  the  applications  of 
Mechanics  to  engineering  problems. %  It  is  issued  uniform  with  "  Stresses  in  Framed  Struc- 
tures "  by  the  same  author,  which  will  be  revised  as  soon  as  possible,  so  as  to  form  the 
second  volume  of  the  series. 

The  present  volume  opens  with  a  carefully  considered  presentation  of  the  fundamental 
principles  of  Kinematics,  Statics,  and  Kinetics.  Then  follow  in  proper  order  the  practical 
applications.  Throughout  the  work  numerous  problems  and  illustrative  examples  are  given 
in  direct  connection  with  each  important  mechanical  principle. 

The  method  of  presentation  is  the  result  of  much  thought  and  teaching  experience. 
The  author  believes  that  this  method  will  commend  itself  to  teacher  and  student.  Thus  the 
first  chapters  are  devoted  to  the  preliminary  discussions  of  mass  and  space  only,  without  refer- 
ence to  time,  motion,  or  force.  Under  this  head  are  treated  not  only  mass,  density,  and 
centre  of  mass,  but  also  moment  of  inertia,  considered  simply  as  a  mass  and  space  quantity. 
The  experience  of  the  author  has  convinced  him  of  the  advantages  of  this  method  of  presen- 
tation. The  student  in  his  preparatory  study  of  Geometry  has  already  become  conversant 
with  those  space  relations  of  which  he  must  make  future  use.  He  should  now  become  equally 
familiar  with  those  mass  and  space  relations  of  which  he  must  also  make  future  use,  and  it  is 
both  proper  and  advantageous  that  this  subject  should  be  treated,  like  Geometry,  as  a  prepara- 
tory study.  Then  the  student  is  in  a  position  to  take  up  not  only  the  subject  of  Kinematics, 
which  deals  with  relations  of  space  and  motion,  but  also  the  subject  of  Kinetics,  which  deals 
with  mass,  space,  and  motion. 

In  the  chapters  on  Kinematics  and  Kinetics  the  method  of  presentation  has  been  much 
abridged  from  the  author's  "  Elementary  Principles  of  Mechanics, "  and  will  be  found  simpler, 
more  logical  and  direct.  The  numerous  examples  introduced  here  will  be  found  of  service 
both  to  teacher  and  student. 

In  the  applications  the  author  has  included  the  results  of  his  own  work  in  this  direction, 
and  he  believes  that  the  professional  reader  will  find  here  new  and  valuable  discussions  of 
engineering  problems,  especially  in  the  chapters  on  Masonry  Walls  and  Dams,  the  Strength  of 
Long  Columns,  the  Swing  Bridge,  the  Metal  Arch,  the  Suspension  System,  and  the  Stone  Arch. 
In  all  these  chapters  the  entire  treatment  and  many  results  are  different  from  those  already 
given  by  the  author  in  "Stresses  in  Framed  Structures"  and  "Elementary  Principles  of 
Mechanics."  By  the  application  of  the  principle  of  least  work  many  new  and  simple  results 
and  methods  have  been  obtained.  Especially  is  this  the  case  as  regards  the  stone  arch. 
Treatises  on  this  subject  are  prolix,  overburdened  with  mathematical  discussions  and  involved 
formulae,  limited  at  best  in  their  application,  diverse  in  their  assumptions,  and  discouraging 

ix 


x  PREFACE. 

alike  to  the  student  and  engineer.     The  method  of  solution  here  given  is  simple,  direct  and 
practical,  and  adapts  itself  with  equal  ease  to  any  form  of  arch  and  any  surcharge. 

The  size  of  this  volume  is  due  to  its  scope  and  plan.  It  treats  with  thoroughness  and 
with  direct  reference  to  practice,  of  several  topics  for  which  the  student  must  otherwise  have 
recourse  to  separate  treatises.  It  thus  offers  in  one  volume  and  in  proper  sequence  Courses 
on  Elementary  Mechanics,  Kinematics,  Statics,  Kinetics,  Framed  Structures,  Graphical 
Statics.  Walls  and  Dams,  Retaining  Walls,  Mechanics  of  Materials,  Swing  Bridge,  Metal 
and  Stone  Arches,  and  Suspension  System. 


GENERAL    CONTENTS. 


INTRODUCTION. 


CHAPTER  I. 

PAGE 

Mechanics.     Kinematics.     Dynamics.     Statics.      Kinetics.      Measurement i 

CHAPTER  II. 
Position.      Terms  and  Definitions.  .  u 


MEASURABLE  RELATIONS  OF  MASS  AND  SPACE. 


CHAPTER  I. 
Mass  and  Density 14 

CHAPTER  II. 
Centre  of  Mass 20 

CHAPTER    III. 
Moment  of  Inertia.  .  .  .  ; 31 


KINEMATICS  OF  A  POINT.     GENERAL  PRINCIPLES. 


CHAPTER  I. 
Linear  and  Angular  Displacement 51 

CHAPTER  II. 
Resolution  and  Composition  of  Linear  Displacements 54 

CHAPTER  III. 
Resolution  and  Composition  of  Angular  Displacements 58 


xii  GENERAL   CONTENTS. 

CHAPTER  IV. 

PACK 

Linear  and  Angular  Speed  and  Velocity 6 1 

CHAPTER   V. 
Linear  and  Angular  Velocity 65 

CHAPTER  VI. 
Linear  and  Angular  Rate  of  Change  of  Speed.     Linear  Acceleration 73 

CHAPTER  VII. 
Angular  Acceleration v 82 

CHAPTER  VIII. 
Moments.      Moment  of  Displacement,  Velocity  and  Acceleration 88 


KINMEATICS  OF  A  POINT.     APPLICATION  OF  PRINCIPLES. 


CHAPTER   I. 
Motion  of  a  Point.      Constant  and  Variable  Rate  of  Change  of  Speed 91 

CHAPTER  II. 
Uniform  Acceleration 100 

CHAPTER  III. 

Motion    under  Variable  Acceleration  in  General.       Central  Acceleration.       Central  Acceleration 

inversely  as  the  Square  of  the  Distance 109 

CHAPTER   IV. 
Central  Acceleration  directly  as  the  Distance.      Harmonic  Motion t 125 

CHAPTER  V. 

Constrained  Motion  of  a  Point 136 


KINEMATICS  OF  A  RIGID  BODY. 


CHAPTER  I. 
Angular  Velocity  and  Acceleration  Couples.     Angular  and  Linear  Velocity  Combined 143 

CHAPTER   II. 
Rotation  and  Translation.      Analytic  Relations 153 


GENERAL  CONTENTS.  xii« 

DYNAMICS.      GENERAL  PRINCIPLES. 


CHAPTER   I. 

PAGE 

Force.     Newton's  Laws  of  Motion 1 69 

CHAPTER  II. 
Resolution  and  Composition  of  Forces 176 

CHAPTER   III. 
Moment  of  a  Forcer     Resolution  and  Composition  of  Moments 181 

CHAPTER  IV. 
Force  Couples.      Effect  of  Force  Couple  on  a  Rigid  Body 185 

CHAPTER  V. 
Centre  of  Parallel  Forces.     Centre  of  Mass 189 

CHAPTER  VI. 
Non-Concurring  Forces  in  General.     Analytic  Equations 192 

CHAPTER  VII. 
Force  of  Gravitation.     Centre  of  Gravity 203 


STATICS.     GENERAL  PRINCIPLES. 


CHAPTER  I. 
Equilibrium  of  Forces.      Determination  of  Mass 209 

CHAPTER    II. 
Work.      Virtual  Work.  . 


215 

CHAPTER  III. 


Static  Friction. 


KINETICS  OF  A  PARTICLE. 


CHAPTER  I.                               : _  ;.  , 
Deflecting  Force 242 

CHAPTER  II. 
Tangential  Force.      Momentum.      Impulse 255 


xiv  GENERAL   CONTENTS. 

CHAPTER   III. 

rxc« 

Work.     Power 260 

CHAPTER  IV. 
Kinetic  Friction 263 

CHAPTER   V. 

Kinetic  and  Potential  Energy.      Law  of  Energy.       Conservation  of  Energy.       Equilibrium  of  a 

Particle 271 

CHAPTER  VI. 
The  Potential. .  .    286 


KINETICS  OF  A  MATERIAL  SYSTEM. 


CHAPTER  I. 
General  Principles 297 

CHAPTER  II. 
Equilibrium  of  a  Material  System 310 

CHAPTER  III. 
Rotation  about  a  Fixed  Axis 315 

CHAPTER   IV. 

Compound    Pendulum.     Centre  of  Oscillation  and   Percussion.     Experimental  Determination  of 

Moment  of  Inertia 336 

CHAPTER  V. 
Impact 342 

CHAPTER  VI. 
Rotation  about  a  Translating  Axis 367 

CHAPTER  VII. 
Rotation  about  a  Fixed  Point 375 

CHAPTER  VIII. 
Rotation  and  Translation 389 


GENERAL  CONTENTS. 


STATICS  OF  RIGID  BODIES. 


CHAPTER  I. 

PACK 

Framed  Structures 397 

CHAPTER  II. 
Graphical  Statics.      Concurring  Forces 404 

CHAPTER   III. 
Graphical  Statics.     Non -Concurring  Forces 412 

CHAPTER    IV. 
Walls.      Masonry  Dams 424 

CHAPTER  V. 
Retaining  Walls.      Earth  Pressure.      Equilibrium  of  Earth 454 


STATICS  OF  ELASTIC  SOLIDS. 


CHAPTER  I. 
Elasticity  and  Strength  for  Tension,  Compression  and  Shear 473 

CHAPTER  II. 
Strength  of  Pipes  and  Cylinders.      Riveting 486 

CHAPTER  III. 
Strength  of  Beams.     Torsion 493 

CHAPTER   IV. 

Work    of   Straining.      Deflection   of  Framed   Structures.      Principle  of  Least  Work.      Redundant 

Members.      Beams  fixed  Horizontally  at  the  Ends 515 

CHAPTER   V. 
Deflection  of  Beams 530 

CHAPTER  VI. 
Shearing  Stress 553 

CHAPTER   VII. 
Strength  of  Long  Columns 559 


xvi  GENERAL  CONTENTS. 

CHAPTER  VIII. 

FACE 

The  Pivot  or  Swing  Bridge 571 

CHAPTER  IX. 
The  Metal  Arch 578 

CHAPTER  X. 
The  Stone  Arch 597 

CHAPTER   XL 
The  Suspension  System ,  ...      609 


TABLE   OF   CONTENTS. 


INTRODUCTION. 


CHAPTER   I. 

MECHANICS.     KINEMATICS.     DYNAMICS.     STATICS.     KINETICS.     MEASUREMENT. 

PACK 

Physical  Science i 

Mechanics.      Kinematics.      Dynamics.     Statics.      Kinetics i 

Measurement.      Unit.     Statement  of  a  quantity.      Derived  unit 2 

Homogeneous  equations - 3 

Unit  of  time,  length,  mass 4 

Unit  of  angle.      Radian 6 

Unit  of  conical  angle.     Square  radian.      Curvature 7 

Table  of  measures , 9 

CHAPTER   II. 

POSITION.     TERMS  AND   DEFINITIONS. 

Point.      Point  of  reference.      Position  of  a  point n 

Plane  polar  co-ordinates.      Space  polar  co-ordinates 1 1 

Cartesian  co-ordinates 12 

Direction  cosines 13 


MEASURABLE  RELATIONS  OF  MASS  AND  SPACE. 


CHAPTER   I. 
MASS  AND   DENSITY. 

Mass  and  space    relations.     Mass.     Unit  of  mass.      Measurement  of  mass 14 

Mass  independent  of  gravity 14 

Notation  for  mass.      Density 15 

Unity  of  density.      Specific  mass , 16 

Determination  of  specific  mass.      Table  of  specific  mass 17 


xvni  TABLE  OF  CONTENTS. 

CHAPTER    II. 

CENTRE  OF  MASS. 

PACK 

Elementary  mass  or  particle 20 

Material  surface,  line.      Centre  of  mass 20 

Position  of  centre  of  mass  in  general 21 

Moment  of  mass,  volume,  area,  line 22 

Centre  of  gravity 22 

Plane  and  axis  of  symmetry. ......    23 

Material  line,  area,  volume 23 

Determination  of  centre  of  mass 24 

CHAPTER    IIL 
MOMENT   OF  INERTIA. 

Moment  of  inertia 31 

Radius  of  gyration 32 

Reduction  of  moment  of  inertia 32 

Moment  of  inertia  relative  to  an  axis 33 

Polar  moment  of  inertia  for  a  plane  area 33 

Moment  of  inertia  relative  to  a  point 34 

Moment  of  inertia  for  any  axis  in  general 34 

Ellipsoid  of  inertia 34 

Principal  axes 35 

Determination  of  moment  of  inertia 38 


KINEMATICS  OF  A  POINT.     GENERAL  PRINCIPLES. 


CHAPTER   I. 
LINEAR   AND  ANGULAR  DISPLACEMENT. 

Kinematics 51 

Path  of  a  point 51 

Angle  described  by  a  point , 51 

Linear  and  angular  displacement 52 

Line  representative  of  linear  and  angular  displacement 52 

Vector  quantities 52 

Displacement  in  general 52 

CHAPTER    II. 
RESOLUTION  AND   COMPOSITION  OF   LINEAR   DISPLACEMENTS. 

Resolution  and  composition  of  linear  displacements 54 

Trian^  ic  and  polygon  of  linear  displacements 54 

Rectangular  components 55 

Components  of  resultant 55 

Relative  displacement 55 


TABLE  OF  CONTENTS.  xix 

PACK 

Notation 56 

Triangle  and  polygon  of  relative  displacements 56 

CHAPTER   III. 
RESOLUTION  AND   COMPOSITION   OF  ANGULAR  DISPLACEMENTS. 

Resolution  and  composition  of  finite  successive  angular  displacements 58 

Resolution  and  composition  of  angular  displacements  in  general 59 

CHAPTER   IV. 
LINEAR  AND   ANGULAR   SPEED  AND  VELOCITY. 

Mean  linear  and  angular  speed 61 

Mean  linear  and  angular  velocity 61 

Line  representatives  of  mean  linear  and  angular  velocity 61 

Distinction  between  mean  speed  and  velocity 62 

Instantaneous  linear  and  angular  velocity 63 

CHAPTER   V. 

LINEAR   AND   ANGULAR   VELOCITY. 

Speed  and  velocity  in  general 65 

Uniform  and  variable  linear  velocity 65 

Uniform  and  variable  angular  velocity 65 

Resolution  and  composition  of  linear  velocity 66 

Rectangular  components  of  linear  velocity 67 

Analytic  determination  of  resultant  linear  velocity 67 

Resolution  and  composition  of  angular  velocity 68 

Rectangular  components  of  angular  velocity 69 

Analytic  determination  of  resultant  angular  velocity 69 

Linear  in  terms  of  angular  velocity 70 

CHAPTER   VI. 

LINEAR   AND   ANGULAR   RATE   OF   CHANGE   OF   SPEED.     LINEAR   ACCELERATION. 

Change  of  speed 73 

Instantaneous  rate  of  change  of  speed 73 

Mean  linear  acceleration 75 

Instantaneous  linear  acceleration 75 

Resolution  and  composition  of  linear  acceleration 76 

Rectangular  components  of  acceleration 77 

Analytic  determination  of  resultant  acceleration 77 

Tangential  and  central  acceleration 77 

Uniform  and  variable  acceleration 78 

The  Hodograph 79 

Axial  and  radial  or  deflecting  and  deviating  acceleration 80 

CHAPTER   VII. 
ANGULAR   ACCELERATION. 

Mean  angular  acceleration 82 

Instantaneous  angular  acceleration 83 


xv  TABLE  Of  CONTENTS. 

FAGR 

Resolution  and  composition  of  angular  acceleration 83 

Rectangular  components  of  angular  acceleration 83 

Analytic  determination  of  resultant  angular  acceleration     83 

Axial  and  normal  angular  acceleration 83 

Uniform  and  variable  angular  acceleration 84 

Linear  in  terms  of  angular  acceleration 84 

Homogeneous  equations 86 

CHAPTER   VIII. 
MOMENTS.     MOMENT  OF   DISPLACEMENT,    VELOCITY,    ACCELERATION. 

Moment  of  a  vector  quantity 88 

Resolution  and  composition  of  moments 88 

Moment  about  an  axis 88 

Moment  of  resultant 89 

Moment  of  displacement 89 

Moment  of  velocity 89 

Moment  of  acceleration 89 

Moment  of  angular  velocity 90 

Moment  of  angular  acceleration 90 


KINEMATICS    OF   A  POINT.     APPLICATION    OF    PRINCIPLES. 


CHAFFER   I. 
MOTION   OF  A   POINT.     CONSTANT   AND   VARIABLE   RATE   OF   CHANGE   OF   SPEED. 

Rate  of  change  of  speed  zero 91 

Rate  of  change  of  speed  constant 91 

Rate  of  change  of  speed  variable 92 

Graphic  representation  of  rate  of  change  of  speed 93 

Rate  of  change  of  angular  speed  zero 96 

Rate  of  change  of  angular  speed  uniform 96 

Rate  of  change  of  angular  speed  variable 97 

Graphic  representation  of  rate  of  change  of  angular  speed 98 

CHAPTER   II. 
UNIFORM   ACCELERATION. 

Uniform  acceleration,  motion  in  a  straight  line 100 

Value  of  g ioo 

Body  projected  vertically  up  or  down 101 

Uniform  acceleration — motion  in  a  curve 103 

"                   "          "  "      "     equation  of  the  path 103 

"               "                   "          ""     "     velocity  in  the  path ; 104 

"               "                  "          "  "     "     time  of  flight  and  range 105 

"               "                   "          "  "     "     displacement 105 

"                   "          "."     "     angle  of  elevation 105 

"               "                   "          "  "     "     envelope  of  all  trajectories 106 


TABLE  OF  CONTENTS.  xxi 

CHAPTER    III. 

MOTION  UNDER  VARIABLE  ACCELERATION  IN  GENERAL.   CENTRAL  ACCELERATION.   CENTRAL 
ACCELERATION  INVERSELY  AS  THE  SQUARE  OF  THE  DISTANCE. 

PAGE 

Motion  under  variable  acceleration  in  general 1 09 

Central  acceleration m 

Central  acceleration  inversely  as  the  square  of  the  distance 112 

"                "                 "           "    "        "       "    "          "       path  rectilinear n^ 

"               "                 "           "    "        "      "    "          "       curved  path 114 

Planetary  motion.      Kepler's  laws 120 

Velocity  of  a  planet 121 

Periodic  time 122 

Value  ofy^  for  planetary  motion 123 

CHAPTER    IV. 
CENTRAL   ACCELERATION   DIRECTLY   AS   THE   DISTANCE.      HARMONIC    MOTION. 

Harmonic  motion 125 

Simple  harmonic  motion 125 

Periodic  time 127 

Epoch,  phase 128 

Compound  harmonic  motion 129 

Resolution  and  composition  of  harmonic  motions 129 

Graphic  representation 133 

Blackburn's  pendulum 135 

CHAPTER   V. 
CONSTRAINED   MOTION   OF   A   POINT. 

Motion  on  an  inclined  plane,  uniform  acceleration 136 

Motion  in  a  curved  path,  uniform  acceleration 137 

Motion  in  a  circle,  uniform  acceleration 138 

Motion  in  a  cycloid,  uniform  acceleration 139 


KINEMATICS  OF  A   RIGID   BODY. 


CHAPTER    I. 

ANGULAR   VELOCITY   AND   ACCELERATION   COUPLE.       ANGULAR   AND   LINEAR    VELOCITY 

AND  ACCELERATION   COMBINED. 

Angular  velocity  couple 143 

Moment  of  anguW  velocity  couple 143 

Angular  acceleration  couple . 1 44 

Angular  and  linear  velocity  combined 144 

Instantaneous  axis  of  rotation 145 

Spin.      Screw-spin 145 

Spontaneous  axis  of  rotation 145 

Angular  and  linear  acceleration  combined z^0 


xxii  TABLE  OF  CONTENTS. 

PAGE 

Instantaneous  axis  of  acceleration 151 

Twist.      Screw-twist 152 

Spontaneous  axis  of  acceleration 152 

CHAPTER   II. 
ROTATION  AND   TRANSLATION.      ANALYTIC   RELATIONS. 

Components  of  motion 153 

Components  of  change  of  motion 153 

Motion  of  a  point  of  a  rigid  body.     General  analytic  equations 153 

Rotation,  centre  of  mass  fixed 154 

Rotation,  any  point  fixed 154 

Rotation,  translating  axis 154 

Resultant  velocity 154 

Resultant  angular  velocity 155 

Moment  of  velocity 155 

Velocity  along  axis 155 

Velocity  normal  to  axis 155 

Instantaneous  axis  of  rotation 156 

Invariant  for  components  of  motion 157 

Change  of  motion  of  a  point  of  a  rigid  body.     Analytic  equations 157 

Tangent  acceleration 157 

Central  acceleration 158 

Deflecting  and  deviating  acceleration 159 

Resultant  acceleration 159 

Resultant  angular  acceleration 159 

Moment  of  acceleration 1 60 

Acceleration  along  axis 160 

Acceleration  normal  to  axis 1 60 

Instantaneous  axis  of  acceleration 1 6 1 

Invariant  for  components  of  change  of  motion 1 6 1 

Euler's  geometric  equations 167 


DYNAMICS.     GENERAL  PRINCIPLES. 


CHAPTER   I. 
FORCE.     NEWTON'S  LAWS  OF   MOTION. 

Dynamics 169 

Material  particle ! 169 

Impressed  force 1 69 

Newton's  first  law  of  motion 169 

Inertia 169 

Force  proportional  to  acceleration 170 

Force  proportional  to  mass 170 

Unit  of  force 170 

Weight  of  a  body 170 


TABLE  OF  CONTENTS.  xxiii 

PACK 

Gravitation  unit  of  force 171 

Momentum  and  impulse 172  i, 

Newton's  second  law  of  motion 172  \, 

Measurement  of  mass ' 173 

Mass  independent  of  gravity *.  .  173 

Notation  for  mass .  . '. 1 74 

Newton's  third  law  of  motion 1 74 

Remark's  on  Newton's  Laws 174 

Stress 174 

Motion  of  centre  of  mass 174 


CHAPTER    II. 
RESOLUTION  AND   COMPOSITION  OF   FORCES. 

Line  representative  of  force 176 

Resolution  and  composition  of  forces 176 

Examples 176 

CHAPTER    III. 
CONCURRING   FORCES.     MOMENT   OF  A  FORCE.     TWO   NON-CONCURRING   FORCES. 

Concurring  forces 179 

Resultant  for  concurring  forces , 179 

Analytic  determination  of  resultant 179 

Moment  of  a  force • 181 

Line  representative  of  moment 181 

Resolution  and  composition  of  moments 181 

Significance  of  force  moment 182 

Unit  of  force  moment 182 

Resultant  of  two  non-concurring  forces 182 

Resultant  of  two  parallel  forces 183 

CHAPTER    IV. 

FORCE  COUPLES.     EFFECT   OF   FORCE  COUPLE  ON   RIGID   BODY. 

Force  couple ' 185 

Moment  of  force  couple 185 

Line  representative  of  force  couple 185 

Resolution  and  composition  of  couples 186 

Resultant  of  force  couple 1 86 

Effect  of  force  couple  on  a  rigid  body 187 

Resultant  for  non-concurring  forces  acting  on  a  rigid  body.  . 187 

CHAPTER   V. 
CENTRE   OF   PARALLEL   FORCES.     CENTRE   OF   MASS. 

Centre  of  parallel  forces 189 

Properties  of  centre  of  mass 190 


xxiv  TABLE  OF  CONTENTS. 

CHAPTER    VI. 

NON-CONCURRING   FORCES  IN  GENERAL.     ANALYTIC   EQUATIONS. 

»AG« 

Resultant  for  non-concurring  forces 1 92 

Effect  of  any  system  of  forces  acting  on  a  rigid  body 192 

Wrench*.     Screw-wrench 193 

Dynamic  components  ot  motion 193 

General  analytic  equations 194 

Resultant  force  at  centre  of  mass 194 

Resultant  couple  or  wrench  at  centre  of  mass 194 

Resultant  couple  or  wrench  at  any  point 195 

Moment  at  centre  of  mass  along  resultant  force 195 

Moment  at  centre  of  mass  normal  to  resultant  force 195 

Force  at  centre  of  mass  along  axis 196 

Force  at  centre  of  mass  normal  to  axis 196 

Position  of  resultant  force  of  the  screw-wrench 196 

Components  of  motion 197 

Screw-wrench 197 

Position  of  force  normal  to  axis 197 

The  invariant 197 

CHAPTER   VII. 
FORCE  OF   GRAVITATION.     CENTRE   OF   GRAVITY. 

Force  of  gravitation 203 

Attraction  of  a  homogeneous  shell  or  sphere 203 

Value  of  constant  of  gravitation 205 

Astronomical  unit  of  mass 206 

Centre  of  gravity 207 


STATICS.     GENERAL  PRINCIPLES. 


CHAPTER   I. 
EQUILIBRIUM  OF   FORCES.     DETERMINATION   OF  MASS. 

Statics 209 

Equilibrium  of  concurring  forces 209 

Static,  molar,  dynamic  and  molecular  equilibrium 210 

Equilibrium  of  non-concurring  forces 211 

Determination  of  mass  by  the  balance 212 

CHAPTER  II. 
WORK.     VIRTUAL  WORK. 

Work 215 

Work  of  resultant 215 


. 

TABLE  OF  CONTENTS.  xxv 

PAGE 

Work,  non-concurring  forces 215 

Principle  of  virtual  work 216 

CHAPTER    III. 
STATIC    FRICTION. 

Friction 220 

Adhesion 220 

Kinds  of  friction 220 

Coefficient  of  friction.  % 221 

Limiting  equilibrium 221 

Coefficient  of  static  sliding  friction,  experimental  determination 221 

Cone  of  friction 222 

Laws  of  static  friction 222 

Limitations  of  the  laws 223 

Values  of  coefficient  of  static  sliding  friction 223 

Static  friction  for  pivots 224 

Solid  flat  pivot 224 

Hollow  flat  pivot ' 225 

Conical  pivot 225 

Pivot  a  truncated  cone 226 

Pivot  with  spherical  end ., 227 

Static  friction  of  axles. 227 

Axle — partially  worn  bearings 228 

Axle — triangular  bearing 228 

Axle — new  bearing 229 

Friction  wheels .  230 

Static  friction  of  cords  and  chains 2-51 

Rigidity  of  ropes 234 

Hemp  ropes ^5 

Wire  ropes ;  ;5 

Static  rolling  friction 235 


KINETICS  OF  A  PARTICLE. 


CHAPTER    I. 
DEFLECTING   FORCE. 

Kinetics 242 

Kinetics  of  a  particle .  .  . ' 242 

Impressed  and  effective  force 242 

D'Alembert's  principle „ 242 

Deflecting  force „ 243 

Simple  conical  pendulum • 244 

Centrifugal  force 245 

Deflecting  force  at  the  earth  surface.  . 246 


xxvi  TABLE  OF  CONTENTS. 

PAGE 

Particle  moving  on  earth  surface 251 

Deviation  of  a  falling  body  owing  to  earth  rotation 252 

CHAPTER    II. 
TANGENTIAL  FORCE.     MOMENTUM.     IMPULSE. 

Tangential  force 255 

Significance  of  momentum 256 

Impulse 257 

Relation  between  impulse  and  momentum 257 

CHAPTER    III. 
WORK.     POWER. 

Work 260 

Unit  of  work 261 

Rate  of  work.     Power 262 

CHAPTER   IV. 
KINETIC   FRICTION. 

Kinetic  friction 263 

Coefficient  of  kinetic  friction 263 

Angle  of  kinetic  friction 263 

Kinetic  friction  of  pivots,  axles,  ropes 263 

Experimental  determination  of  coefficient  of  kinetic  sliding  friction 263 

By  sled  and  weight 264 

By  sled  on  inclined  plane 264 

By  friction  brake 265 

Friction -brake  test 266 

Work  of  axle  friction 266 

Coefficients  of  kinetic  sliding  friction 267 

Efficiency.     Mechanical  advantage 267 

CHAPTER  V. 

KINETIC    AMD   POTENTIAL   ENERGY.     LAW   OF   ENERGY.     CONSERVATION   OF   ENERGY. 
EQUILIBRIUM   OF   A    PARTICLE. 

Kinetic  energy 271 

Kinetic  energy  of  a  rotating  body 272 

Potential  energy 274 

Total  energy 275 

Law  of  energy 275 

Conservation  of  energy 276 

Equilibrium  of  a  particle 277 

Stable  and  unstable  equilibrium  of  a  particle 277 

Change  of  potential  energy 278 

For  uniform  force 278 

For  central  force 278 

For  central  force  constant 2  79 

For  central  force  proportional  to  distance 279 

For  central  force  inversely  proportional  to  distance  ....    280 


TABLE  OF  CONTENTS.  xxvil 

CHAPTER    VI. 

THE    POTENTIAL. 

PAGB 

The  potential 286 

Principle  of  the  potential 286 

Equipotential  surface 287 

Lines  offeree 287 

Tubes  of  force 287 

Gravitation  potential 287 

Determination  of  potential.      Examples 290 


KINETICS  OF  A  MATERIAL  SYSTEM. 


CHAPTER   I. 

GENERAL   PRINCIPLES. 

Material  system 297 

Internal  and  external  forces 297 

Impressed  and  effective  forces 297 

D'Alembert's  principle 297 

Velocity  of  centre  of  mass.     Momentum  of  a  system 298 

Acceleration  of  centre  of  mass 299 

Motion  of  centre  of  mass 299 

Conservation  of  centre  of  mass 299 

Conservation  of  momentum 300 

|  Moment  of  momentum 300 

Moment  of  momentum  for  a  system 300 

\.  Acceleration  of  moment  of  momentum 301 

>  Conservation  of  moment  of  momentum 302 

Invariable  axis  and  plane 302 

Conservation  of  areas , 303 

Kinetic  energy  of  a  system 303 

Potential  energy  of  a  system 303 

Law  of  energy 303 

Conservation  of  energy 303 

Perpetual  motion 303 

Law  of  conservation  of  energy  general 303 

Examples 305 

CHAPTER    II. 
EQUILIBRIUM   OF   A   MATERIAL   SYSTEM. 

Equilibrium  of  a  material  system 310 

Stable  equilibrium 311 

Principle  of  least  work - 311 

Stability  in  rolling  contact 312 


TABLE  OF  CONTENTS. 

CHAPTER     III. 

ROTATION  ABOUT  A   FIXED  AXIS. 

MM 

Rotation  about  a  fixed  axis.     Effective  forces > 315 

Moments  of  effective  forces 316 

Origin  of  the  term  moment  of  inertia 317 

Momentum  of  a  rotating  body 318 

Moment  of  momentum 318 

Pressures  on  fixed  axis 319 

Conservation  of  moment  of  momentum. . . 320 

Kinetic  energy 320 

Analogy  between  equations  for  rotation  and  translation 321 

Reduction  of  mass 322 

Examples 322 

CHAPTER    IV. 

COMPOUND   PENDULUM.     CENTRE   OF   OSCILLATION   AND    PERCUSSION.     EXPERIMENTAL   DETERMI- 
NATION OF   MOMENT   OF   INERTIA.     EXPERIMENTAL   DETERMINATION   OF  g. 

Simple  pendulum 336 

Compound  pendulum 337 

Centre  of  oscillation 337 

Significance  of  the  term  radius  of  gyration 338 

Centre  of  percussion 339 

Experimental  determination  of  moment  of  inertia 340 

Experimental  determination  of  g 341 

CHAPTER   V. 
IMPACT. 

Impact 342 

Direct  central  impact.      Non-elastic 342 

"          "  "  Perfectly  elastic 344 

Coefficient  of  elasticity 346 

Modulus  of  elasticity 347 

Direct  central  impact.      Imperfect  elasticity 347 

Earth  consolidation 350 

Pile-driving ^ 351 

Oblique  central  impact 352 

Friction  of  oblique  central  impact 35  ^ 

Strength  and  impact 356 

Impact  of  beams 358 

Impact  of  rotating  bodies 360 

Impact  of  an  oscillating  body 361 

Ballistic  pendulum 362 

Eccentric  impact 363 

CHAPTER    VI. 
ROTATION   ABOUT  A   TRANSLATING   AXIS. 

Effective  forces 367 

Moments  of  effective  forces 367 


TABLE  OF  CONTENTS.  xxix 

PAGE 

Momentum • 368 

Moment  of  momentum 368 

Pressures  on  axis 369 

Conservation  of  moment  of  momentum 369 

Kinetic  energy 370 

Instantaneous  axis , 371 

Examples. 371 

CHAPTER   VII. 
ROTATION   ABOUT   A   FIXED   POINT. 

Effective  forces 375 

Moment  of  effective  forces 375 

Momentum 376 

Moment  of  momentum 377 

Pressure  on  fixed  point .  . . 377 

Conservation  of  moment  of  momentum 377 

Invariable  axis 378 

Kinetic  energy 378 

Examples 379 

CHAPTER   VIII. 
ROTATION   AND   TRANSLATION. 

Effective  forces 389 

Moments  of  effective  forces 389 

Momentum ". .  . ,  390 

Moment  of  momentum 390 

Conservation  of  moment  of  momentum 390 

Invariable  axis 391 

Kinetic  energy _ •. .  . .  391 

Spontaneous  axis 392 

Instantaneous  axis 392 

Examples 392 


STATICS  OF  RIGID   BODIES. 


CHAPTER   I. 
FRAMED   STRUCTURES. 

Stress 397 

Tensile  stress  and  force _ 397 

Compressive  stress  and  force 397 

Shearing  stress  and  force 397 

Framed  structures 400 

Determination  of  stresses '. 400 

By  resolution  of  forces 400 

By  moments 401 


xxx  TABLE  OF  CONTENTS. 

PACK 

Superfluous  members t    402 

Criterion  for  superfluous  members 402 

CHAPTER   II. 
GRAPHICAL  STATICS.     CONCURRING   FORCES. 

Graphical  statics 404 

Concurring  co-planar  forces 404 

Notation  for  framed  structures 405 

Character  of  the  stresses 405 

Application  to  a  frame 406 

Apparent  indetermination  of  stresses 407 

Remarks  upon  method 408 

Choice  of  scales 408 

CHAPTER    III. 
GRAPHICAL  STATICS.     NON-CONCURRING   FORCES. 

Non-concurring  forces 412 

Equilibrium  polygon 413 

Graphic  construction  for  centre  of  parallel  forces 414 

Properties  of  equilibrium  polygon 414 

Application  to  parallel  forces 415 

CHAPTER   IV. 
WALLS.     MASONRY   DAMS. 

Definitions 424 

Weight  and  friction  of  masonry 424 

Stability  of  masonry  joint 425 

Greatest  unit  pressure 425 

Middle-third  rule 426 

Stability  of  a  wall  in  general ) 426 

Stability  for  sliding 1 427 

Stability  for  rotation 427 

Stability  for  pressure 428 

High  and  low  wall 428 

Design  of  low  wall 429 

Design  of  high  wall 429 

Design  of  wall  in  general 430 

Water-pressure 430 

Ice-  and  wave-pressure 432 

Stability  of  a  dam 432 

Design  of  dam.     Trapezoid  section 434 

Low  dam,  bottom  base 434 

Low  dam,  top  base 434 

Low  dam,  back  batter  angle 434 

Low  dam,  economic  section 435 

High  dam,  bottom  base 435 

High  dam,  top  base • 435 

High  dam,  back  batter  angle 435 

High  dam,  economic  section 436 


TABLE  OF  CONTENTS.  xxxi 

PAGE 

Design  of  dam.      Economic  section 440 

Arch  dam 450 

CHAPTER  V. 
RETAINING    WALLS.     EARTH-PRESSURE.     EQUILIBRIUM   OF   EARTH. 

Retaining  wall , 454 

Point  of  application  of  earth-pressure 454 

Magnitude  and  direction  of  earth -pressure.      Graphic  determination 454 

Magnitude  and  direction  of  earth-pressure.      Analytic  determination 458 

Case  i.   Earth-surface  horizontal 461 

Case  2.   Earth-surface  horizontal.      Back  vertical 462 

Case  3.   Earth-surface  horizontal.      Back  ang'le  90°  —  ^ 462 

Case  4.   Earth-surface  inclined  at  angle  of  repose 463 

Case  5.   Earth-surface  inclined  at  angle  of  repose.      Back  vertical 463 

Values  of  angle  of  friction,  coefficient  of  friction,  density  of  earth 464 

Cohesion  of  earth 466 

Equilibrium  of  a  mass  of  earth 467 

Angle  of  rupture 468 

Coefficient  of  cohesion. 468 

Height  of  slope. 468 

Angle  of  slope 469 

Curve  of  slope 469 


STATICS  OF  ELASTIC  SOLIDS. 


CHAPTER  I. 
ELASTICITY   AND   STRENGTH    FOR   TENSION,    COMPRESSION   AND   SHEAR. 

Elasticity 473 

Prismatic  body 473 

Stress  and  force 473 

Strain 473 

Law  of  elasticity , 474 

Set  and  shock 475 

Determination  of  elastic  limit  and  ultimate  strength 475 

Coefficient  of  elasticity 476 

Strain  due  to  weight 479 

Stress  due  to  change  of  temperature 480 

Coefficients  of  expansion 480 

Working  stress.      Factor  of  safety 481 

Variable  working  stress 48  i 

Combined  tension  or  compression  and  shear.  .    484 

CHAPTER   II. 
STRENGTH   OF   PIPES   AND   CYLINDERS.     RIVETING. 

Strength  of  pipes  and  cylinders 486 

Theory  and  practice  of  riveting 487 


xxxii  TABLE  OF  CONTENTS. 

PACk 

Kinds  of  riveted  joints 487 

Size  and  number  of  rivets 488 

Pitch  of  -rivets 489 

Rivet  table 490 

CHAPTER   III. 
STRENGTH  OF   BEAMS,  PINS  AND  EYEBARS.     TORSION. 

Flexure  or  binding  stress 493 

Assumptions  upon  which  theory  of  flexure  is  based ' 493 

Neutral  axis  of  cross-section 493 

Bending  moment 494 

Resisting  moment 496 

Designing  and  strength  of  beams 499 

Crippling  load.     Coefficient  of  rupture 499 

Comparative  strength 502 

Beams  of'uniform  strength '. 503 

Theory  of  pins  and  eyebars 506 

Bearing 506 

Diameter  of  pin 506 

Maximum  bending  moment 5*7 

Sizes  for  pins 508 

Torsion 509 

Neutral  axis 509 

Twisting  and  resisting  moment . 510 

Coefficient  of  rupture  for  torsion 510 

Coefficient  of  elasticity  for  shear  determined  by  torsion 511 

Work  of  torsion 512 

Transmission  of  power  by  shafts 512 

Combined  flexure  and  torsion 512 

CHAPTER  IV. 

WORK   OF   STRAINING.      DEFLECTION  OF   FRAMED   STRUCTURES.      PRINCIPLE   OF  LEAST   WORK. 
REDUNDANT  MEMBERS.     BEAMS   FIXED   HORIZONTALLY. 

Work  of  straining 515 

Work  and  coefficient  of  resilience 515 

Deflection  of  a  framed  structure 516 

Principle  of  least  work 517 

Redundant  members 520 

No  economy  due  to  redundant  members 521 

Work  of  bending 521 

Beams  fixed  horizontally  at  ends 522 

Case  i.  Beams  fixed  horizontally  at  one  end.        Concentrated  load 522 

Case  2.       "         "             "                                    Uniform  load 524 

Case  3.       "         "             "           "  both  ends.       Concentrated  load 525 

Case  4.       "         "             "           "     "       "             Uniform  load 527 

Bending  and  tension  or  compression  combined 528 

CHAPTER   V. 
DEFLECTION  OF   BEAMS. 

Deflection  of  a  beam , 530 

Beam  fixed  at  one  end  ;  loaded  at  the  other 533 

"        "     "     "     "       uniformly  loaded 535 


TABLE  OF  CONTENTS.  Xxxiii 

PAGH 

Application  to  metal  springs 537 

Beam  supported  at  both  ends.     Concentrated  load 542 

"  "          "      "       "          Uniform  load 544 

"  "          "  one  end,  fixed  at  other.     Uniform  load 544 

"  "          "     "  '"        "     "       "        Concentrated  load 545 

"     fixed  at  both  ends.     Uniform  load 547 

"       "     "     "       "         Concentrated  load 548 

CHAPTER  VI. 
SHEARING   STRESS. 

Shearing  stress  in  beams 553 

Work  of  shear  in  beams 554 

Influence  of  shear  on  deflection -. 554 

Determination  of  coefficient  of  elasticity  for  shear 555 

Influence  of  shear  on  reaction 556 

Maximum  internal  stresses  in  a  beam .• 557 

CHAPTER  VII.      . 
STRENGTH   OF  LONG  COLUMNS. 

The  ideal  column 550 

Theory  of  the  ideal  column 559 

Deportment  of  the  ideal  column 562 

Experimental  verification 563 

Euler's  formula 563 

Diagram  for  ideal  column 564 

Actual  column 564 

Practical  formulas  for  long  columns 565 

Straight-line  formula 565 

Parabola  formula 565 

Rankine" s  formula 567 

Gordon's  formula 568 

Merriman's  formula 568 

Allowable  unit  stress 569 

CHAPTER  VIII. 
THE  PIVOT   OR   SWING   BRIDGE. 

Pivot  or  swing  bridge 571 

Centre  span  without  bracing 571 

General  formulas ' 574 

Solid  beam,  uniform  cross-section 576 

CHAPTER    IX. 
THE  METAL  ARCH. 

Kinds  of  metal  arch 578 

Framed  arch  hinged  at  crown  and  ends 578 

Framed  arch  hinged  at  ends  only 579 

Temperature  thrust 581 

Solid  arch  hinged  at  ends  only   583 


xxxiv  T/IBLE  OF  CONTENTS. 

rAC« 

Solid  semi-circle  hinged  at  ends  only 5^5 

Temperature  thrust,  solid  arch  hinged  at  ends 5^5 

Framed  arch  hinged  at  ends 5^6 

Temperature  stress 591 

Solid  arch  fixed  at  ends 592 

Temperature  thrust 595 

CHAPTER   X. 
THE  STONE  ARCH. 

Definitions 597 

Reduced  surcharge 598 

Pressure  curve 598 

Condition  of  stability 599 

Determination  of  H,  M  and  V 600 

The  straight  arch 607 

CHAPTER   XI 
THE  SUSPENSION   SYSTEM. 

Suspension  system 609 

Horizontal  pull  of  cable .' 610 

Shape  of  cable 6 1  o 

Length  of  suspenders 610 

Length  of  cable  segment ' 6 10 

Stress  in  cable  segment 61 1 

Deflection  of  cable  due  to  temperature 611 

Deflection  of  truss  for  uniform  load 612 

Temperature  load  for  truss 612 

Old  theory  of  suspension  system 612 

New  theory  of  suspension  system 620 

Work  on  suspenders 620 

Work  on  cable 620 

Work  on  truss .  621 


GREEK  ALPHABET. 

In  all  mathematical  works  Greek  letters  are  used.  The  reader  should  therefore 
familiar  with  these  letters,  so  that  he  can  write  them  and  recognize  them  at  sight  and  < 
them  by  their  names. 

Letter.                        Transliteration.  Name. 

a  a  Alpha 

ft  b  Beta 

r,  Y  g  Gamma 

J,  <f  d  Delta 

6  e  short  Epsilon 

C  z  Zeta 

if  e  long  Eta 

0  th  Theta 

*  i  Iota 

K  k  Kappa 

A,  \  1  Lambda 

/i  m  Mu 

r  n  Nu 

$  x  Xi 

o  o  short  Omicron 

U,  «•  p  Pi 

p  T  Rho 

2,  <r  s  Sigma 

r  t  Tau 

Ttv  u  Upsilon 

0  ph  Phi 

*  ch  Chi 

^  ps  Psi 

/2,  a?  o  long  Omega 


MECHANICS. 


INTRODUCTION. 


CHAPTER   I. 

MECHANICS.    KINEMATICS.    DYNAMICS.    STATICS.    KINETICS.    MEASUREMENT. 

UNITS. 

Physical  Science. — We  live  in  a  world  of  matter,  space  and  time.  We  do  not  know 
what  these  are  in  themselves,  and  we  cannot  explain  or  define  any  one  of  them  in  terms  of 
the  others. 

Thus  we  recognize  matter  in  certain  states  which  we  call  SOLID,  LIQUID,  or  GASEOUS. 
We  distinguish  also  different  kinds  of  matter,  such  as  iron,  wood,  glass,  water,  air,  etc., 
which  we  call  SUBSTANCES.  We  also  recognize  limited  portions  of  matter  of  definite  shape 
and  volume,  such  as  a  pebble,  a  rain-drop,  a  planet,  etc.,  which  we  call  BODIES.  But  what 
matter  is  in  itself  we  do  not  know. 

We  also  recognize  matter  as  occupying  space,  and  we  note  successive  events  as  occupy- 
ing time.  But  what  space  and  time  are  in  themselves  we  do  not  know. 

We  also  recognize  FORCE  as  causing  change  of  motion  of  matter,  or  change  of  the 
volume  or  shape  of  bodies.  But  what  force  is  in  itself  we  do  not  know. 

Yet,  although  we  thus  know  nothing  of  matter,  space,  time  and  force  in  themselves,  we 
can  and  do  investigate  them  in  their  measurable  relations,  and  such  investigation  is  the  object 

Of  all  PHYSICAL  SCIENCE. 

Mechanics — Kinematics  and  Dynamics — Statics  and  Kinetics. — That  branch  of 
physical  science  which  treats  of  the  measurable  relations  of  space  alone  is  called  geometry. 

That  which  deals  with  the  measurable  relations  of  space  and  time  only — that  is,  with 
pure  motion— is  called  KINEMATICS  (Kivrjfjia,  motion).  To  the  ideas  of  geometry  it  adds 
the  idea  of  motion. 

That  which  deals  with  the  measurable  relations  of  force,  and  of  those  measurable  rela- 
tions of  space,  time  and  matter  involved  in  the  study  of  the  change  of  motion  of  material 
bodies  under  the  action  of  force,  is  called  DYNAMICS  (Svvaftis,  force).  To  the  ideas  of 
kinematics  it  adds  the  idea  of  force. 

We  may  divide  dynamics  into  two  parts.  That  portion  which  treats  of  material  bodies 
at  rest,  under  the  action  of  balanced  forces,  is  called  STATICS.  That  portion  which  treats  of 
the  change  of  motion  of  material  bodies  under  the  action  of  unbalanced  forces  is  called 

KINETICS. 

I 


, 

2  MECHANICS-INTRODUCTION.  [CHAP.  I. 

The  term  MECHANICS  is  used  to  include  the  general  principles  of  both  kinematics  and 
dynamics. 

Measurement. — Since,  then,  we  have  to  do  in  all  that  follows  with  the  measurable 
relations  of  force,  matter,  space  and  time,  the  subject  of  the  measurement  of  these  quantities 
should  first  engage  our  attention. 

Unit. — In  order  to  measure  any  quantity  whatever,  we  must  always  compare  its 
magnitude  with  the  magnitude  of  another  quantity  of  the  same  kind.  The  quantity  thus 
taken  as  a  standard  of  comparison  is  called  the  UNIT  of  measurement. 

Thus  the  unit  of  length  must  itself  be  some  specified  length,  as,  for  instance,  one  foot, 
one  yard,  one  centimeter  or  one  meter.  The  unit  of  time  must  be  a  specified  time,  as  one 
second.  The  unit  of  mass  must  be  a  specified  mass,  as  one  pound  or  one  gram  or  one 
kilogram. 

The  units  of  mass,  length  and  time  are  called  FUNDAMENTAL  UNITS,  because  not  derived 
from  any  others. 

Statement  of  a  Quantity. — The  complete  statement  of  a  quantity  requires,  there- 
fore, a  statement  of  the  unit  adopted  and  also  a  statement  of  the  result  of  comparison  of  the 
magnitude  of  the  quantity  with  the  magnitude  of  the  unit. 

The  result  of  this  comparison  is  always  a  ratio  between  the  magnitudes  of  two  quantities 
of  the  same  kind  and  is,  therefore,  always  an  abstract  number. 

This  ratio  or  abstract  number  is  called  the  NUMERIC. 

Thus  we  say  3  feet,  4  seconds,  5  pounds.  In  each  of  these  cases  we  state  both  the  unit 
and  the  numeric,  or  ratio  of  the  magnitude  of  the  quantity  to  that  of  the  unit.  Thus  3  feet 
denotes  a  quantity  whose  magnitude  is  three  times  the  magnitnde  of  one  foot. 

So  for  any  quantity.  In  general,  if  L  stands  for  any  length  and  [Z]  stands  for  the  unit 
of  length,  we  have  L  =  /[Z,],  or  the  length  equals  /  times  the  unit  of  length.  Here  /  is  the 
numeric  and  is  an  abstract  number. 

Again,  if  T  is  a  certain  interval  of  time,  and  [  J"]  stands  for  the  unit  of  time,  we  have 
T  —  f[T],  or  the  time  equals  /  times  the  unit  of  time.  Here  t  is  the  numeric  and  is  an 
abstract  number. 

So  also  if  M  is  a  certain  mass  and  [M]  stands  for  the  unit  of  mass,  we  have  M  —  »i[J7], 
or  the  mass  equals  m  times  the  unit  of  mass.  Here  m  is  the  numeric  and  is  an  abstract 
number. 

Derived  Unit. — A  unit  of  one  kind  which  is  derived  by  reference  to  a  unit  of  another 
kind,  is  called  a  DERIVED  UNIT. 

Thus  the  unit  of  area  may  be  taken  as  a  square  whose  side  is  one  unit  of  length,  or  one 
square  foot.  The  unit  of  volume  may  be  taken  as  a  cube  whose  edge  is  the  unit  of  length, 
or  one  cubic  foot.  The  unit  of  velocity  may  be  taken  as  one  unit  of  length  per  unit  of  time, 
or  one  foot  per  second. 

Such  units  are  derived  units,  while  the  units  of  mass,  space  and  time,  not  being  thus 
derived  from  any  others,  are  FUNDAMENTAL  units. 

Dimensions  of  a  Derived  Unit — A  statement  of  the  mode  in  which  the  magnitude 
of  a  derived  unit  varies  with  the  magnitudes  of  the  fundamental  units  which  compose  it,  is  a 
statement  of  the  DIMENSIONS  of  the  derived  unit. 

Thus  let  [A]  denote  the  unit  of  area,  and  [L~\  the  unit  of  length.  Then  if  A  =  <*[A~\  is 
the  area  of  a  square  whose  side  is  L  =  /[/,],  where  a  and  /are  abstract  numbers,  we  shall 
have  a[A]  =  P[L]\ 

Now  we  shall  have  the  numeric  equation  a  =  /*,  or  the  number  of  units  of  area  equals 
the  square  of  the  number  of  units  of  length,  provided  we  have  [A]  =  [A]2,  or  the  unit  of 
area  equal  to  the  square  of  the  unit  •><"  length. 


CHAP.  I.]  DIMENSIONS  OF  A  DERIVED   UNIT.  3 

The  statement  [^4]  =  [£]2  is  a  statement  of  the  dimensions  of  the  unit  of  area. 

Again,  let  [Z]  denote  the  unit  of  length,  [7^]  denote  the  unit  of  time,  and  [V"\  denote 
the  unit  of  velocity.  Then  if  D  ~  d\L\  is  any  displacement  and  T=  t[T~\  is  the  time  and 
V=  z>[F]  is  the  mean  velocity,  we  have 


We  shall  then  have  the  numeric  equation  v  =  —,  or  the  number  of  units  of  velocity  is  equal 

to  the  number  of  units  of  displacement  divided  by  the  number  of  units  of  time,  provided  we-. 
have 


that  is,  provided  the  unit  of  velocity  is  the  unit  of  length  per  unit  of  time. 
This,  then,  is  a  statement  of  the  dimensions  of  the  unit  of  velocity. 
Meaning  of  "  Per."  —  It  will  be  observed  that  the  statement 


m 

is  read,  "  the  unit  of  velocity  is  equal  to  the  unit  of  length  per  unit  of  time."  The  word 
per  is  indicated  by  the  line  of  division  between  [Z]  and  [jT]. 

Now  we  can  divide  the  numeric  d  by  the  numeric  t  and  write  v  =  —  ,  because  these  are 

abstract  numbers.  But  it  would  be  nonsense  to  speak  of  dividing  length  by  time,  or  a  unit 
of  length  by  a  unit  of  time.  We  therefore  avoid  such  a  statement  by  the  use  of  the  word 
per.  If,  then,  we  give  to  the  symbol  of  division  this  new  meaning,  we  can  treat  it  by  the 
rules  which  apply  to  the  old  meaning,  and  thus  avoid  the  invention  of  a  new  symbol  by  using 
an  old  one  in  a  new  sense.  The  sign  of  division  stands  then  for  actual  division  so  far  as 
numerics  are  concerned,  but  so  far  as  units  are  concerned  it  stands  for  the  word  per. 

Whenever,  then,  the  word  PER  is  used  it  can  be  replaced  by  the  sign  of  division. 

Homogeneous  Equations  —  The  letters  in  all  equations  or  statements  of  the  relations 
of  quantities  always  stand  for  the  numerics,  and  the  units  are  always  understood  but  not 
written  in. 

Thus  such  an  equation  as  v  =  —  or  d  =  vt  is  a  numeric  equation,  where  the  units  are 

understood  and  must  be  supplied  when  interpreting  the  equation.  When  the  units  are 
thus  supplied,  all  terms  on  both  sides  of  any  equation,  which  are  combined  by  addition  or 
subtraction,  must  always  denote  quantities  of  the  same  kind.  Such  an  equation  is  called 
HOMOGENEOUS. 

If  any  numeric  equation  is  not  thus  homogeneous,  it  is  incorrectly  stated. 

It  is  also  evident  that  all  algebraic  combinations  of  such  homogeneous  equations  must 
always  produce  homogeneous  equations.  If  not,  some  error  must  have  been  made  in  the 
algebraic  work.  Thus  in  the  equation  d  =  vt,  if  we  supply  the  omitted  units,  we  have 


4  .  MECHJNICS.-1NTRODUCT1ON.  [CHAP.  I. 

The  equation  is  therefore  homogeneous,  since  the  unit  of  length  is  to  be  understood  in  both 
terms,  and  we  have  a  distance  equal  to  a  distance. 

Again,  suppose  the  result  of  some  investigation  is  expressed  by 

ioz/. 


Then,  without  any  reference  whatever  to  the  various  steps  by  which  this  result  may  have  been  obtained,  we 
can  say  at  once  that  it  is  incorrect. 

For  if  we  insert  the  units  we  have 


Zdfeet  +  2  /  seconds  =  \<yv  feet  per  sec. 

We  see  at  once  that  the  quantities  in  each  term  are  not  of  the  same  kind.     The  equation  is  not  homogene- 
ous and  is  absurd. 

If,  however,  we  had 


this  equation  is  homogeneous,  because  when  we  insert  the  units  we  have 

3ti[L]  +  2/[7-]z/A]  =  MMgjjlTi     or     yt[L\  +  2tv[L]  = 

Here  all  the  terms  are  quantities  of  the  same  kind  and  the  equation  is  homogeneous.     It  reads  now, 

ylfeet  +  itv  feet  =  iovt  feet. 

The  relations  expressed  by  it  are  not  absurd.     It  does  not  follow  that  it  is  correct.     It  may  still  have  been 
incorrectly  deduced.     But  on  its  face  it  is  not  absurd,  while  ^d  +  2/  =  io-v  is  manifestly  so. 

The  student  should  make  it  a  rule  to  test  in  this  manner  any  equation  the  truth  of  which  is  suspected, 
;is  it  may  often  save  the  trouble  of  examining  in  detail  the  entire  investigation  by  which  it  has  been  deduced. 
If,  however,  it  stands  this  test,  then  the  derivation  must  be  examined  also,  if  error  is  still  suspected. 

Unit  of  Time.  —  The  unit  of  time  ordinarily  adopted  in  dynamics  is  the  SECOND  or 
some  multiple  of  the  second. 

It  is  the  time  of  vibration  of  an  isochronous  pendulum  which  vibrates  or  beats  86400 
times  in  a  mean  solar  day  of  24  hours,  each  hour  containing  60  minutes  and  each  minute  60 
seconds  (24  X  60  X  60  =  86400). 

The  sidereal  day  contains  86164.09  of  these  mean  solar  seconds. 

Unit  of  Length.  —  The  unit  of  length  ordinarily  adopted  in  dynamics  is  the  FOOT  on  the 
METER  or  some  multiple  of  these. 

Unit  of  Mass.  —  The  unit  of  matter  or  mass  ordinarily  adopted  in  dynamics  is  the 
POUND  or  the  KILOGRAM. 

Standard  Unit.  —  All  units  adopted  are  defined  by  reference  to  certain  STANDARD 
units.  A  standard  unit,  in  general,  should  possess,  so  far  as  possible,  a  permanent  magni- 
tude unchanged  by  lapse  of  time  and  unaffected  by  the  action  of  the  elements  or  by  change 
of  place  or  temperature.  It  should  be  capable  of  exact  duplication  and  should  admit  of 
direct  and  accurate  comparison  with  other  quantities  of  the  same  kind. 

Standard  Unit  of  Time.  —  The  standard  unit  of  time,  is  the  period  of  the  earth's 
rotation,  or  the  sidereal  day.  Thir-  has  been  proved  by  Laplace,  from  the  records  of  celes- 


CHAP.  I.]  MEASUREMENT -STANDARD   UNIT.  5 

tial  phenomena,  not  to  have  changed  by  so  much  as  one  eight-millionth  part  of  its  length  in 
the  course  of  the  last  two  thousand  years. 

The  length  of  the  solar  day  is  variable,  but  the  mean  solar  day,  which  is  the  exact  mean 
of  all  its  different  lengths,  is  the  period  already  mentioned,  which  furnishes  the  second  of 
time.  It  is  1.00273791  of  a  sidereal  day. 

The  second  can  therefore  be  defined,  with  reference  to  the  standard  unit  of  time,  as  the 
time  of  one  swing  of  a  pendulum  so  adjusted  as  to  make  86400  oscillations  in  1.00273791 
of  a  sidereal  day. 

Standard  Units  of  Length. — The  English  standard  unit  of  length  is  the  length  of 
a  standard  bronze  bar,  deposited  in  the  Standards  Department  of  the  Board  of  Trade  in 
London. 

Since  such  a  bar  changes  in  length  with  its  temperature,  the  length  is  taken  at  the  speci- 
fied temperature  of  62°  Fah. 

The  length  of  this  bar  at  this  temperature  is  the  English  standard  unit  of  length,  and 
is  called  the  STANDARD  YARD.  Accurate  copies  of  this  standard  are  distributed  in  various 
places,  and  from  these  all  local  standards  of  length  are  derived.* 

The  foot  is  defined  as  one  third  the  length  of  the  standard  yard  at  62°  Fah. 

The  French  standard  of  length  is  the  meter,  and  is  the  length  of  a  bar  ot  platinum  at 
the  temperature  of  melting  ice,  or  o°  C.  This  bar  is  preserved  at  Paris.  Its  length  was 
intended  to  be  the  ten-millionth  part  of  a  quadrant  of  the  earth's  meridian  through  Paris. 

The  quadrant  of  the  meridian  through  Paris  is  10001472  standard  meters,  according  to 
Colonel  Clarke's  determinations  of  the  size  and  figure  of  the  earth,  which  are  at  present  the 
most  authoritative,  and  thus  the  standard  Paris  meter  is  slightly  less  than  the  length  upon 
which  it  was  founded.  The  material  bar  is  therefore  the  standard,  just  as  is  the  case  with 
the  English  standard. 

The  relation  of  the  meter  to  the  meridian  was  intended  as  a  means  of  reproduction  in 
case  of  destruction  of  the  standard,  but  in  such  case  the  standard  would  probably  be  repro- 
duced from  the  best  existing  copies. 

This  was  actually  the  case  with  the  original  English  standard,  which  was  destroyed  by 

fire  in    1834.      It  had  been  originally  defined  as  having  at  62°  Fah.   a  length  of   

39-1393 

of  the  length  of  a  pendulum  vibrating  seconds  in  the  latitude  of  London  at  the  sea-level. 
But  this  provision  for  its  restoration  was  repealed  and  a  new  standard  bar  was  constructed 
from  authentic  copies  of  the  old  one. 

The   English  inch,  or  the  36th  part  of  the  length  of  the  standard  yard,  is  very  nearly 

equal  to  the  five-hundred-millionth  part  of  the  length  of  the  earth's  polar  axis  ( — — -~\. 

The  utility  of  the  standard,  however,  does  not  depend  upon  any  such  earth  relations, 
the  only  value  of  which  is  for  reproduction  in  case  of  destruction — a  value  which,  as  we  have 
seen,  is  practically  disregarded. 

The  ultimate  standards  are  therefore  the  actual  bars. 

Standard  Units  of  Mass. — The  English  standard  unit  of  mass  is  a  piece  of  platinum 
deposited  in  the  Office  of  the  Exchequer  at  London  and  called  the  "Imperial  Standard 
Pound  Avoirdupois." 

The  French  standard  unit  of  mass  is  a  piece  of  platinum  preserved  at  Paris  and  called 
the  kilogram. 

•>< 
*  The  English  standard  yard  is  I  part  in  17230  shorter  than  the  U.  S.  copy. 


6  MECHANICS  -INTRODUCTION.  [CHAP.  1. 

Unit  of  Angle.  —  There  are  two  units  of  angle  in  use,  the  DEGREE  and  the  RADIAN. 
The  degree  is  that  angle  subtended  at  the  centre  of  any  circle  by  an  arc  equal  in  length 

to  -  —  part  of  the  circumference  of  that  circle.      It  is  subdivided  sexagesimally  into  degrees 

(°),  minutes  ('),  and  seconds  (").  The  seconds  are  subdivided  decimally.  Minutes  and 
seconds  of  time  are  distinguished  by  being  written  min.,  sec. 

The  radian  is  that  angle  subtended  at  the  centre  of  any  circle  by  an  arc  equal  in  length 
to  the  radius.  It  is  subdivided  decimally. 

If  then  the  length  of  any  arc  is  s[L'],  or  s  units  of  length,  and  the  length  of  the  radius 
is  r[£],  or  r  units  of  length,  and  if  the  angle  subtended  at  the  centre  is  a  radians,  we  have 

s[L]        s 

=  ~>     °r     '"=*• 


The  number  of  radians  in  any  angle  is  then  found  by  dividing  the  number  of  units  of 
length  in  the  subtending  arc  by  the  number  of  units  of  length  in  the  radius,  and  this  number  is 
independent  of  the  partictdar  unit  of  length  adopted,  whether  feet  or  centimeters. 

If  the  subtending  arc  is  the  entire  circumference,  the  number  of  radians  is  —  —  =  2n. 

Hence    2ft   radians    correspond    to    360    degrees,    or    I    radian   corresponds   to  -  -  =  — 

27*  n 

=  57-29578  degrees  =  57°  17'  44".  8. 

Any  angle  expressed  in  radians  may  then  be  converted  into  degrees  by  multiplying  the 

180° 
number  of  radians  by  -  =  57.29578  degrees  =  I  radian. 

Any  angle  expressed  in  degrees  may  be  converted  into  radians  by  multiplying  the 
number  of  degrees  by—  —  =0.0174533  radians  =  I  degree. 

I  oO 

. 

Examples.—  (I)  Express  I2n  34'  56"  in  terms  of  radians  ;  and  3  radians  in  terms  of  degrees. 
ANS.  0.2196  radians;   171°  53'  I4".424. 

(2)  The  radius  of  a  circle  is  10  feet  ;  what  is  the  angle  subtended  at  the  centre  by  an  arc  of  3  feet? 

, 

ANS.     —  radian,  or  17°  u    19  .44. 
10 

(3)  How  much  must  a  rail  30  feet  lone;  be  bent  in  order  to  fit  into  a  curve  of  half  a  mile  radius  ? 

ANS.  Tfrr   radian,  or  o°  39'  3"-92. 
88 

(4)  Express  45  degrees  in  terms  of  radians,  and  4  j  radians  in  terms  of  degrees. 

ANS.   *  radians  =  0.7854  radians;  257°  49'  51  ".636. 

(5)  The  angle  subtended  at  the  centre  of  a  circle  by  an  arc  whose  length  is  1.57  feet  is  7j°/  what  is  ////• 
radius  f 


i8o  I5?r 

(6)    What  is  the  sin  ^  raff  tans  ;  cos  g  radians  ;  co*      radians  ;  tan  —  radians? 


ANS.  0.5; 


C:IAP.  I.]  MEASUREMENT— UNIT  OF  CONICAL   ANGLE.  ^ 

(7)  Express  in   degrees   and  in   radians   the   angle  made   by   the  hands  of  a  clock   at  35  minues  past 
3  o'clock. 

ANS.  102.5  degrees ;  1  -79  radians. 

Unit  of  Conical  Angle. — Let  the  area  of  any  portion  of  the  surface  of  a  sphere  be 
A[A],  or  A  units  of  area,  and  let  the  square  of  the  radius  be  r\A\,  or  r*  units  of 
area. 

If  lines  are  drawn  from  the  centre  C  of  the  sphere  to  every  point  of  the  area,  they  form 
a  cone,  and  the  angle  subtended  at  the  centre  C  by  the  area  we  'call  a  CONICAL 
ANGLE.  . 

The  conical  angle  subtended  at  the  centre  of  a  sphere  by  a  portion  of  its  surface  whose 
area  is  equal  to  the  square  of  its  radius  we  call  a  SQUARE  RADIAN.  If  we  denote  the 
conical  angle  subtended  by  the  area  A  by  a  square  radians,  we  have 


The  number  of  square  radians  in  any  conical  angle  is  tJius  found  by  dividing  the  number 
of  units  of  area  in  the  subtending  area  by  the  number  of  units  of  area  in  the  square  of  the 
radius,  and  this  number  is  independent  of  the  unit  of  area  adopted. 

If  the  subtending  area  is  the  entire  surface  of  the  sphere,  the  number  of  square  radians  is 

2     —  4/r.      Hence  the  surface  of  a  sphere  subtends  a  conical  angle  of  4?r  square  radians. 

[The  terms  solid  angle  and  solid  radian  are  usually  employed  in  place  of  conical  angle 
and  square  radian  as  defined,  but  as  they  seem  in  no  way  descriptive,  we  have  employed  the 
latter  terms  as  more  expressive.] 

Curvature. — The  direction  of  a  plane  curve  at  any  point  is  that  of  the  tangent  to  the 
curve  at  this  point. 

Thus  the  direction  of  the  curve  AB  at  the  point 
A  is  that  of  the  tangent  A  T. 

The  change  of  direction  between  any  two  points 
of  a  plane  curve  is  the  angle  between  the  tangents 
at  these  two  points,  and  is  called  the  INTEGRAL 
CURVATURE. 

Thus  the  angle  a,  or  change  of  direction  be- 
tween the  tangents  at  A  and  B,  is  the  integral  curvature  for  the  curve  between  A  and  B. 

The  integral  curvature  for  any  portion  of  a  plane  curve  divided  by  the  length  of  that 
portion  is  the  mean  curvature. 

Thus  if  the  length   from  A   to  B  of  the  curve  is  s[L],  or  s  units  of  length,  the  mean 

curvature  is    rr-\>      Since  or  is  given   in  radians,  the  unit  of  curvature  is  one  radian  per  unit 

of  length  of  arc.      When  we  say,  therefore,  that  the  mean  curvature  is  -  ,  we  mean  —  radians 

per  unit  of  length  of  arc. 

The  limiting  value  of  the  mean  curvature  when  the  two  points  are  indefinitely  near  is 
called  the  curvature. 


8  MECHANICS-INTRODUCTION.  [CHAP.  I. 

The  curvature,  therefore,  is  the  limiting  rate  of  change  of  direction  per  unit  of  length  of 
arc.  Its  unit  is  one  radian  per  unit  of  length  of  arc. 

Curvature  of  a  Circle. — If  the  curve  is  a  circle,  the  angle  at  the  centre  between  the 
radii  at  A  and  B  will  be  equal  to  the  angle  a  between  the  tangents  at  A  and  B. 

We  have  then  «  =  -  radians.     The  mean  curvature  is  then  —  =  —  radians  per  unit  of 

r  s       r 

length  of  arc. 

Since  this  is  independent  of  s,  the  curvature  at  every  point  of  a  circle  is  constant  and 

equal  to  the  mean  curvature  for  any  two  points,  viz.,  -  radians  per  unit  of  length  of  arc. 

Curvature  of  any  Curve. — For  any  curve  whatever  a  circle  can  always  be  described 
whose  curvature  is  the  same  as  that  of  the  given  curve  at  the  given  point.  This  is  the  circle 
of  curvature  of  the  curve  at  that  point.  Its  radius  is  the  radius  of  curvature  of  the  curve  at 
that  point. 

If  then  p  is  the  radius  of  curvature  of  a  curve  at  any  given  point,  the  curvature  of  the 

curve  at  that  point  is  -  radians  per  unit  of  length  of  arc. 

Since  curvature  then  depends  only  upon  the  radius  of  curvature,  the  circle  is  the  only 
curve  whose  curvature  is  constant. 

1 i )  A  circle  has  a  radius  of  10  feet.      What  is  its  curvature  ? 
Ans.    ^  radian  per  foot  of  arc,  or  5°. 73  per  foot  of  arc. 

(2)  If  the  radius  is  10  yards,  what  is  the  curvature  ? 

Ans.  -jij.  radian  per  yard  of  arc,  or  5°. 73  per  yard  of  arc,  or  -fa  radian  per  foot  of  arc,  or 
i°  91  per  foot  of  arc. 

Dimensions  of   Unit  of   Curvature. — If    C   is    the    curvature    and    c   the   number  of 

a[a] 

units  of  curvature,  we  have  by  definition  c[C]  =    fry.  where  [C]  is  the  unit  of  curvature, 

and  [or]  is  the  unit  of  angle,  [L]  the  unit  of  length,  and  «,  s  the  number  of  units  of  angle  and 
length. 

a  ["1 

We  shall  always  have  c  =  -,  provided  we  take  [C]  =  ryr,  that  is,  provided  the  unit  of 

curvature  is  equal  to  the  unit  of  angle  divided  by  the  unit  of  length. 

This  is  a  statement  of  the  dimensions  of  the  unit  of  curvature. 

The  unit  of  curvature  is  then  one  unit  of  angle  per  unit  of  length  oS  arc,  as,  for  instance, 
one  radian  per  foot  of  arc,  or  one  degree  per  foot  of  arc. 

A  railway  curve  has  a  length  of  one  mile,  the  curvature  is  uniform,  and  the  integral 
curvature  is  jo  degrees.  What  is  the  curve,  the  curvature,  and  the  radius  of  curvature  / 

Ans.   A  circle;   0.5236  radian  per  mile  arc;    1.9  miles  radius. 

Tables  of  Measures. — We  shall  deal  in  the  course  of  this  work  with  many  other  derived 
units,  which  will  be  explained  as  they  occur.  It  will  be  useful  to  collect  here  for  conven- 
ience of  reference  a  number  of  such  units. 


CHAP.  I.] 


MEASUREMENT— TABLES  OF  MEASURES. 


I.     MEASURES    OF    SPACE. 
A.  LENGTH. 


) 

CENTI£ 

vIETERS 

10 

) 

INC 

1 

HES 

\ 

TABLE  1. 
i  centimeter  =  0.3937079  inch 

(  39-37079  inches 
(  3.2809  feet 
(0.62137  mile 


I  meter 


i  kilometer     = 


(0.535987  nautical  mile 


TABLE  2. 

I  inch  =  2.539954  centimeters 
i  foot  =  30.479449          " 
I  yard  =  0.91438347  meter 
I  mile  =  1.60935  kilometers 
I  nautical  mile, 
or  6080. 26  ft. 


=  1.85327  kilometers 


The  following  are  approximate : 

The  centimeter  is  about  f  inch.     The  meter  is  about  3  ft.  3f  inches. 
The  decimeter  is  about  4  inches.      One  kilometer    is  about  f  of  a  mile. 
Distance  from  pole  to  equator  is  about  10000000  meters. 
Earth's  polar  radius  is  about  500000000  inches. 


TABLE  3. 

i  sq.  centimeter  =  0.155006  sq.  inch 
I  sq.  meter  =  10.7643  sq.  feet 

i  sq.  hectometer,  ) 

[  =  2.47114  acres 
or  i  hectare        ) 

i  sq.  kilometer     =  0.38611  sq.  mile 


B.  AREA. 

TABLE  4. 

I  sq.  inch  =  6.45137  centimeters 
I  sq.  foot  =  928.997  " 

I  sq.  yard  —  0.836097  meter 
I  acre          =  0.404672  hectare 
I  sq.  mile  =  2.58989  sq.  kilometers 


TABLE  5. 

i  cubic  centimeter  =  0.0610271  cu.  inch 
I  liter,  or  i  cubic  |  =  61.0271  cu.  inches 

decimeter  )  =   1.76172  pints 

i  cubic  meter  =  35.3166  cubic  feet 


C.  VOLUME. 

TABLE   6. 

i  cubic  inch  =  16.3866  cubic  centimeters 
I  cubic  foot   =  28.3153  liters 
I  cubic  yard  =  0.764513  cubic  meter 
I  pint  =  0.567627  liter 

i  gallon  =   4.54102  liters 


II. 

TABLE   7. 
I  centigram  =  0.154323  grain 

i  gram  =  {  '5-4323  grains 

(0.035373902. 
I  kilogram     =  2.20462  Ibs. 


MEASURES  OF    MASS. 

TABLE   8. 

I  grain  =  0.064799  gram 

i  oz.  =  28.3496  grams 

i  lb.,  or  16  oz.        =  0.453593  kilogram 
i  ton,  or  2240  Ibs.  =  1016.05  kilos 


io  MECHANICS-INTRODUCTION.  [CHAP.  I. 

I  gram  =  mass  of  I  cubic  centimeter  of  pure  water  at  4°  C. 

I  kilogram  =  I  liter  of  pure  water  at  4°  C. 

I  gallon  =  277.274  cubic  inches.      The  gallon  contains  10  Ibs.  of  pure  water  at  62°  F. 

I  cubic  foot  of  water  contains  about  1000  oz.,  or  62  £  Ibs. 

The  pint  contains  20  fluid  oz. 

Acceleration  of  gravity  at  London  =  32.182  feet-per-second  per  second  =  980.889  cen- 
timeters-per-second  per  second.  Average  value  32^  feet-per-second  per  second  or  980.3 
centimeters-per-second  per  second. 

i  dyne  =  force  which  will  give  a  mass  of  I  gram  an  acceleration  of  I  centimeter-per- 
second  per  second  =  about  -g£T  weight  of  gram  =  weight  of  about  I  milligram. 

i  poundal  =  force  which  will  give  a  mass  of  I  pound  an  acceleration  of  I  foot-per- 
second  per  second  =  about  weight  of  £  oz. 


CHAPTER    II. 


POSITION.     TERMS  AND  DEFINITIONS. 


It  is  there- 


position. 


Point. — A  mathematical  POINT  has  neither  length,  breadth,  nor  thickness, 
fore  without  dimensions  and  indicates  position  only. 

Point  of  Reference. — When  we  speak  of  a  point  as  having  position,  some  other  point  or 
points  must  always  be  assumed,  by  reference  to  which  the  position  is  given.  Such  a  point  is 
a  POINT  OF  REFERENCE.  It  is  also  called  a  POLE,  or  ORIGIN. 

Position,  then,  is  always  relative.      We  know  nothing  of  "absolute 

Thus  the  position  of  the  point  C  is  known  with  respect  to  A 
when  we  know  the  length  of  the  line  AC  and  the  angle  BAC  or  the 
direction  of  the  line  AC.  The  points  A  and  B  are  points  of  reference, 
by  means  of  which  C  is  located. 

Position  of  a  Point. — The  position  of  a  point  with  reference  to     t 
other  assumed  points  is  then  known  when  we  have  sufficient  data  to 
locate  it.      These  data  give  rise  to  two  methods  of  location  : 

ist,  by  polar  co-ordinates; 

2d,  by  Cartesian  co-ordinates,  so  called  because  first  employed  by  Descartes. 

Plane    Polar   Co-ordinates. — The  data  necessary  for  locating  a  point  by  polar  coordi- 
nates, when  the  point  is  situated  in  a  given  plane,  consist  of  a  distance  and  an  angle;    if  the 
point  is  not  in  a  known  plane,  of  a  distance  and  two  angles. 

Thus,  if  the  point  P,  in  the  plane  of  this  page,  is  to  be 
located,  we  first  assume  a  line  OA  in  the  plane,  as  a  line  of  refer- 
ence. Then  the  position  of  P  with  reference  to  O  is  given  by 
the  angle  A  OP  and  by  the  distance  OP. 

The  assumed  point  O  is  called  the  POLE ;    OA  is  the  LINE  OF 
A      REFERENCE;    the  distance  OP  is  called  the  RADIUS  VECTOR,  and 

its  magnitude   is   usually  denoted   by  r;    the    angle    AOP    is  the  DIRECTION  ANGLE;    its 
magnitude  is  denoted  by  0,  and  it  is  measured  around  from  OA  to  the  left. 

The  polar  co-ordinates  for  a  point  in  a  given  plane  are  therefore  r  and  0,  or  a  distance 
and  an  angle.  These  are  PLANE  polar  co-ordinates. 

Space  Polar  Co-ordinates. — If  the  point  P  is  not  in  a  given  plane,  we  assume  as  before 
a  pole  O,  and  a  reference  line  OA  in  space.  Through  this  line  we  assume  any  plane,  as  the 
plane  of  this  page,  OABC,  and  let  OB  be  the  intersection  of  this  plane  with  a  plane  OPB, 
perpendicular  to  it  and  passing  through  OP.  The  location  of 
P  is  then  given  by  the  length  OP  or  the  radius  vector  r,  the 
angle  A  OB  or  0,  and  the  angle  BOP  or  0. 

The  polar  co-ordinates  for  a  point  not  in  a  given  plane  are 
therefore  r,  0,  and  ft,  or  a  distance  and  two  angles.  These  are 
SPACE  polar  co-ordinates. 

If  O  is  a  point  on  the  earth's  surface,  and  the  reference  line 
OA    is  a  north   and   south  line   in  the  plane  of  u*c   ii 
astronomical  altitude  of  the  point  P. 


the   angle   0  would  be  the 


MECHANICS-INTRODUCTION. 


[CHAP.  II. 


! 

» 

'           2            D 

Y 
1 

p 

« 

B 

+rt 

—y 

6 

3 

4 

-y 

Cartesian  Co-ordinates --Plane. — The  data  necessary  for  locating  a  point  by  Cartesian 
co-ordinates,  if  the  point  is  in  a  known  plane,  consists  of  two  distances,  parallel  to  two 

assumed  lines  of  reference  in  that  plane,  passing 
through  the  point  of  reference,  which  is  called  the 
ORIGIN.  The  assumed  lines  of  reference  are 
usually  taken  at  right  angles. 

Thus  if  the  point  P  is  known  to  be  in  the 
plane  of  this  page,  we  assume  any  origin  O  and 
draw  two  reference  lines  OX  and  OY  through  O 
in  this  plane  and  at  right  angles.  These  two  lines 
are  called  the  AXES  OF  CO-ORDINATES,  the 
horizontal  one  the  axis  of  x,  or  the  x  axis,  the  other 
the  axis  of  y,  or  the  y  axis.  The  distance  BP 
or  OA  is  denoted  by  x  and  called  the  ABSCISSA  of 
the  point  P.  The  distance  AP  is  denoted  oy  y  and  called  the  ORDINATE  of  the  point  P. 

The  abscissa  x  is  positive  to  the  right,  negative  to  the  left  of  the  origin,  while  the  ordi- 
nate  y  is  positive  when  laid  off  above,  and  negative  when  below,  the  origin. 

Any  point  in  the  plane  is  thus  located  with  respect  to  O.  If  a  point  is  in  the  first  quad- 
rant, its  co-ordinates  are  -{-  ,r,  -j-  y\  if  in  the  second  quadrant,  —  x  and  -\-  y\  if  in  the  third 
quadrant,  —  x  and  —  y\  if  in  the  fourth  quadrant,  -j-  x  and  —  y. 

The  axis  OX  is  always  taken  so  that  OA  =  -f-  x  is  towards  the  rig/it  when  we  look  along 
O  Y  from  O  towards  Y. 

When  the  point  is  in  a  known  plane,  the  co-ordinates  are  PLANE  co-ordinates. 
Cartesian  Co-ordinates — Space. — If  the  point  P  is  not  in  a  known  plane,  we  take  three 
axes  through  the  origin,  all  at  right  angles.      Two  of  these  we  denote  by  .A' and  Fas  before; 
the   third,  at  right  angles  to  the  plane  of  XY,  we 
call  the  axis  of  z  or  the  z  axis. 

Thus  the  position  of  the  point  Pis  given  by  the 
distance  OA  =  x,  the  distance  AB  or  OC  —  y,  and 
the  distance  BP  —  2. 

These  are  the  SPACE  co-ordinates  of  the  point  P. 
The  signs  prefixed  to  the  co-ordinates  indicate 
the  quadrant  in  which  the  point  is  located  as  before. 
Thus  -+-  **  4-  jand  ±2  denote  a  point  in  the  first 
quadrant  either  above  or  below  the  plane  of  XY\ 
-  x%  +/,  ±  z,  a  point  in  the  second  quadrant  above 
or  below  the  plane  of  XY\  —  x,  —  y,  ±.  z,  and 
+  •*•  —  }'>  ±  -.  points  in  the  third  and  fourth  quad- 
rants  above  or  below  the  plane  of  XY. 

The  plane  of  XYls  usually  taken  as  horizontal, 
so  that  OZ  points  to  the  zenith,  and  the  axis  OX  is 

always  taken  so  that   OA   -  -  +  x  is  towards  the  right  when  we   look  along   O  Y,  f  row    t 
towards    Y.     Thus  if  XOY  is  the  plane   of  the  horizon,  Z  is  the  zenith.      If,  then,  O\      - 
towards  the  north  point  of  the  horizon,  OX  is  towards  the  east. 

Angles  about  O  measured  in  the  plane  XY  are  always  measured  round  from  OX  tow- 
ards OY;  in  the  plane  YZ  from  OY  towards  OZ\  in  the  plane  ZX  from  OZ  towards  OX, 
as  indicated  by  the  arrows  in  the  figure. 


CHAP.  II.]  DIRECTION  COSINES.  13 

Direction  Cosines.  —  If  we  join  the  origin  O  and  the  point  P  by  a  line  and  denote  the 
angle  of  OP  with  the  X  axis  by  a,  with  the  F  axis  by  ft,  and  with  the  Z  axis  by  y,  we 
have  the  relations 

x  =.  OP  cos  a,  y  =  OP  cos  ft,  z  =  OP  cos  y. 

These  cosines  are  called  the  DIRECTION  COSINES  of  OP,  and  they  always  have  the  same  sign 
as  the  co-ordinates  x,  y  and  z  respectively. 

Since  OP  is  the  diagonal  of  a  parallelopipedon,  we  have 

OP2  =  x2  4-  j2  -f-  ,s2  —  OP\cos*<x  4-  cos2/?  4-  cos2/). 
Hence 

cos2*  4-  cos2/?  -f  cos2/  =  i  ..........     (i) 

If,  therefore,  any  two  of  the  direction  cosines  are  given,  the  magnitude  of  the  third 
can  always  be  found. 

Since,  by  trigonometry, 

COS  20C  —  2  COS2*  —   I,  COS  2ft  =  2  COS2/?  —   I,  .  COS  2y  =  2  COsV  —  If 

we  have 

cos  2  a  -f-  cos  2ft  4~  cos  2y  =  2  (cos2  a  -f-  cos2/?  -}-  cosV)  —  3» 
or,  from  (i), 

cos  2a  4-  cos  2ft  -f-  cos  2y  =  —  i  .........     (2) 

From  (2)  if  any  two  direction  cosines  are  given  in  magnitude  and  sign,  the  third  can  be 
found  in  magnitude  and  sign. 

Again,  since,  from  trigonometry, 

cos  (a  4-  ft}  =  cos  a  cos  ft  —  sin  a  sin  ft,       cos  (a  —  ft)  =:  cos  a  cos  ft  -\-  sin  a  sin  /?, 
we  have 

cos  (at  4~  ft}  cos  0*  ~  /^)  =  cos2**  cos2/?  —  sin2**  sin2/?. 


Since,  by  trigonometry, 

sin2<*  =i  —  cos2or,          sin2/?  =  I  —  cos2y#, 
we  have 

cos  (a  4-  ft}  cos  (a  -/?)=-  i  4-  cos2*  4-  cos2/?. 
Hence,  from  (i), 

cos  (<*  4-  /?)  cos  (a  —  /?)  4-  cosV  =  O  .....      ;       .      (3) 

We  shall  have  occasion  to  refer  to  these  equations  hereafter. 


MEASURABLE  RELATIONS  OF  MASS  AND  SPACE. 



CHAPTER  I. 

. 

MASS  AND  DENSITY. 

Mass  and  Space  Relations. — As  we  have  seen  (page  i),  we  deal  in  mechanics  with 
measurable  relations  of  matter,  space  and  time.  Space  relations  alone  constitute  geometry, 
and  a  knowledge  of  geometry  is  therefore  essential  before  taking  up  the  subject  of 
mechanics. 

There  are  certain  measurable  relations  of  matter  and  space  alone  which  are  equally 
essential,  and  it  will  be  found  advantageous  for  the  student  to  be  familiar  with  these  also, 
before  taking  up  the  subject  of  mechanics.  We  therefore  give  these  relations  first. 

Mass — Unit  of  Mass.— The  MASS  of  a  body  is  proportional  to  the  quantity  of  matter 
the  body  contains.  We  do  not  know  what  matter  is,  but  if  we  take  some  one  definite 
unchanging  body  as  a  standard,  it  is  possible  to  compare  other  bodies  with  it.  That  is,  we 
can  say  that,  whatever  matter  may  be,  a  given  body  has  twice  or  three  times,  etc.,  as  much 
matter  as  the  standard. 

Mass  then,  like  position,  rest  and  motion,  is  relative. 

The  unit  of  mass  adopted  is  the  standard  pound  or  the  standard  kilogram.  These  units, 
as  we  have  seen  (page  6),  are  definite  unchanging  bodies  and  therefore  contain  definite 
unchanging  quantities  of  matter. 

Measurement  of  Mass. — We  shall  see  later,  when  we  consider  the  mutual  relations  of 
matter  and  force  (page  212),  that  when  two  bodies  exactly  balance  in  an  accurate  equal- 
armed  balance,  then  these  two  bodies  must  contain  equal  quantities  of  matter  and  therefore 
have  equal  mass. 

By  means  of  the  balance,  then,  we  can  readily  duplicate  standard  masses  and  deter- 
mine the  mass  of  any  given  body  relatively  to  the  mass  of  any  assumed  standard  mass. 

The  student  will  at  once  recognize  this  as  in  accord  with  daily  experience.  Thus 
we  say  that  the  mass  of  a  body  is  2,  3  or  4  Ibs.  when  it  exactly  balances  2,  3  or  4  standard 
Ibs.  In  such  case  the  quantity  of  matter  contained  by  the  body  is  2,  3  or  4  times  that  con- 
tained by  the  standard. 

Mass  Independent  of  Gravity — The  mass  of  a  body  thus  determined  is  in  common 
language  called  the  "  weight  of  the  body."  Thus  we  speak  of  a  "  weight  of  four  pounds." 

In  the  language  of  mechanics  this  is  incorrect.  The  "  weight  "  of  a  body  is,  strictly  speak- 
ing, the  force  wit/i  which  tin-  ciirtlt  attracts  it.  This  force  varies  with  the  locality  and  the  height 
above  sea-level.  It  is  therefore  a  variable  quantity  for  the  same  body. 

14 


CHAP.  I.]  NOTATION  FOR.   MASS.  15 

But  the  mass  or  quantity  of  matter  of  course  must  remain  the  same  whatever  the  local- 
ity. Two  bodies  of  equal  mass,  which  therefore- balance  on  an  equal-armed  balance  in  any 
locality,  would,  as  we  shall  see  later  (page  2  12),  balance  in  any  other  locality.  Gravity  is  thus 
made  use  of  in  determining  mass,  because  weight  is  always  proportional  to  mass.  But  the 
mass  thus  determined  is  independent  of  gravity. 

When,  therefore,  we  speak  of  a  "  mass  of  4  Ibs. "  we  mean  a  definite  invariable  quan- 
tity of  matter.  When  we  speak  of  a  "weight  of  4  pounds"  we  mean  the  force  of  attraction 
of  the  earth  for  a  mass  of  4  Ibs.  at  some  specified  locality. 

The  operation  of  finding  the  mass  of  a  body  by  an  equal-armed  balance  is  then  not, 
strictly  speaking,  "  weighing"  a  body.  It  is  "  balancing"  a  body.  The  operation  deter 
mines  the  mass  and  not  the  weight.  It  gives  the  same  result  for  any  locality,  whereas  the 
weight  varies  with  the  locality. 

To  determine  the  weight  proper,  we  should  use  a  spring  or  dynamometer.  This  would 
show  a  different  weight  in  different  localities,  but  the  suspended  mass  would  be  always  the 
same. 

Notation  for  Mass. — When  the  mass  of  a  body  is  2,  3  or  4  pounds,  it  is  customary  to 
write  it  2  Ibs.,  3  Ibs.,  4  Ibs.  Here  the  abbreviation  "  Ib."  stands  for  the  Latin  word  "  libra  " 
(balance),  and  thus  indicates  that  the  mass  has  been  determined  by  balancing. 

To  avoid  confusion,  it  will  be  well  for  the  student  always  to  adhere  to  this  notation. 
Thus  "  4  Ibs."  means  a  mass,  while  "  4  pounds  "  means  the  weight  of  4  Ibs.  at  some  speci- 
fied locality.  The  expression  "4  Ibs."  should  really  read  "4  libras,"  but  as  this  word 
"  libra"  has  never  come  into  common  use,  no  one  would  understand  what  is  meant,  and 
"  4  Ibs."  must  therefore  be  read  "  four  pounds."  But  as  the  abbreviation  "Ibs."  is  in  com- 
mon use,  we  can  at  least  make  the  distinction  to  the  eye,  if  not  to  the  ear. 

In  the  French  system  the  distinction  is  complete,  for  we  speak  of  "  4  grams  "  when  we 
mean  mass  and  4  dynes  when  we  mean  force,  grams  being  measured  by  the  balance,  and 
dynes  by  the  spring  or  "  dynamometer." 

We  shall  always  denote  the  mass  of  a  body  by  the  letter  m  with  a  dash  above  it, 
thus:  m. 

Density. — The  number  of  units  of  mass  of  a  body  divided  by  its  number  of  units  of 
volume,  or  the  -mass per  unit  of  volume,  is  the  MEAN  DENSITY  of  the  body. 

The  mean  density  gives,  then,  the  number  of  Ibs.  in  a  cubic  foot,  or  the  number  of  grams 
in  a  cubic  centimeter. 

The  density  at  a  given  point  of  a  body  is  the  ratio  of  mass  to  volume  of  an  indefinitely 
small  volume  of  the  body  at  that  point.  If  this  is  the  same  at  all  points,  the  body  is  HOMO- 
GENEOUS, or  the  density  is  UNIFORM.  If  it  varies,  the  density  is  variable  and  the  body  is 
non-homogeneous. 

The  density  of  a  body  in  a  given  state  is  the  mass  per  unit  of  volume  of  any  portion  of 
the  body  in  that  state. 

When  the  length  of  a  body  is  great  relatively  to  its  other  dimensions,  the  mass  per  unit 
of  length  is  called  its  mean  LINEAR  DENSITY. 

For  a  thin  body  the  mass  per  unit  of  area  is  called  its  mean  SURFACE  DENSITV. 

If  m  is  the  mass  of  a  homogeneous  body  and  V  its  volume  and  d  its  density,  we  have  then 


or  density  equals  mass  per  unit  of  volume. 


1 6  MEASURABLE  RELATIONS  OF  MASS  AND  SPACE.  [CHAP.  I. 

Unit  of  Density. — If  [J/J  is  the  unit  of  mass  and  m  the  number  of  units  of  mass,  [F] 
the  unit  of  volume  and  Fthe  number  of  units  of  volume,  [D]  the  unit  of  density  and  6  the 
number  of  units  of  density,  we  have 


We  shall  have 

tf_jin 

provided  we  take 


tn 

The  unit  of  density,  then,  is  one  unit  of  mass  per  unit  of  volume,  as  one  Ib.  per  cubic 
foot,  or  one  gram  per  cubic  centimeter. 

Specific  Mass The  density-ratio  of  a  body  relatively  to  that  of  some  specified  stand- 
ard substance  is  properly  called  its  SPECIFIC  MASS.  It  is  often  called  "  specific  gravity,"  as  a 
consequence  of  not  distinguishing  between  weight  and  mass.  The  ideas  are  different,  but 
the  numerical  values  the  same,  since  the  weight  of  a  body  is  proportional  to  its  mass. 

The  standard  substance  taken  is  water.  If  y  is  the  density  or  mass  of  a  unit  of  volume 
of  water,  and  8  the  density  or  mass  of  a  unit  of  volume  of  any  other  body,  then  the  specific 
mass  e  is  given  by 

-I <«> 

Since  6  =  -,,•  where  m  is  the  mass  and  V  the  volume  of  the  body,  we  have 


Since  y  is  the  mass  of  a  unit  of  volume  of  water,  yV  is  the  mass  of  a  volume  of  water 
equal  in  volume  to  the  body.  Hence  the  specific  mass  of  any  body  is  equal  to  the  ratio  of  its 
mass  to  the  mass  of  an  equal  volume  of  water. 

In  the  English  system  the  mass  of  one  cubic  foot  of  pure  water  at  4°C.,  or  the  point  of 
maximum  density,  is  nearly  1000  ounces,  or  62.5  Ibs.  (more  exactly  998.6  ounces).  The 
density  of  water  is  then  about  62.5  Ibs.  per  cubic  foot,  or 

_  62. 5  Ibs.. 
"  i  cub.  ft. 

If,  then,  Fis  one  cubic  foot,  we  have,  from  (3), 

m  Ibs. 


where  m  is  the  mass  in  Ibs.  of  one  cubic  foot  of  any  body. 


CHAP.  I.]  DETERMINATION  OF  SPECIFIC  MASS.  ^^ 

In  the  C.G.S.  system,  the  mass  of  one  cubic  centimeter  of  pure  water  at  4°  C.  is  very 
nearly  one  gram,  and  was  intended  to  be  so  exactly.  The  density  of  water  by  this  system 
is  then 

I  gram 
=  i  cub.  c." 

If,  then,  Fis  one  cubic  centimeter,  we  have,  from  (3), 

m  grams      — 

6  —    ^  m 

I  gram 

where  m  is  the  mass  in  grams  of  one  cubic  centimeter.  That  is,  the  mass  in  grams  of  one 
cubic  c'entimeter  gives  at  once  the  specific  mass,  while  in  the  English  system  the  mass  in 
Ibs.  of  one  cubic  foot  mnst  be  divided  by  62.5.  Or  inversely  the  specific  mass  of  anybody 
gives  at  once  the  mass  in  grams  of  one  cubic  centimeter,  while  it  must  be  multiplied  by  62.5 
to  obtain  the  mass  in  Ibs.  of  one  cubic  foot. 

Determination  of  Specific  Mass. — A  body  totally  immersed  in  water  displaces  its  own 
volume  of  water.  It  is  a  well-known  physical  fact  that  a  body  so  immersed  is  buoyed  up  by 
a  force  equal  to  the  weight  of  the  volume  of  water  displaced. 

If,  then,  a  body  is  weighed  in  air  and  then  weighed  again  while  wholly  immersed  in 
water,  the  weight  in  air  is  proportional  to  the  mass  of  the  body,  and  the  loss  of  weight  is 
proportional  to  the  mass  of  the  displaced  water.  When  very  great  accuracy  is  required 
the  body  should  be  weighed  in  a  vacuum,  or  else  allowance  must  be  made  for  the  buoyant 
force  of  the  air.  But  in  all  cases  of  practical  mechanics  this  is  an  unnecessary  refinement. 

To  determine  the  specific  mass,  then,  we  have  only  to  divide  the  weight  of  the  body  by 
its  loss  of  weight  in  water,  or,  since  weight  is  proportional  to  mass,  we  divide  the  number  of 
units  of  mass  of  the  body  by  the  number  of  units  of  mass  of  an  equal  volume  of  water. 

Table  of  Specific  Mass. — In  the  following  table  the  density-ratios,  or  specific  mass,  or 
so-called  "specific  gravity"  relative  to  water,  of  a  few  substances  are  given. 

The  exact  value  in  many  cases  will  depend  on  the  temperature  and  the  mechanical 
process,  such  as  hammering,  etc.,  to  which  the  bodies  have  been  subjected  in  their  manu- 
facture. 


Air  at  o°  C 0.0012759 

Alcohol  at  o°  C 0.791 

Turpentine  at  o°  C o.  870 

Ice 0.92 

Sea-water  at  o°  C 1.026 

Crown   glass .  .  .  .  , 2.5 

Flint  glass 3.0 

Aluminum 2.6 

Zinc 7.0 


Tin 7-4 

Iron 7.7 

Copper 8.8 

Silver 10.5 

Lead 11.4 

Mercury  at  o°  C 13. 596 

Gold.    19.3 

Platinum 21.5 


Examples. — (i)  The  mass  of  a  piece  of  limestone  is  310  grams.       When  immersed  in  water  it  balances  a 
mass  of  188.5  grams.     Find  the  specific  mass. 

ANS.  The  mass  of  an  equal  volume  of  water  is  310—  188.5  =  121.5  grams.     Hence  the  specific  mass  is 
310 


1 8  MEASURABLE  RELATIONS  OF  MASS  AND  SPACE.  [CHAP.  I 

(2)  In  ordtr  to  find  the  specific  mass  of  a  piece  of  oak,  a  piece  of  wire  which  lost  10.5  grams  when  ivci^  hcd 
in  water  was  wrapped  around  the  wood,  the  mass  of  which  was  426.5  grams.      The  compound  mass  was  484.5 
grams  lighter  in  the  water  than  in  the  air.     Find  the  specific  mass. 

ANS.  The  loss  of  the  wood  alone  was  484.5  —  10.5  =  474  grams.    Hence  the  specific  mass  is^2 --'-?-  =  0.9. 

(3)  An  iron  vessel  filled  with  mercury  has  a  mass  0/300  Ids.  and  when  weighed  in  water  shows  a  loss  of  40 
Ibs.    If  the  specific  mass  of  the  iron  is  7.7  and  of  the  mercury  13.6,  find  the  mass  of  the  -vessel  and  of  the  mercury. 

S  m 

ANS.     Since  specific  mass  e  =  -,  where  5  is  density  and  y  is  density  of  water,  and  since  5  =  — ,  where  m 

is  mass  and  v  is  volume,  we  have  e  =  — ,  or  v  =  — . 

Let  nil  be  the  mass  of  the  iron  and  m«  of  the  mercury,  and  m  the  combined  mass. 

Then  for  the  volume  of  the  iron  we  can  write  v\  =       -,  for  the  volume  of  the  mercury  z/»  =  - — -*,  and  for 

m 

the  combined  volume  v  = 

Since  the  combined  volume  is  equal  to  the  sum  of  the  volumes,  we  have 

°*L +*!=»,   or     "!  +  ?-'  =  5.  .      I 


Since  the  combined  mass  is  equal  to  the  sum  of  the  masses,  we  have 

m,  +  m,  =  m. (2) 

Combining  (i)  and  (2)  we  obtain 

-  _—  !_  -L 

m,  =  m  — — —,        m»  =  m — . 

ii  ii 

fl  ft  €t          €\ 

In  the  present  case  we  have  e  =  5— ,  e,  =  7.7,  «?,  =  13.6  and  m  =  500  Ibs.     Hence 

m,  =  57{|  Ibs.,  m,  =  442^  Ibs. 

NOTE. — This  is  called  the  problem  of  Archimedes,  because  first  solved  by  him  for  an  alloy  of  gold  and 
silver. 

Its  application  to  alloys  or  chemical  compositions  is,  however,  limited,  as  in  general  in  such  cases 
there  is  a  change  of  volume,  so  that  the  combined  volume  is  not  equal  to  the  sum  of  the  volumes  of  the 
components,  and  equation  (i)  does  not  hold. 

(4)    To  find  the  specific  mass  of  a  mixture,  given  the  volume  or  mass  and  specific  mass  of  each  constituent. 

ANS.  We  must  assume  the  volume  of  the  mixture  equal  to  the  sum  of  the  volumes  of  the  constituents. 
This  is  not  invariably  the  case,  especially  where  there  is  chemical  union. 

mi,  m,.,  m3,  etc.,  be  the  masses  of  the  constituents; 
*i.   f»,  ^9,    "      "     "    specific  masses  of  the  constituents ; 
v\,  vt,  vt, volumes  of  the  constituents. 

Let  m,  v  and  «=  be  the  mass,  volume  and  specific  mass  of  the  mixture.  Let  y  be  the  density  or  mass  of  a 
unit  of  volume  of  water. 

Then  nii  +  m,  +  ms  +  etc.  =  m. 

But     m,  =  ftrvt.     5,  =  f.v,,.      etc.     Hence. 

y  +  etc.  =  evy,   or     e  =  g|Z/»  +  f*>*  +  etc«t 
We  have  also  v  =  z/,  +  v,  +  Vt  +  etc.    Therefore 

€  ~     v^  +  v,  +  etc. W 


CHAP.  I.]  EXAMPLES. 


Again  we  have  z/»  =  — i-,    v,  =  — ,     etc.     Hence 


_  mi  +  ma  +  etc. 


. 
-  +  --  +  etc. 


(5)  A  fiat  bar  of  iron  4f  inches  wide  and  §  inch  thick  has  a  linear  density  of  9.91  Ibs.  per  foot.     Find  the 
mass  of  a  bar  of  iron  i  inch  square  and  i yard  long. 

ANS.  10  Ibs. 

(6)  From  the  preceding  example  state  a  rule  for  finding  the  mass  per  foot  of  a  bar  of  iron  of  constant  area 
of  cross-section  ;  also  for  finding  the  section  if  the  mass  per  foot  is  given. 

ANS.   To  find  the  mass  per  foot  in  Ibs.,  multiply  the  area  of  cross-section  in  square  inches  by  10  and 
divide  by  3. 

To  find  the  area  of  cross-section  in  square  inches,  multiply  the  mass  per  foot  in  Ibs.  by  3  and  divide  by  10. 

(7)  If  a  railroad  rail  weighs  jo  Ibs.  per  foot  of  length,  -what  is  its  area  of  cross-section^f 
ANS.  15  sq.  inches. 

" 


CHAPTER  II. 

CENTRE  OF  MASS. 

Elementary  Mass  or  Particle. — We  can  consider  the  entire  volume  V  of  any  body  as 
subdivided  into  an  indefinitely  large  number  of  indefinitely  small  ELEMENTS,  each  of  equal 
volume  v,  so  that  V  —  2v.  We  can  carry  this  subdivision  as  far  as  we  please,  until  each 
equal  elementary  volume  can  be  treated  mathematically  as  a  point. 

If,  then,  6  is  the  density  of  such  an  elementary  volume  v,  its  mass  is  m  =  dv,  and  the 
entire  mass  m  of  the  body  is  then  m  =  26v  =  2m. 

We  thus  consider  the  body  as  composed  of  elementary  masses  or  PARTICLES  of  masses 
;//,  ,  ;//2,  MS,  etc.,  if  the  body  itself  is  not  homogeneous,  or  of  elementary  masses  or  particles 
each  of  mass  m  if  the  body  is  homogeneous. 

This  conception  lies  at  the  basis  of  all  our  mathematical  treatment,  and  the  student 
should  especially  note  that  it  involves  no  theory  whatever  as  to  the  actual  constitution  of 
matter  or  as  to  what  matter  really  is,  i.e.,  as  to  whether  a  body  is  really  composed  of 
"  atoms  "  or  "molecules"  or  "  particles,"  or  as  to  whether  matter  is  "continuous"  or 
"  discontinuous." 

We  commit  ourselves,  then,  to  no  theory  when  we  say  that,  whatever  matter  is,  any 
body  can  be  divided  into  equal  portions  so  small  that  each  portion  can  be  considered  as 
homogeneous  and  treated  as  a  point.  Whenever  we  speak  of  a  "  particle,"  it  is  to  such  a 
portion  that  we  have  reference. 

Whatever,  then,  the  constitution  of  -matter  may  really  be,  we  can  consider  and  treat  a 
body  as  composed  of  an  indefinitely  large  number  of  indefinitely  small  homogeneous  elements 
or  material  particles  of  equal  volume,  so  small  that  each  may  be  treated  as  a  point.  We  denote 
the  mass  of  such  an  element  by  m. 

Material  Surface. — A  MATERIAL  SURFACE,  then,  is  a  body  whose  uniform  thickness  is 
small  compared  to  its  other  two  dimensions.  We  can  consider  the  entire  area  A  of  such  a 
surface  as  made  up  of  indefinitely  small  elements  of  area  a,  so  that  A  =  2a.  Let  6  be  the 
surface  density  (page  1 5)  of  such  an  element.  Then  its  mass  is  m  =  da,  and  the  entire  mass 

m  of  the  material  surface  is  m  =  2Sa. 

Material  Line. — A  MATERIAL  LINE  is  a  body  whose  length  is  large  compared  to  its 
other  dimensions,  and  whose  sectional  area  at  right  angles  to  the  length  is  constant.  We 
can  consider  the  length  L  of  such  a  line  as  made  up  of  indefinitely  small  elements  of  length 
/,  so  that  L  =  2f.  Let  6  be  the  linear  density  (page  15)  of  such  an  element.  Then  its 
mass  is  m  =  dl,  and  the  entire  mass  m  of  the  material  line  is  m  =  2dl. 

Centre  of  Mass — We  consider,  then,  a  material  body,  surface  or  line  as  composed  of  an 
indefinitely  large  number  of  indefinitely  small  particles. 

The  centre  of  mass  of  such  a  body,  surface  or  line  is  that  point  whose  distance  from  any 
plane  is  equal  to  the  average  distance  of  all  the  particles  from  that  plane. 


CHAP.  II.  ] 


CENTRE  OF  MASS. 


Thus  suppose  a  body  composed  of  particles  whose  masses  are  mv  m.2,  my  etc. 
the  first  to   contain  a  number  nl  of  smaller  particles  of 
equal  mass  ;«,  the  second  a  number  n2  of  the  same  mass 
m,  and  so  on.      Then 


21 

Suppose 


m,  —  njn, 


etc. 


. *m 


Let  xlt  xz,  x^,  etc.,  be  the  distances  of  m^ ,  m2 ,  ;;/3  ,  etc.,      m," 
from  the  plane   FZ,  each  x  being  taken  with  its  proper 
sign.      Let  the  entire  number  of  equal  particles  of  mass 
m  be  N,  so  that  the  total  mass  of  the  body  is 

m  =  Nm. 

Then  we  have  for  the  average  distance  of  all  the  particles 

from  the  plane  FZ,  that  is  for  the  distance  of  the  centre  of  mass  x  from  the  plane 


N 
If  we  multiply  numerator  and  denominator  by  mt  we  have 

mzxz  -f  etc. 

"  — 


_ 
•*  ~~ 


- 
m  m 

In  the  same  way  we  obtain  for  the  distance  y  of  the  centre  of  mass  from  the  plane 


and  for  the  distance  z  of  the  centre  of  mass  from  the  plane  XY 


Position  of  Centre  of  Mass  in  General  —  If,  then,  z;  is  the  volume  of  an  element  of  a 
body,  $  the  density,  and  m  the  mass,  we  have  m  =  6v.  The  entire^mass  m  is  m  =  26v,  and 
we  have,  from  the  preceding  equations, 


m 


in 


(I) 


if  the  body  is  homogeneous,  ft  is  constant  for  every  element.      It  therefore  cancels  out,  and 
we  have  for  a  homogeneous  body 


where  V  is  the  entire  volume. 


22  MEASURABLE  RELATIONS  OF  MASS  AND  SPACE.  [CHAP.  II. 

In  the  same  way  if  a  is  the  area  of  an  element  of  a  material  surface,  8  the  surface  density, 
and  m  the  mass,  we  have  m  =  da.  The  entire  mass  m  is  m  =  28a,  and  we  have  for  a 
material  surface 


28  ax  2  day  28az 

*=~'  ='         Z=-- 


If  the  surface  is  homogeneous,  8  cancels  out  and  we  have 

-      2ax          -      2ay  -      2az 


where  A  is  the  entire  area. 

So  also,  if  ds  is  the  length  of  an  element  of  a  material  line,  and  8  the  linear  density,  we 
have  m  =  S.ds.  The  entire  mass  is  m  =  26.ds,  and  we  have  for  a  material  line 

28.ds.x  -      28.ds.y  -      2S.ds.z  /TTT^ 

*  =  :».*-•     '  =  -20^'     *  =  -^5T (III) 

If  the  line  is  homogeneous,  8  cancels  out  and  we  have 

-  _  2ds.x  -  _  2ds.y  -  _  2ds.z  ,  * 

s  s  s 

where  s  is  the  entire  length. 

Moment  of  Mass,  Volume,  Area,  Line. — We  call  the  product  of  any  mass  m,  volume 
V,  area  A  or  line  S  by  the  distance  of  its  centre  of  mass  from  any  plane,  the  MOMENT  of  that 
mass,  volume,  area  or  line,  relative  to  that  plane. 

We  can  then  express  all  the  preceding  equations  by  the  simple  statement  that  the 
moment  of  the  total  mass,  volume,  area,  or  length  relative  to  any  plane  is  equal  to  the  algebraic 
sum  of  the  moments  of  all  the  elementary  masses,  volumes,  areas  or  lines  relative  to  the  same 
plane. 

In  taking  the  algebraic  sum,  distances  are  to  be  taken  with  their  proper  signs,  positive 
in  the  directions  OX,  OY,  OZ,  negative  in  the  opposite  directions. 

Centre  of  Gravity. — We  shall  see  hereafter  (page  190)  that  the  centre  of  mass  of  a  body 
coincides  with  the  point  of  application  of  the  resultant  of  a  system  of  parallel  forces. 

The  attraction  of  the  earth  for  a  body  is  the  resultant  of  a  system  of  forces  acting  upon 
the  particles  of  a  body,  each  directed  towards  the  centre  of  the  earth. 

We  have  thus  a  system  of  forces  not  strictly  parallel.  But  practically  the  deviation  from 
parallelism  is  insignificant,  since  the  greatest  dimension  of  any  body  on  the  earth  with  which 
we  have  to  do  is  insignificant  in  comparison  with  the  diameter  o*f  the  earth.  Hence  the 
resultant  force  of  gravity  passes  practically  through  the  centre  of  mass.  This  resultant  is  the 
weight  of  the  body.  The  weight  of  the  body  acts  practically,  therefore,  at  the  centre  of  mass. 

To  determine  the  centre  of  mass  by  experiment,  we  have  then  only  to  suspend  it  suc- 
cessively from  two  different  points  of  its  surface.  The  two  lines  of  suspension,  if  produced, 
must  intersect  practically  at  the  centre  of  mass. 

The  centre  of  mass  is  therefore  often  called  the  "centre  of  gravity."  This  term  is, 
however,  strictly  speaking,  incorrect.  Centre  of  mass  is  not  dependent  at  all  upon  gravity 
any  more  than  mass  itself.  We  can  make  use  of  gravity  in  finding  it,  just  as  we  do  in 
measuring  mass  (page  14). 


CHAP.  II.]  ORIGIN  AT   THE  CENTRE  OF  MASS.  23 

The  term  "  centre  of  gravity"  can  only  be  properly  applied  t6  that  point  at  which  if  t^lie 
entire  mass  of  the  body  were  concentrated,  it  would  attract  and  be  attracted,  for  all  positions 
of  the  body,  just  the  same  as  the  body  itself.  In  this  sense,  as  we  shall  see  hereafter 
(page  207),  only  a  few  bodies  possess  a  centre  of  gravity.  But  all  bodies  possess  a  centre  of 
mass.  The  two  conceptions  are  therefore  entirely  distinct. 

Origin  at  the  Centre  of  Mass.  —  In  taking  the  algebraic  sums  of  the  elementary 
moments  2wx,  *2my,  2mz,  we  must,  as  already  noted,  take  x,  y,  z  for  each  element  with 
their  proper  signs.  If,  then,  we  take  the  origin  of  co-ordinates  at  the  centre  of  mass,  we 
must  have  x  =  o,  y  =  o,  z  =  o.  Hence  in  this  case 


—  o,  "2my  =  o,  2mz  =  o  .......     (i) 

If  we  take  polar  co-ordinates  and  take  the  pole  at  the  centre  of  mass,  we  have 

2mr  =  o,      ............     (2) 

where  r  is  the  distance  of  any  particle  from  the  pole. 

That  is,  the  algebraic  sum  of  the  moments  of  all  the  elements  relative  to  the  centre  of 
mass  is  zero. 

Plane  and  Axis  of  Symmetry.  —  A  body  is  symmetrical  with  respect  to  a  plane  when 
for  each  element  on  one  side  of  the  plane  there  is  an  equal,  similarly  situated  element  on 
the  other  side.  In  such  case  the  centre  of  mass  is  in  this  plane. 

A  body  is  symmetrical  with  respect  to  an  axis  when  it  is  symmetrical  with  respect  to 
two  planes  passing  through  that  axis.  In  such  case  the  centre  of  mass  is  in  that  axis. 

If  a  body  is  symmetrical  with  respect  to  two  axes,  the  centre  of  mass  is  at  their  inter- 
section. 

Material  Line,  Area,  Volume.  —  There  is  of  course  a  certain  inconsistency  in  speaking 
of  the  centre  of  mass  of  geometrical  lines,  areas  and  volumes.  The  expression  is,  however, 
allowable,  since  we  are  understood  to  mean  a  material  line,  area  or  volume,  8  being  linear, 
surface  or  body  density,  as  defined  on  page  15.  In  the  case  of  a  homogeneous  line,  area 
or  volume,  #  'cancels  out,  as  we  have  seen  in  the  equations  of  page  22,  and  we  can  then 
allowably  speak  of  the  centre  of  mass  of  the  line,  area  or  volume  itself. 

Determination  of  Centre  of  Mass.  —  We  have  already  seen  how  to  determine  the  posi- 
tion of  the  centre  of  mass  by  experiment  (page  22). 

The  position  of  the  centre  of  mass  for  many  homogeneous  bodies  may  often  be  told  at 
once,  by  applying  the  principle  of  symmetry  just  given. 

Thus  the  centre  of  mass  of  a  homogeneous  straight  line  is  at  its  middle  point.  For  a 
homogeneous  circle,  or  circular  area,  or  sphere,  or  spherical  shell  it  is  at  the  centre  of  figure. 
For  a  homogeneous  parallelogram  it  is  at  the  intersection  of  the  two  lines  through  the 
middle  points  of  opposite  sides,  or  at  the  intersection  of  the  two  diagonals,  since  these  are 
lines  of  symmetry.  So  also  for  a  homogeneous  parallelopipedon  the  centre  of  mass  is  at  the 
intersection  of  its  diagonals,  or  any  two  lines  through  the  centre  of  mass  of  opposite  sides. 
For  a  homogeneous  circular  cylinder  with  parallel  ends,  at  the  middle  point  of  the  axis. 

For  a  homogeneous  triangle  the  centre  of  mass  is  at  the  intersection  of  any  two  lines 
from  an  apex  to  the  middle  point  of  the  opposite  side,  since  these  are  lines  of  symmetry. 

For  bodies  in  general  we  may  find  the  centre  of  mass  by  applying  the  equations  of  page 
22,  or  more  generally  by  the  application  of  the  Integral  Calculus  to  effect  the  summations 
indicated. 

It  is  not  necessary  or  desirable  to  occupy  space   here  with  such   applications.      It  is  of 


MEASURABLE  RELATIONS   OF  MASS  AND  SPACE. 


[CHAP.  II. 


far  more  importance  that  the  student  should  understand  clearly  what  the  centre  of  mass  is, 
as  defined  on  page  20,  and  what  use  is  made  of  it,  and  how  it  enters  into  mechanical  prob- 
lems, as  will  be  explained  later  (pages  175  and  191),  than  that  he  should  spend  much  time 
in  finding  its  position. 

In  the  following  examples  we  show  how  to  apply  the  equations  of  page  22  to  a  few 
cases. 


Examples. — (i)  Find  the  centre  of  mass  of  a  homogeneous  circular  arc. 

ANS.  Let  ADD  be  a  homogeneous  circular  arc  with  centre  at  O',  and  let  the  axis  of  X  pass  through  (7 

and  the  centre  D  of  the  arc. 

A  Then  O'D  is  an  axis  of  symmetry,  and  the  centre  of  mass  O  is  on 

this  axis. 

Let  the  chord  AB  =  c,  and  the  length  of  arc  ADB  =  s,  and  r  be  the 
radius.  Take  an  indefinitely  small  element  ad  oi  length  ds  whose  cen- 
tre of  mass  is  at  e,  and  let  ab  be  the  vertical  projection  of  ad  =  ds.  Drop 
eN  perpendicular  to  O'X.  Then  we  have,  by  similar  triangles, 

ds:ab::r\  O'N ; 

-X     or,  since  O'N  =  x,  the  abscissa  of  the  poi  nt  e,  we  have 

x .  ds  =  r  x  ab. 

By  equation  (3),  page  22,  we  have  then  for  the  distance  O'O  =  x  oi 
the  centre  of  mass 

-  _  2.t .  ds       r'Sab 


But  2<i6  is  the  chord  AB  =  c.     Hence 


~  _  rc 
s 


Hence  the  centre  of  mass  O  of  a  circular  arc  ADB  is  in  the  axis  of  symmetry  O' D  at  a  distance  x  =  O'O 
from  the  centre  O',  which  is  a  fourth  proportional  to  the  arc,  the  radius  and  the  chord,  or 


s  :  r  :  :  c  :  x. 


If  6  is  the  central  angle  AO'B  in  radians,  c  =  2r  sin  — ,  s  =  rQ  and 

2 


For  a  semicircles  =  itr,  c  =  2r  or  6  =  it,  sin  — =  i  and 


For  a  circle  s  =  zitr,  c  =  o,  or  fl  =  2*.  sin  —  =  o  and  x  =  o,  or  the  centre  of  mass  is  at  the  centre  of  the 


•ircle. 


BY  CALCULUS.— The  distance  ab  =  <tv  \  we  have  then 

ds  :  dy  ::  r  .  ....     or    xds  —  rdy. 
By  equation  (3),  page  22, 


CHAP.  II.] 


CENTRE  OF  MASS-EXAMPLES. 


(2)  Find  the  centre  of  mass  for  a  homogeneous  triangle. 

ANS.  Let  ABC  be  the  triangle,  the  base  AC  =  b,  and  the  altitude  BP 

By  the  principle  of  symmetry  the  centre 
of  mass  O  is  at  the  intersection  of  any  two 
lines  from  an  apex  to  the  middle  point  of  the 
opposite  side. 

By  geometry,  then,  the  distance  y  of  the 
centre  of  mass  above  the  base  is 

A  Ml>    P  C       A  PPM  C 

To  find  the  distance  x  =  Ap,  let  M  be  the  middle  point  of  the  base,  and  draw  BM  and  drop  the  perpen- 
diculars Op  and  BP.     Then  we  have 

AP  —  h  cot  A, 
and  for  the  distance  MP 

MP  =  h  cot  A  — — ,     or     MP  —  —  —  h  cot  A, 

according  as  A  is  the  smaller  or  larger  base  angle. 
We  have  then  Alp  =  \MP,  or 


h  cot  A       b 
— or 


and  hence  in  both  cases,  if  A  is  acute, 

-       b 


h  cot  A      b  +  h  cot  A 


BY  CALCULUS. — Take  an  elementary  strip  parallel  to  the  base  and  at  a  distance  y  above  it.  Let  the 
length  of  this  strip  be  x.  Its  thickness  will  be  dy,  and  its  area  xdy.  We 
have  then 

h  h 


But  we  have  A  =  — •,  and,  by  proportion, 


Substituting  these  values  of  A  and  x,  we  have 

k  h 

~  — 2  /'  ^        —  ^  r 


~  f (/t - 


b      h  cot  A 

~3+        3 


(3)  Find  the  centre  of  mass  of  a  homogeneous  trapezoid. 

ANS.  Let  A  BCD  be  the  trapezoid.     Denote  the  top  or  smaller  base  BC  by  £,,  arid  the  bottom  or  larger 
base  AD  by  bi,  and  the  altitude  by  h. 

We  have  then  for  the  area  of  the  trapezoid  ' 

/f  r>/~/-i  (^'  "I"    b-i\h 

area  ABLD  •—  —        — • 


We  can  divide  the  trapezoid  into  two  triangles,  ABD 
and  BDC,  by  the  diagonal  BD. 

The  area  of  the  triangle  ABD  is 


and,  as  just  proved  (example  2),  the  distance  of  its  centre  of  mass  above  AD  is  \h,  and  on  the  right  of  A, 
bi  +  h  cot  A 


The  area  of  the  triangle  BDC  is 


26  MEASURABLE  RELATIONS  OF  MASS  AND  SPACE.  [CHAP.  II. 

and  the  distance  of  its  centre  of  mass  above  AD  is  f/*,  and  on  the  right  of  A,  from  equation  (2),  example  (2), 

fa  +  fa  —  h  cot  A      fa  +  fa  +  2  h  cot  A 
h  cot  A  +  — - —  — - — 

We  have  then  for  the  distance/  of  the  centre  of  mass  O  above  AB 

area  ABCD  x  /  =  area  ABD  x  —  +  area  BDC  x  —  ht 
and  for  the  distance  ~x  of  the  centre  of  mass  O  on  the  right  of  A 
area  ABCD  x  x  =  area  ABD  x 


*  +  A  +  area  BDC  x  *'  +  *         *  C0t  A. 


Inserting  the  values  for  the  areas,  we  obtain,  if  A  is  acute, 

-_  (2fa  +  fa)h 


-_£.*  +  Ma  +  V  +  A  (20.   +  fa)  COt  /? 


For  a  parallelogram  b\  =  fa  and  y  ——h,  x  —  — I-  —  h  cot  A.    For  a  rectangle  A  =  90°  and/  =  — ^,  JT  =  —  b. 

Graphical  Construction. — ist.   Draw  lines  Dm\,  Dm*  from  apex  D  to  the  middle  points  m\  and  nit  of 

the  opposite  sides.  Then  the  centre  of  mass  Ot  of  the  triangle 
ABD  is  on  the  line  Dtn\  at  a  distance  DO\  equal  to  f  of  Dm\. 
The  centre  of  mass  O*  of  the  triangle  BDE  is  on  the  line  Din* 
at  a  distance  DO*  equal  to  f  of  Dm*.  If  we  join  O\  and  O* 
thus  found,  the  centre  of  mass  must  be  on  the  line  O\O*. 

"  Draw  m*m»  through  the  middle  points  of  the  parallel 
sides.  This  line  is  an  axis  of  symmetry,  since  it  passes  through 
the  centre  of  all  elements  parallel  to  BC  and  AD.  The  centre 
of  mass  must  therefore  lie  in  the  line  tut»ti.  It  is  therefore  at 
the  intersection  C'of  m*m%  and 

2d.  We  have,  from  the  values  of  x  and/  just  obtained, 
fa  +  x  _  2bi  +  fa  +  A  cot  A 

y  h 

Now  if  we  lay  off  CG  parallel  to  AD  and 
make  it  equal  to  b* ,  and  lay  off  ///"parallel 
to  #6"  and  make  it  equal  to  b\  ,  and  join  F 
and  G,  the  line  FG  must  pass  through  O. 
For  we  have,  by  proportion, 

0i  +  *  _  fa  +  h  cot  A  +  fa  +  fa 

y  h 

Hence  the  intersection  of  FG  with  the  axis  of  symmetry  m,m,  gives  the  centre  of  mass  O. 
3d.  Another  convenient  construction  is  as  follows  : 

Draw  the  diagonals  AC  and  DB  intersecting  at  /.     Lay  off  along  AC  the  distance  Ae  =  1C,  and  along 
bf  c  DB  the  distance  Db  =  IB. 

Bisect  the  diagonals  at  ;//,  and  w3,and  join  n>,<*  a   d 
fft-f.     The  intersection  O  is  the  centre  of  mass. 

/'roo/.—Ry   construction    the    sum    of    the   horizoi  t;  ! 

•m*        ^  projections  of  Ae  and  Db  is  equal  to  l>, ,  and  the  points  c  ;:.  >J 

b  are  both  at  the  same  distance  above  AD. 
Hence  the  line  eb  is  horizontal  and 


We  have  then,  by  construction,  wiw,  also  horizontal  and 


CHAP.  II.]  CENTRE  OF  MASS— EXAMPLES. 

We  have  also,  by  proportion,  if  ji  is  the  vertical  from  /on  BC, 

• —  =  • — - — -,     or    y\  = T~' 

y\       h  — y\  bi  +  o\ 

By  similar  triangles,  \iy  is  the  distance  of  O  above  AD, 

eb  m  \rn-i 


y-y*    ~-v 

->    + 


Inserting  values, 


&*-6i      _  W*  -  h) 

bJi     ~     h 


(261 


y- 


just  as  already  found. 

(4)  Find  the  centre  of  mass  of  a  homogeneous  trapezium. 

ANS.   Let  ABCD  be  the  trapezium.     Denote  AD  by  6,,  £Cby  di ,  AB  by  d  and  DC  by  a.     Also  let 
ft  be  the  angle  of  b\  with  horizontal,  and  take  AD  horizontal. 

We  have  then  for  the  area  of  the  trapezium 

area  ABCD  =  M  *'"  A  +  ***  S''"  C.  ^bl 

We  can  divide  the  trapezium  into  two  triangles,  ADS 
and  BDC,  by  the  diagonal  BD. 

The  area  of  the  triangle  ABD  is 


area  ABD  = 


V/sin  A, 


the  distance  of  its  centre  of  mass  above  AD  is  —       — ,  and  on  the  right  of  A  (example  (2)) 


d  co 


The  area  of  the  triangle  BDC  is 


area  BDC 


b\a  sin  C . 


the  distance  of  its  centre  of  mass  above  AD  is 


d  sin  A  +  «  sin 


,  and  on  the  right  of  A, 


cos  fi  -\-  bi  +  id  cos 


We  have  then  for  the  distance  y  of  the  centre  of  mass  O  above  AB 
area  ABCD  x  y  =  area  & 


^sin  .7  </sin 

x +  area  BDC  x  - 


and  for  the  distance  x  of  the  centre  of  mass  O  on  the  right  of  A 

area  ABCD    x  x  =  area  ABD  x  b*  +  d  cos  -j  +  area  ^Z)C  x 

3 
Inserting  the  values  of  the  areas,  we  have 


*.  cos 


-_  fan  siri  C(d  sin  A  +  a  sin  Z?) 


-  b^  a  sin  C  cos  ft 


sin  C  +  ^Wsin 


y4  +  ^/cos 
b\a  sin  C) 


+  a  sin 


*/  cos 


sin  ^  +  zfaa  sin 


For  a  trapezoid,  /5  =  o,  d  s\\\  A  =  a  sin  /?  =  a  sin  C  =  h, 
and-T  and  "x  reduce  to  the  values  already  found. 

Construction  —  Draw  the  diagonals  BD  and  CA,  find  the 
centre  of  mass  Oi  of  the  triangle  ABC  and  the  centre  of  mass 
O,  of  the  triangle  .4CZ>.  Then  the  centre  of  mass  O  must 
be  on  the  line  OiOt.  Again,  find  the  centre  of  mass  O3  of  the 
triangle  DEC.  and  O,  of  Z>/?.4.  The  centre  of  mass  0  must  be 
on  the  line  O3O*  .  It  is  therefore  at  the  intersection  of  OiO* 
D"  and  6>3<94  . 


MEASURABLE  RELATIONS  OF  MASS  AND  SPACE. 


[CHAP.  II. 


(5)  Find  the  centre  of  mass  of  a  homogeneous  circular  sector, 
A 


ANS.  Let  O'ACBO'  be  the  sector  with  centre  at  O'  and  radius 
a  A  =r. 

The  sector  can  be  divided  into  an  indefinitely  large  number  of 
indefinitely  small  triangles.  The  centre  of  mass  of  each  triangle  is  at 
a  distance  of  Jr  from  O'.  These  centres  form  then  a  homogeneous 
circular  arc  AiCiBi ,  and  the  centre  of  mass  of  the  sector  is  the  centre 
of  mass  of  this  arc. 

F*rom  example  (i),  then,  the  centre  of  mass  O  lies  upon  the  radius 
of  symmetry  O'C  and  at  a  distance  O'O  =  x  from  the  centre  given  by 

— _  chord  AiJSt     2_ 
arc  A,L\B,  '  Jr' 

Let  c  be  the  length  of  chord  AB  of  the  sector,  and  s  the  length 
of  arc  ACB  of  the  sector,  then 


chord 


and  we  have 


arc  AtC\Bt 


—      2     re 


If  0  is  the  central  angle  AO'B  in  radians,  c  =  2r  sin  — ,  s  =  r&  and 


For  the  semicircle,  c  =  2r,  s  =  itr  or  6  =  it,  sin  —  =  i ;  hence 


For  a  quadrant 
For  a  sextant 


Student  should  solve  by  Calculus. 

(6)  Find  the  centre  of  mass  for  a  homogeneous  segment  of  a  circle. 

ANS.  Let  ACB  be  the  segment.  The  centre  of  mass  O  is  in  the 
radius  of  symmetry  O'C. 

Let  the  radius  O 'A  =  r,  and  the  length  of  arc  ACB  =  s.  Then 
the  area  of  the  sector  O'ACBO'  is 

A,  =  -. 

The  distance  O'Oi  of  the  centre  of  mass  0,  of  the  sector  has  just 
been  found  in  example  (5)  to  be 

(70,  =  — , 

y 

where  c  is  the  length  of  chord  AB. 

The  height  of  the  triangle  0AB  is.  I  r'  —  — . 


Its  area  is  then 


The  distance  O'O*  of  the  centre  of  mass  0,  of  the  triangle  is 


CHAP.  II.] 


CENTRE  OF  MASS— EXAMPLES. 


29 


The  area  A  of  the  segment  is  A  =  (Ai  —  At).    Let  O'O  =  x  be  the  distance  of  its  centre  of  mass.     Then 
we  have  the  moment  of  the  sector  equal  to  the  sum  of  the  moments  of  the  triangle  and  segment,  or 

Ai  .  O'Oi  =  At  .  O'Ot  +  (Ai  —  A*)x~. 
Substituting  the  values  of  A\t  A3,  O'Oi,  and  O'Ot,  we  obtain 


12  A 


If  6  is  the  central  angle  AO'B  in  radians,  we  have  s  =  rQ,  c  =  2r  sin  -,  and  hence 

c* 


For  a  semicircular  segment,  Q  =  if,  cos  -=  o,  c  =  2r,  and  hence  x  =  — ,  just  as  we  have  already  found 

in  example  (5). 

Student  should  solve  by  Calculus. 

(7)  Find  the  centre  of  mass  for  a  homogeneous  circular  ring. 

ANS.  Let  the  outer  radius  O'At  be  r\ ,  and  the  inner  radius 
O'At  be  r ,. 

Let  the  length  of  the  outer  arc  A,Ci£t  be  si ,  and  its  chord 
AiBi  be  ci\  the  length  of  the  inner  arc  AtCiB*  be  s*,  and  its  chord 
AtBt  be  ct. 

Then,  from  example  (5),  we  have  for  the  distance  of  the  centre 
of  mass  of  the  sector  O'AiBiO' 


and  its  area 

2 
The  distance  of  the  centre  of  mass  of  the  sector  O  'AtB*O '  is 

Xt  =  —   : 

and  its  area  is 


The  area  of  the  ring  is 


Let  O'O  =  x  be  the  distance  of  its  centre  of  mass. 

Then  we  have  the  moment  of  the  outer  sector  equal  to  the  sum  of  the  moments  of  the  inner  sector  and 
ring,  or 

Aixi  =  AtXt  +  (A  i  —  A*)x. 

Substituting  the  values  of  Ai,  At,  x\,  x*,  we  obtain 


_ 
- 


or,  since  Si  =  riQ,  where  6  is  the  central  angle  A\OBi  in  radians,  and  ci  =  2ri  sin  -, 

.    0 
4  *'"  r,,. ,  _ 


x  = 


(8)  Find  the  centre  of  mass  for  the  homogeneous  surface  of  a  cylinder,  or  of  any  prism. 

ANS.  The  centre  of  mass  for  the  homogeneous  surface  of  a  cylinder  is  at  the  middle  point  of  the  axis. 
For  all  the  elements  of  the  surface,  obtained  by  taking  slices  parallel  to  the  base,  have  their  centres  of  mass 
upon  the  axis.  The  centre  of  mass  of  the  surface  is  then  the  centre  of  mass  of  the  axis. 


MEASURABLE  RELATIONS  OF  MASS  AND  SPACE. 


[CHAP.  II. 


For  the  same  reason  the  centre  of  mass  for  the  homogeneous  surface  of  a  prism  is  at  the  middle  of  the 
line  which  unites  the  centres  of  mass  of  its  bases. 

(9)  Find  the  centre  of  mass  for  the  homogeneous  surface  of  a  right  cone  or  pyramid. 

ANS.  The  centre  of  mass  for  the  homogeneous  surface  of  a  right  cone  is  in  the  axis  at  two-thirds  its 
length  from  the  apex.  For  the  curved  surface  can  be  divided  into  an  indefinite  number  of  indefinitely  small 
triangles.  The  centres  of  mass  of  all  these  triangles  form  a  circle  at  a  distance  of  two-thirds  of  the  axis 
from  the  apex,  whose  centre  of  mass  is  in  the  axis. 

The  same  holds  for  any  right  pyramid. 

(10)  Find  the  centre  of  mass  of  a  solid  homogeneous  prism  or  cylinder  with  parallel  bases. 

ANS.  The  centre  of  mass  of  a  solid  homogeneous  prism  is  at  the  middle  of  the  line  joining  the  centres  of 
mass  of  its  bases.  For  by  passing  planes  parallel  to  the  bases  we  divide  it  into  equal  slices  whose  centres  of 
mass  lie  in  the  axis. 

The  same  holds  for  a  cylinder. 

(M)  Find  the  centre  of  mass  for  a  homogeneous  pyramid  or  cone. 

ANS.  Let  ABCD  be  a  homogeneous  triangular  pyramid.  Take  b  at  the 
middle  point  of  BC,  and  draw  Ad  and  Db.  Take  a  point  a  on  Ad,  so  that 
ab  =  \Ab,  and  a  point  e  on  Db,  so  that  eb  =  %Db.  Join  ae,  and  draw  Da 
and  Ae.  Then  Da  and  Ae  are  axes  of  symmetry  and  the  centre  of  mass  O  is 
at  their  intersection.  But  ae  is  parallel  to  AD  and  equal  to  \AD,  and  the 
triangle  aOe  is  similar  to  AOD.  Hence  aO  =•  OD,  or  \OD  =  3  .  aO,  and 
>C  therefore  aD  =  4  .  aO,  or  aO  =  £  .  aD. 

The  centre  of  mass  is  then  on  the  line  joining  a  vertex  with  the 
centre  of  mass  of  the  opposite  base,  and  at  a  distance  from  the  vertex  of  £ 
the  length  of  this  line. 

Since  any  pyramid  or  cone  is  composed  of  triangular  pyramids  with  a 
common  vertex,  the  same  holds  true  for  the  centre  of  mass. 
We  can  therefore  determine  the  centre  of  mass  of  a  pyramid  or  cone  by  passing  a  plane  parallel  to  the 
base  at  a  distance  from  the  base  of  \  the  altitude,  and  finding  the  centre  of  mass  of  this  section,  or  the 
point  where  it  is  pierced  by  the  line  from  the  apex  to  the  centre  of  mass  of  the  base. 


CHAPTER  III. 

MOMENT  OF  INERTIA.* 

Moment  of  Inertia.  —  The  product  of  any  indefinitely  small  elementary  mass  or  area  or 
volume  by  the  square  of  its  distance  from  any  given  point,  line  or  plane  is  called  the 
MOMENT  OF  INERTIA  of  the  elementary  mass  or  area  or  volume  relative  to  that  point,  line  or 
plane. 

If,  then,  a  is  an  elementary  area  and  r  its  distance  from  any  point,  line  or  plane,  arz  is 
the  moment  of  inertia.  If  m  is  the  mass  of  a  particle,  mrz  is  the  moment  of  inertia.  If  v  is 
the  volume  of  an  element,  vr*  is  the  moment  of  inertia. 

The  summation  for  any  volume,  area  or  mass  of  all  these  products  is  the  moment  of 
inertia  of  the  entire  volume,  area  or  mass,  relative  to  the  assumed  point,  line  or  plane. 
Thus  for  any  area  the  moment  of  inertia  is  2arz,  for  any  volume  2vrz,  for  any  mass  2mr*. 

The  point  chosen  is  always  the  centre  of  mass  of  the  entire  body,  mass,  volume,  or 
area,  unless  otherwise  specified.  The  plane  or  line  chosen  is  always  a  plane  or  line  through 
this  centre  of  mass,  unless  otherwise  specified. 

We  use  the  term  "  centre  of  mass  of  an  area"  or  of  a  volume  in  the  sense  already 
defined  (page  23). 

The  line  relative  to  which  the  moment  of  inertia  is  determined  is  the  AXIS. 

We  always  denote  the  moment  of  inertia  relative  to  the  centre  of  mass,  or  an  axis  or 
plane  through  the  centre  of  mass,  by  /. 

We  have  then  for  the  moment  of  inertia  of  an  area  relative  to  the  centre  of  mass,  or  any 
axis  or  plane  through  the  centre  of  mass, 

f 

For  the  moment  of  inertia  of  a  volume  we  have 

I 
For  the  moment  of  inertia  of  a  mass, 


Significance  of  the  Term.  —  The  term  "  moment  of  inertia"  is  not  well  chosen  according 
to  the  terminology  of  modern  science.  As  we  see  from  the  definition,  it  has  nothing  to  do 
with  "inertia"  (see  page  169).  It  has  been  proposed  to  call  it  ''second  moment  "  of  area 
or  mass.  The  term  "  moment  of  inertia  "  is,  however,  of  such  general  use  that  it  is  hardly 
worth  while  to  try  to  introduce  such  %  change.  It  is  sufficient  for  the  student  to  note  that 
it  is  the  arbitrary  name  for  a  certain  quantity.  This  quantity  occurs  so  often  in  mechanical 
problems  that  it  is  desirable  to  have  a  special  name  for  it,  and  to  discuss  it  thoroughly  in 
advance  of  its  use. 

*  This  chapter  may  be  omitted  here  if  thought  desirable,  but  should  be  taken  before  Kinetics  of  a  Material 
System,  page  297. 

31 


3 2  MEASURABLE  RELATIONS  OF  MASS  AND  SPACE.  [CHAP.  III. 

Radius  of  Gyration. — That  distance  at  which,  if  the  entire  volume,  area  or  mass  were 
concentrated  in  a  point,  the  moment  of  inertia  would  be  the  same  as  for  the  volume,  area  or 
body  itself,  is  called  the  RADIUS  OF  GYRATION  for  the  volume,  area  or  body. 

We  denote  the  radius  of  gyration  relative  to  the  centre  of  mass,  or  a  line  or  plane 
through  the  centre  of  mass,  by  k. 

We  have  then,  by  our  definition  and  notation,  for  a  body  of  mass  m, 


for  an  area  A, 


or          =    .  = 

m        m 


T-  At*  &-1  -2aV* 

-A--/T 

for  a  volume  V, 


Moment  of  Inertia  and  Radius  of  Gyration  in  General.  —  Unless  otherwise  specified  the 
term  moment  of  inertia  always  signifies  the  moment  of  inertia  for  a  body  of  mass  m.  The 
letter  /  always  means  this  moment  of  inertia  for  the  centre  of  mass,  or  an  axis  or  plane 
through  the  centre  of  mass.  The  letter  k  denotes  the  corresponding  radius  of  gyration. 

Any  other  point,  axis  or  plane  is  called  an  ECCENTRIC  point,  axis  or  plane.  The 
moment  of  inertia  is  denoted  in  such  case  by  /',  and  the  corresponding  radius  of  gyration 
by£'. 

Everything  in  what  follows  which  holds  for  /  and  k,  I'  and  k'  ',  as  thus  defined,  fora  body 
of  mass  m,  holds  good  also  for  a  surface  of  area  A  or  a  volume  V,  unless  otherwise  specified. 

Reduction  of  Moment  of  Inertia.  —  If,  then,  /  is  the  moment  of  inertia  for  any  axis 
through  the  centre  of  mass,  and  /'  the  moment  of  inertia  for  any  parallel  axis  at  the  distance 
d,  we  can  easily  prove  the  relation 


That  is,  the  moment  of  inertia  of  a  body  relative  to  any  eccentric  axis  is  equal  to  the 
moment  of  inertia  relative  to  a  parallel  axis  through  the  centre  of  mass  plus  the  product  of  the 
mass  by  the  square  of  tlie  distance  between  the  two  axes. 

The  same  holds  for  the  moment  of  inertia  of  a  surface  if  we  replace  m  by  the  area  A  of 
the  surface,  or  for  a  volume  if  we  replace  m  by  the  volume  V.  We  have  then 

7'  =  7  +  ^,         T'=I+V(P. 

This  is  called  the  theorem  of  moment  of  inertia  for  parallel  axes.  By  means  of 
it  we  can  find  /'  for  any  axis,  if  7  for  a  parallel  axis  through 
the  centre  of  mass  and  the  distance  d  between  these  axes  are 
given.  Conversely,  we  can  find  7  if  7'  and  d  are  given. 

The  proof  is  simple.      Let  m  be  the  mass  of  a   particle 
of  a  body  whose  distance  from  an  axis  through  the  centre  of 
mass  O  of  the  body  is  r.      Let  r'  be  its  distance  from  a  parallel 
axis  at  O',  the  distance  O'  O  between  the  axes  being  d.     Let  V  be  the  angle  of  r  with  O'O. 
Then  we  have  for  any  body 

r  -  2mrf  *     and     7  =  2tnr*. 


CHAP.  III.] 

But  we  have 
Hence 


REDUCTION  OF  MOMENT  OF  INERTIA. 


d*  ±  2rdcos0. 


'*  =  2mr*  +  d*2m  ±  2d2mr  cos  0. 


33 


But  mr  cos  6  is  the  moment  of  m  relative  to  the  axis  through  the  centre  of  mass  O. 
Therefore,  as  we  have  seen  (page  23), 


2mr  cos  6  =  O» 


We  have  then 


2mr'z  =  2mr*  +  d*  2m, 
or,  since  2m  =  m,  the  entire  mass  of  the  body 


The  same  holds  for  an  area  if  we  replace  m  by  A  and  2m  by  2a  =  A,  or  for  a  volume 
if  we  replace  m  by  V  and  2m  by  2v  =  V. 
We  have  then  also 

&'*  =  P  -f  dz. 

It  is  evident,  then,  that  the  moment  of  inertia  relative  to  any  axis  through  the  centre  of 
mass  is  less  than  for  any  parallel  axis,  and  the  radius  of  gyration  relative  to  any  axis  through 
the  centre  of  mass  is  less  than  for  any  parallel  axis. 

Moment  of  Inertia  relative  to  an  Axis.  —  Let  O'X  be  any  axis  and  XO'  Y,  XO'  Z  any 
two  rectangular  planes  passing  through  that  axis. 

Then  for  any  particle  of  a  body  of  mass  m  whose 
co-ordinates  are  x,y,  #,  we  have  the  moment  of  inertia 
relative  to  O'X 

mr*  =  my*  -f~  m^  —  m(y*  -f~  &*•}. 

Summing  the  moments  of  inertia  for  all  the  par- 
ticles of  the  entire  body,  we  have  for  the  moment  of 
inertia  of  the  body  relative  to  the  axis  O'X 

2mr*  =  2myz  +2mz*  =  2[m(y*+*f)]. 

But  2mrz  is  the  moment  of  inertia  of  the  body 
relative  to  the  axis  O'X.      We  denote  it,  therefore,  by 
I'x.     Also,  2my*  is  the  moment  of  inertia  of  the  body 
relative  to   the  plane   ZX,  and  2mz*  relative  to  the        '. 
plane  XY.     We  denote  them,  therefore,  by  /«and  I'xy.      X 
We  can  therefore  write 


In  the  same  way  we  have 


or,  the  moment  of  inertia  of  any  body  with  reference  to  a  line  is  equal  to  the  sum  of  the 
moments  of  inertia  for  any  two  rectangular  planes  passing  through  that  line. 

Polar  Moment  of  Inertia  for  a  Plane  Area.  —  For  any  plane  area,  as  XO'  Y,  we  have 
lxy  =  o,  and  hence  /.,'  =  /«  and  I,  =  1^  ;   hence 


MEASURABLE  RELATIONS  OF  MASS  AND  SPACE. 


[CHAP.  III. 


or,  the  moment  of  inertia  of  any  plane  area  relative  to  an  axis  perpendicular  to  the  plane  is 
equal  to  the  sum  of  the  moments  of  inertia  for  any  two  rectangular  lines  in  the  plane  through 
the  foot  of  the  perpendicular. 

The  moment  of  inertia  for  a  plane  area  relative  to  an  axes  at  right  angles  to  the  plane 
is  called  the  POLAR  moment  of  inertia  relative  to  that  line. 

Moment  of  Inertia  Relative  to  a  Point.— Let  O'  be  any  point,  O'X,  u        O'Z  three 
co-ordinate  planes. 

Then  for  any  particle  of  mass  m  whose  co-or- 
dinates are  x,  y,  z,  we  have  for  the  moment  of 
inertia  with  reference  to  O' 

mr*  =  mx*  -f-  my*  -\-  mz*. 

Summing  the  moments  of  inertia  for  all  the 
particles,  we  have  for  the  moment  of  inertia  of  the 
body  with  reference  to  O' 


=  /o  is  the  moment  of  inertia  of  the  body 
reference    to    the     point    O',    and   2mx*  =•  I'r, 
=  I'zx  ,   2ms2  =  I'xy  ,  are  the  moments  of  inertia 
of  the  body  with  reference  to  the  co-ordinate  planes 
FZ,  ZX,  XV.     Hence 

/o  =  /;+/;,+/;,. 

But  we  have  just  seen  that  /,',  +  /««  =  I'x>      Hence 


That  is,  the  moment  of  inertia  with  reference  to  any  point  is  equal  to  the  sum  of  the  moments  of 
inertia  for  any  three  rectangular  planes  through  that  point  ; 

or,  is  equal  to  the  sum  of  the  moment  of  inertia  for  any  line  through  the  point  and  a  plane 
through  the  point  at  right  angles  to  this  line. 

Moment  of  Inertia  for  an  Inclined  Axis,  in  General.— Let  M  be  the  mass  of  any 
particle  of  a  body  whose  co-ordinates  are 
x,  j,  z  for  any  assumed  origin  O'  and  co-ordi- 
nate axes  O'X,  O'  Y,  O'Z,  and  let  O'o  be  any 
line  through  the  origin  O1 ',  making  the  angles 
it,  ft,  y  with  the  co-ordinate  axes.  Let  r  be 
the  perpendicular  from  m  upon  O'o. 

Then  we  have  the  distance 

O'o  =  x  cos  at  -\-  y  cos  /?  -f-  z  cos  y ; 
also,  the  square  of  the  distance  O'm, 


Hence          r2  =  ~O7n    —  Wo  ,  or 

r2  =  (x3  -f  y  +  z*)  -  (x  cos  a  -f  y  cos  ft  -f-  z  cos  y)2. 

The  moment  of  inertia  of  a  particle  of  mass  m  relative  to  any  axis  O'o  is,  then, 
mr*  =  mix*  -f  f  -f  j3)  -   tnix  cos  a  -f-  y  cos  ft  -\-  z  cos  yf. 


CHAP.  III.]  MOMENT  OF  INERTIA—  PRINCIPAL  AXES. 

. 
Summing  the  moments  of  inertia  for  all  the  particles  of  the  body  we  have,  since 

cos2  a  -\-  cos2/?  -J-  cos2^  =  I, 
2(mr*)  =  2\in(**  +  f  +  z*}  (cos2  a  +  cos3  /?-fcos2  y)]  -  2\_m(x  cos  a  +  y  cos  £  +  z  cos 

Multiplying  out  and  reducing,  we  can  write  this 
2(mr*)  =  2[m(y*  +  ^2)]  cos2  a  -f  2[w(^  -f  ^2)]  cos2  /?  +  2[m(x*  +  /)]  cos2  y 

-  2  cos  or  cos  ft  2(mxy)  —  2  cos  /3  cos  y  2(myz)  —  2  cos  7  cos  a 
But,  as  we  have  seen,  page  33, 


are  the  moments  of  inertia  of  the  body  relative  to  the  axis  of  X,  Y,  and  Z.     Also,  2(mr2}  is 
the  moment  of  inertia  /'  relative  to  the  axis  O'o.      Hence 

I 
/'  =  I'x  cos2  a  -{-  Iy  cos2  ft  -\-TM  cos2  y  —  2  cos  a  cos  ft  ^(mxy)  —  2  cos  /?  cos 

—  2   cos  7  cos  «  2(mzx) 


Equation  (i)  gives  the  moment  of  inertia  of  a  body  for  any  axis  O'o  making  any  given 
angles  a,  ft,  y  with  the  co-ordinate  axes. 

Principal  Axes.  —  If  for  any  origin  O'  we  take  the  rectangular  axes  X,  V,  Z  in  such 
directions  that  for  any  body 


=  o,  myz  =  O,          *2mzx  =  O, 

these  axes  are  called  PRINCIPAL  AXES  for  the  point  O'  ,  the  moments  of  inertia  I'x,  I'y,  I'z 
relative  to  these  principal  axes  are  called  PRINCIPAL  MOMENTS  OF  INERTIA,  and  the  corre- 
sponding radii  of  gyration  k'x,  k'y,  k',  are  called  PRINCIPAL  RADII  OF  GYRATION. 

If  we  take  the  origin  at  the  centre  of  mass,  the  principal  axes  are  called  CENTRAL  PRIN- 
CIPAL AXES,  the  corresponding  moments  of  inertia  /,,  fy,  It  are  central  principal  moments 
of  inertia,  and  the  corresponding  radii  of  gyration  kx,  ky,  ks  are  central  principal  radii  of 
gyration. 

Properties  of  Principal  Axes.  —  The  three  conditions  for  principal  axes  are  then 


=  o,  myz  =  o,  mzx  =  o. 

Introducing  these  conditions  in  (i)  we  have 

. 

/'  =  I'x  cos2  a  -f  I'y  cos2  ft  4-  /;  cos2  y  ........     (2) 

•. 

That  is,  the  moment  of  inertia  of  a  body  relative  to  any  line  is  equal  to  the  sum  of  the  prod- 
ucts obtained  by  multiplying  the  moments  of  inertia  for  the  principal  axes  for  any  point  of  the 
line,  by  the  square  of  the  cosines  of  the  angles  which  the  line  makes  with  these  principal  axes. 

For  any  plane  material  area,  taking  the  plane  as  that  of  ZX,  we  have  for  any  point  of 
the  area  given  by  x  and  z,  y  =  o.  Hence,  no  matter  where  the  origin  nor  what  direction 
the  axes  of  x  and  y  may  have,  we  have 

=  o,          2myz  =  o. 


36  MEASURABLE  RELATIONS  OF  MASS  AND  SPACE.  [CHAP.   Ill 

Also,  if  a  body  has  a  plane  of  symmetry,  then,  taking  this  as  the  plane  of  ZX,  we  have  for 
any  origin  in  the  plane,  for  any  particle  given  by  x,  z  and  -f-  y  above  the  plane,  an  equal 
particle  given  by  x,  z  and  —  y  below.  Hence,  no  matter  where  the  origin  nor  what  direc- 
tion the  axes  of  x  and  y  may  have,  we  have 

2ntxy  =  o,         2myz  =  o. 
Introducing  these  conditions  in  (i)  we  obtain 

/'  —  Ix'  cos8  a  -f-  Iy  cos2  ft  +  I*  cos2  Y  —  2  cos  Y  cos  «^  mzx. 

This  reduces  to  (2)  when  ft  =  o,  and  hence  a  =  90,  y  ==  90,  and  we  have  in  such  case 
P  =  Iy.  That  is,  the  perpendicular  to  the  plane  is  a  principal  axis  for  the  point  of 
intersection. 

It  is  not,  however,  in  general  a  principal  axis  for  any  other  point  of  the  perpendicular. 
For  if  we  take  some  other  point  on  y  at  a  distance  d  from  the  origin,  then  if  the  axis  of  y  is 
a  principal  axis  for  this  point  also,  we  must  have 

2mx  (y  —  d)  =  o,          2mz  (y  —  d)  =  o. 

These  conditions  can  only  be  satisfied  when  2mx  =  o  and  2my  =  o,  that  is  when  the 
perpendicular  passes  through  the  centre  of  mass. 

Hence,  if  any  two  of  the  three  conditions  for  principal  axes  are  fulfilled,  we  have  either 
a  plane  area  or  a  plane  of  symmetry. 

Any  perpendicular  to  a  plane  area  or  a  plane  of  symmetry  is  a  principal  axis  for  the  point 
of  its  intersection  with  the  plane.  If  this  point  of  intersection  is  the  centre  of  mass,  the  per- 
pendicular is  a  principal  axis  for  any  of  its  points. 

A  line  cannot  be  a  principal  axis  at  more  than  one  of  its  points,  unless  it  passes  through 
the  centre  of  mass,  in  which  case  it  is  a  principal  axis  at  every  one  of  its  points. 

Let  a  body  have  two  planes  of  symmetry  at  right  angles.  Taking  one  plane  as  the 
plane  of  ZX,  we  have,  as  already  proved, 

2mxy  =  o,          ~2myz  =  o, 

and  any  perpendicular  to  this  plane  is  a  principal  axis  for  its  point  of  intersection.  Taking 
the  other  plane  as  the  plane  of  YZ,  we  have 

=  o,          "2,mzx  —  o, 


'and  any  perpendicular  to  this  plane  is  a  principal  axis  for  its  point  of  intersection.      We 
have  then  all  three  conditions 


=  o,          ^myz  =  o,          2mzx  =  o 

fulfilled,  and  hence  for  two  rectangular  planes  of  symmetry  the  principal  axes  for  any 
point  on  the  line  of  intersection  of  these  two  planes  are  this  line  of  intersection  and  the  two 
perpendiculars  to  it  at  the  point  in  each  plane. 

If  a  body  has  three  planes  of  symmetry  at  right  angles,  their  point  of  intersection  is  the 
centre  of  mass  (page  23). 

Hence  for  three  rectangular  planes  of  symmetry  any  line  of  intersection  is  a  principal 
avis  for  any  of  its  points.  The  principal  axes  for  any  point  on  a  line  of  intersection  are  this 
line  and  the  tivo  perpendiculars  to  it  at  the  point  in  each  plane.  The  three  lines  of  intersection 
are  principal  axes  for  the  centre  of  mass,  or  central  principal  axes. 

Ellipsoid  of  Inertia.  —  Since  the  mass  m  of  a  body  multiplied  by  the  square  of  its  radius 
of  gyration  k  for  any  axis  gives  the  moment  of  inertia  /  for  that  axis,  we  have,  by 
dividing  equation  (2)  by  m, 


CHAP.  III.] 


ELLIPSOID  OF  INERTIA, 


37 


In  the  preceding  Fig.,    page  34,  take  a  point  P  on  the  line  O'o   at  any  convenient 
distance  O 'P '=  I  from  O' '.     Denote  the  product  Ik'  by  j2,  so  that 


lk'=s\     hence 


/2  cos2  a      /2  cos2  /3 


Substituting  in  (3)  we  have 


Let  the  co-ordinates  of  the  point  P  be  x* ',  y' ',  z' .     Then 


and  we  have 


S* 


=   I 


(4) 


'2        £'2       £'2 

*  Ky  Kt 


The  equation  of  an  ellipsoid  referred  to  its  major  and  minor  axes  is 

2*+^ +Z»"I§ 

where  ^4,  B  and  (7  are  the  semi-axes. 

We  see,  then,  that  (4)  is  the  equation  of  an  ellipsoid  whose  semi-axes  are  principal  axes 
given  by 


where  s  may  have  any  convenient  value. 

Hence    if    we  lay   off    on    every    line    O'o   through    the  origin   O'  a  distance   O' P  —  I 

—  -r, ,  where  the  distance  s  may  be  taken  any 

convenient  distance,  all  the  points  P  thus  de- 
termined will  lie  on  the  surface  of  an  ellip- 
soid whose  equation  is  (4),  whose  axes  are  the 
principal  axes  for  the  point  O',  and  whose 
semi-axes  are  given  by  the  values  of  A,  B 
and  C  just  found. 

The  square  of  the  reciprocal  of  any  semi 
diameter  /  is,  then, 

l-^l2 
!?~  s*  ' 

This  ellipsoid  is  called  the  ELLIPSOID  OF  INERTIA  for  the  point  O',  because  the  square 
of  the   reciprocal   of   any  semi-diameter    O'P=  I  multiplied  by  the  mass  m  of  the  body, 

or  -^  .  m/£'2,  is  proportional  to  the  moment  of  inertia  /'  =  m/£'2  of  the  body  relative  to  an 

axis  coinciding  with  this  semi-diameter,  and  if  we  take  s=  i,  will  be  equal  to  this  moment 
of  inertia. 


MEASURABLE  RELATIONS  OF  MASS  AND  SPACE. 


[CHAP.  III. 


The  axes  of  this  ellipsoid  arc  principal  axes  for  the  point  O'. 

We  see  at  once  that  the  principal  moments  of  inertia  must  include  the  greatest  and  least 
of  all  the  moments  of  inertia  for  the  point  O',  the  least  corresponding  to  the  longest 
semi-diameter,  O'A,  and  the  greatest  to  the  least  semi-diameter,  either  O'B  or  O'C. 

For  any  point  O',  then,  there  must  be  at  least  one  set  of  rectangular  axes,  OA,  OB,  OC, 
which  arc  principal  axes. 

Determination  of  Moment  of  Inertia. — The  moment  of  inertia  of  a  body  can  be 
determined  experimentally,  as  will  be  explained  hereafter  (page  340).  For  many  bodies  it  is 
readily  determined  by  a  simple  application  of  the  Integral  Calculus. 

It  is  not  necessary  or  desirable  to  occupy  space  here  with  such  applications.  It  is  of 
much  greater  importance  that  the  student  should  understand  clearly  what  the  moment  of 
inertia  of  a  body  is,  as  defined  on  page  31,  and  what  use  is  made  of  it  and  how  it  enters 
into  mechanical  problems,  as  will  be  explained  later  (page  321). 

In  the  following  examples  we  show,  how  the  moment  of  inertia  may  be  found  for  a 
number  of  simple  cases,  with  and  without  the  use  of  the  Calculus. 


Examples.  —  (l)  Find  the  moment  of  inertia  for  a  homogeneous  straight  line. 

ANS.   Let  the  \\neab  coincide  with  the  axis  of  Y,  and  let  O  be  its  centre  of  mass.    Then  by  the  principle 

of  symmetry  OX,  OY,  OZ  are  central    principal  axes  (page  35). 

jY  Let  the  length  of  the  line  be  L,  and  S  the  linear  density.    Then  the 

mass  m  is 

m  =  dL. 


Divide  the  line  above  and  below  O  into  an  indefinitely  large  number 
n  of  short  elementary  lines.  The  mass  of  each  is  <5—  ,  and  the 

2/1 

distances  of  these  elementary  masses  above  and  below  OX  or  OZ 

are   —  ,  2—  -,  3  —  ,  etc.    Multiply   the  mass  of  each  element  by  the 
"2n      2fi      "Zn 

square  of  its  distance,  and  summing  the  products,  we  have  for  the 
moment  of  inertia  /*  or  /,  relative  to  the  central  principal  axes 
OX  and  O  Y 


m/.' 

--(i   +  4  +  9  +  .  .  .«'). 


..£• 


The  sum  of  the  series 


i  +  4  +  9  +  ...«»  = 


2«*   +   3«*  +   « 


As  n  increases,  this  sum  approaches  the  limit  -.     Hence  for  n  indefinitely  large 

/.-/.-*£ 

The  moment  of  inertia  /»  for  the  axis  O  Y  is  ft  =  o. 

These  are  the  moments  of  inertia  for  tin*  central  principal  axes. 

Let  Of  be  any  inclined  axis  in  the  plant*  AT  making  the  angles  a.  ft,  y  with  the  central  principal  axes. 
Then  cos  y  =  o  and  cos  <r=  sin  ft.  We  have  then  from  eqna'ion  (2).  page  35,  the  moment  of  inertia  /  for 
any  axis  through  the  centre  of  mass  making  the  angle  ft  with  the  line 


/  =  /,  sin*  ft  =  --  sin'  ft. 


CHAP.  III.] 


MOMENT  OF  INERTIA— EXAMPLES. 


39 


For  a  parallel  axis  OT  through  any  point  O'  of  the  line,  that  is  for  an  axis  through  any  point  of  the  line, 
making  the  angle  ft  with  the  line,  if  y  is  the  distance  O'O  from  the  point  O'  to  the  centre  of  mass,  we  have, 
from  page  32, 


/'  =  /  4- 
We  can  write  this  last  in  the  form 


sin'/S  =  m  (  -- 


ntf/3. 


(O 


* «"'  ft- 


From  this  we  see  at  once  that  the  moment  of  inertia  of  a  homogeneous  line  for  any  axis  whatever  />  the 
same  as  for  a  system  of  three  material  particles  consisting  of  one  sixth  the  mass  of  the  line  at  each  end  and  two 
thirds  the  mass  of  the  line  at  the  centre. 

BY  CALCULUS. — The  mass  of  any  element  of  the  line  is  Sdy.  The  moment  of  inertia  of  the  element 
relative  to  OI  is  then  Sdy  x  y  sin2  ft.  Hence,  since  m  =  <5Z, 


/•+T  =/. 

/  =  J  8y*dy  .  sin'  ft  =  ^-  sin'  ft. 


(2)  Find  the  moment  of  inertia  for  a  system  consisting  of  three  parallels  equidistant  homogeneous  straight 
lines  of  equal  length  I,  the  ends  being  in  parallel  straight  lines,  the  mass  of  the  centre  line  being  ;;/,  and  of 
each  of  the  outer  lines  m\. 

ANS.   Let  aa' ,  bb' ,  cc'  be  the  lines  of  length  /,  of  mass  mi ,  m\  and  ;«a  respectively,  and  let  d  be  the  length 
of   the   line    ab   through    the   ends,    and    A    be   the 
angle  abb' .  _  ; 

The  centre  of  mass  O  of  the  system  is  evidently 
at  the  middle  of  the  centre  line. 

The  moment  of  inertia  of  aa'  relative  to  any 
axis  through  its  centre  of  mass  in  the  plane  of  the 
lines  making  the  angle  0  with  aa'  is,  from  ex- 
ample (i), 


111  1    7  1         •        9 

-/'srn' 


.    d 


For  a  parallel  axis  through  the  centre  of  mass  O,  since  the  distance  between  the  axes  is  —  sin  (A  —  6), 
we  have  (page  32) 


We  have  the  same  result  for  the  line  bb'  .     For  the  line  cc'  we  have 


The  moment  of  inertia  of  the  system  relative  to  any  axis  in  the  plane  of  the  lines  through  the  centre  o' 
mass  O,  making  the  angle  9  with  the  linos,  is  then 


1       sin'  (A  -0}  +  ^/'sin'0. 


FOI  any  parallel  axis  in  the  plane  at.  a  distance  6  we  have,  then  (page  32), 


(2 


MEASURABLE  RELATIONS   OF  MASS  AND  SPACE. 


[CHAP.  III. 


We  can  write  this  in  the  equivalent  form 


L. 


We  see  from  this  that  the  moment  of  inertia  for  the  system  is  the  same  as  for  a  particle  of  mass  ;«,  at 

the  middle  of  each  outer  line,  a  particle  of  mass 

-Uii  +  r~\  at  each  end  of  the  centre  line,  and  a 

Particle  of  mass  -(;«»  —  mi)  at  the  centre  of  mass 

of  the  system. 

SPECIAL  CASES.— If  mt  =  4#/,,  we  have  m\  at 
the  middle  of  each  outer  line,  m\  at  each  end  of 
the  centre  line,  and  2;«i  at  the  centre  of  mass  of 
the  system. 

If  ///,  =  2;«i,  we  have  m\  at  the  middle  of  each 
outer  line,  |»r,  at  each  end  of  the  centre  line  and  f;«,  at  the  centre  of  mass  of  the  system. 

If  Wi  =*  m\,  we  have  m\  at  the  middle  of  each  outer  line  and  ]fin\  at  each  end  of  the  middle  line. 
(3)  Find  the  moment  of  inertia  for  a  homogeneous  parallelogram  and  rectangle. 

ANS.  We  have  seen  that  the  moment  of  inertia  for  a  homogeneous  straight  line  for  any  axis  is  the  same 
as  for  a  particle  of  one  sixth  its  mass  at  each  end  and  of  four  sixths  its  mass  at  the  middle  point. 

Since  the  parallelogram  ABCD  is  generated  by  a  straight  line  AD  moving  parallel  to  itself,  the  particles 
at  A  and  D  will  generate  lines  AB  and  DC,  the  mass  of  each 

being  one  sixth  the  mass  of  the  parallelogram,  and  the  particle  ^ -C 

at  d  will  generate  the  line  dc,  whose  mass  is  four  sixths  the  /  / 

mass  of  the  parallelogram.  /  / 

We  have  then  a  system  of  three  parallel  lines,  the  mass  d/. . yc 

m\  of  each  outer  line  being  m\  =  ^m  and  the  mass  of  the  centre  /  / 

line  being  m,  =  |m,  where  m  is  the  mass  of  the  parallelogram.          /  / 

We  have   then   m*=.\)n\,  and,  as  just   proved    in   the      A —  — g 

preceding  example,  the  moment  of  inertia  of  the  parallelogram 

is  the  same  as  for  a  particle  of  mass  m\  =  ^m  at  the  middle  of  each  outer  line  and  at  the  ends  of  the  centre 

line,  and  a  par  I  tele  of  |m  at  the  centre  of  mass. 

For  the  axis    Jb  through    the   centre  of   mass 
parallel  to  the  base-  AB  =  b  we  have  then 

Ib  =  ™rf»  sin"  A. 

For  the  axis  /<*  through  the  centre  of  mass  parallel 
to  the  side  AD  =  <i  we  have 

»  „ 

/,/=  — P  sin1  A. 
12 

For  the    axis  /*  through  the  centre  of  mass  per- 
pendicular to  the  base  AB  we  have 


Ik  =  —  (if  +  d*  cos1  A). 
For  the  axis  OZ  through  the  centre  of  mass  0'at  right  angles  to  the  plane  we  have  (page  34) 

/.  =  /*  +  /*  =  ?  (p  +  <n 

For  any  axis  in  the  plane  through  the  centre  of  mass  O  making  the  angle  6  with  the  base  AB  we  have 


/  =  -  P  sin1  6  +  ~  d*  sin3  (A  -  0). 


CHAP.  III.]  MOMENT  OF  INERTIA— EXAMPLES. 

For  any  parallel  axis  in  the  plane  at  a  distance/  (page  33) 

/'  =  ™6*  sin2  0  +  —d*  sin3  (A  —  6)  +  m#a. 


The  axis  OZ  is  a  principal  axis  (page  36),  but  the  axes  for  Ib  and  Ih  are  not  principal  axes. 
PRINCIPAL  AXES. —  If  5  is  the  surface  density,  the  mass  m  of  the  parallelogram  is 

m  =  Sbd  sin  A. 
Hence  we  have 


m8        .  , 

Ib=tf6*-^~«          /d  = 


f'W 


If,  then,  we  inscribe  an  ellipse  in  the  parallelogram  tangent  to 
the  sides  at  their  middle  points,  the  square  of  the  reciprocal  of 
any  semi-diameter  is  proportional  to  the  moment  of  inertia  for 
the  coincident  axis.  Hence  this  ellipse  is  the  central  ellipse  of 
inertia  (page  37),  and  if  /  is  the  length  of  any  semi-diameter  and  / 
the  moment  of  inertia  for  the  coincident  axis,  we  have 


(3) 


P  = 


485V 


Let  OX,  OY  be  the  central  principal  semi-axes  denoted  by  /A,  /y,and  Ix,  Iy  the  corresponding  moments 
of  inertia.     Then 


485V/ 


48  5*7; 


Now  in  an  ellipse  the  area  of  the  circumscribed  parallelogram  is  equal  to  four  times  the  product  of  the 
principal  semi-axes,  or 

A  =  ™. 


Inserting  the  values  of  lx  and  ly  we  have,  since  Ix  —  m&%  and  Iy  = 


where  kx  and  ky  are  the  central  principal  radii  of  gyration. 
We  have  also  (page  34) 

/.  =  /,  +  /.     or 


Solving  these  two  equations  we  obtain  for  the  central  principal  axes 


We  thus  know  the  central  principal  moments  of  inertia  Ix  =  m/^i,  /y  = 
If  the  principal  axis  OX  makes  the  angle  a  with  the  base  AB,  we  have 


/&  =  Jx  cos*  a 


Hence 


MEASURABLE  RELATIONS  OF  MASS  AND  SPACE. 

d*  sin*  A        ,, 
/.  _  /  ~~7^  *> 


[CHAP.  III. 


We  thus  know  the  position  of  the  principal  axis  of  A' and  the  principal  moments  of  inertia  Ix,  Iy 
The  moment  of  inertia  for  any  axis  through  the  centre  of  mass  is  then  given  by 

/  =  /,  cos*  a  +  Iy  cos'  ft  +  I,  cos'  y. 
For  a  rectangle  we  have  A  =  90,  and  hence  for  any  axis  equation  (2)  becomes 

12 


For  the  axis  Oft,  p  =  o,  6  =  o  and 


For  the  axis  OId,  p 


For  the  axis  OZ 


o,  ff  =  90  and 

fd=  —  - 


In  this  case  m  =  8M,  and  we  have  from   our  general  formula  k}  =  — ,   or  Iy  = .      Hence  cos7  «  =  o, 

or  tt=  90°.     Hence  the  axes  of  Ib,  Id  and  /,  are  central  principal  axes,  as  they  should  be,  since  the  planes 
of  I6fd  and  IdOZ  are  planes  of  symmetry  (page  36). 

BY  CALCULUS.— The  mass  of  an  element  parallel  to  AB  at  a-  distance  y  from   OX  is  m  =  &bdy.     Its 
moment  of  inertia  for  an  axis  Of  making  the  angle 
6  with  AB  is,  from  (i), 

vib*  sin1  6  », 


and  transferring  to  a  parallel  axis  through  the  centre 

of  mass  O  at  the  distance  ^  S-  ,  we  have 

sin  A 


ml?  sin"  0 
~~'\2 


Hence 


sin'  (A  -  0) 


•  Js 

-/:: 


-  0) 


or,  since  &bd  sin 


in, 


I  =  77  [^a  sin'  0  +  rf»  sin8  (A  -  6)], 
as  already  found. 

(4)  T-YW  ///<?  moment  of  inertia  for  a  homogeneous  triangle, 

ANS.  We  have  just  seen  in  example  (3)  that  for  a  homogeneous  parallelogram  the  moment  of  inertia 

r  *or  anY  ax!s  ls  tne  same  as  for  a  partic/e  of  one  sixth  the  mass  of 
the  parallelogram  at  the  middle  point  of  each  side,  and  a  particle 
of  two  sixths  the  mass  of  the  parallelogram  at  the  centre  of  mass. 
If,  then,  we  draw  the  diagonal  AC.  we  divide  the  parallel- 
ogram into  two  equal  triangles,  and  if  m  is  the  mass  of  each 
triangle,  we  have  Jffi  at  the  middle  point  of  each  side. 

Hence  for  any  axis  the  moment  of  inertia  of  a  homo- 
geneous trianele  />  ///••  ™w,«  ,™  fnr  a  particle  cf  one  third  the 
mass  of  the  triangle  at  the  middle  point  of  each  side. 


CHAP.  III.] 


MOMENT  OF  INERTIA— EXAMPLES. 


Let  ABC  be  a  triangle  whose  angles  are  A,  B,  C  and  sides 
a,  b,  c.  Take  any  axis  AI'a  through  the  vertex  A  in  the  plane  of 
the  triangle,  making  the  angle  6  with  the  base  b.  Drop  the  perpen- 
diculars pi  and  pt  iron)  B  and  C,  the  perpendicular  p3  from  the 
middle  point  of  the  side  a,  and  the  perpendicular^*  from  the  centre 
of  mass  O. 

Then  we  have 


pi  =  c  sin  (6  —  A), 


=  b  sin  0, 


-I-  p,       b  sin  0  +  c  sin  (0  —  A) 


If  pi  is  less  than  -  sin  0,  we  have 


or        = 


If  /i  is  greater  than  -  sin  0,  we  have 


In  all  cases,  then, 


-sin0  +  1^,  -^  sin  0V      or    /  =  !(/,  +  £sin  0). 


=  -  [  b  sin  0  +  c  sin  (9  —  A)  ]. 


If,  then,  we  take  one  third  of  the  mass  m  at  the  middle  point  of  each  side,  we  have  for  the  moment  of  inertia 
for  the  axis  I'a 


l'a  =  -£  [F  sin"  0  +  c*  sin"  (8  —  A)  +  be  sin  0  sin  (8  -  A)]. 
Reducing  to  a  parallel  axis  through  the  centre  of  mass  O  by  subtracting   m^'  (page  33),  we  have 


sin"0  +  f  sina  (9  —  A)  —  fc  sin  0  sin  (0  —  A)] 


(3) 


For  any  axis  in  the  plane  of  the  triangle,  then,  if  p  is  the  perpendicular  distance  from  the  centre  of  mass  to 
the  axis, 


/'  = 


78 


J«  sin'  0  +  c*  sin8  (0  -  A)  -  be  sin  0  sin  (0  -  A)\  +  mp\ 


For  the  axis  /*  through   the  centre  of  mass  O  parallel  to  the 
base  b  we  have  p  =  o,  0  =  o,  and  equation  (3)  becomes 


A. 


For  the  axis  IH  through  the  centre  of  mass  perpendicular  to 
the  base  b  we  have  p  =  o,  6  =  90°,  and  equation  (3)  becomes 

Ik  -  —  [P  +  c1  cos'  A  —  be  cos  A\ 
18 

For  the  axis  /<-  through  the  centre  of  mass  parallel  to  the  side 
c  we  have^  =  o  9  =  A,  and  equation  (3)  becomes 


44  MEASURABLE  RELATIONS  OF  MASS  AND  SPACE.  [CHAP.  III. 

For  the  axis  OZ  through  the  centre  of  mass  at  right  angles  to  the  plane  of  the  triangle  we  have  (page  34) 


(P  +  c  *  -  be  cos  A), 
18 


or.  since  2  be  cos  A  =  b*  +  c*  —  a\    we  can  write 

*-* 


The  axis  OZ  is  a  principal  axis,  but  the  axes  of  h  and  A 
are  not  principal  axes. 

Principal  Axes.— For  the  axis  Db  through  the  vertex  B  and 
middle  point  of  AC  we  have^  =  o  and  0  —  A  equal  to  the 
angle  ABb,  and  hence 


c  sin  (8  —  A)  —  -  sin  6, 


and  equation  (3)  becomes 


IB  =  —  P  sin1  6. 


We  also  have  Bb  sin  8  =  c  sin  A,  or 

sinsO  = 
Hence,  substituting  this  value  of  sin*  6, 


In  the  same  way  we  have  for  the  axis  Cc 


and  for  the  axis  Aa 


c*  sin'  A      c*  sin*  A 


9 

sin3  A 


0V 


216 


sin1  A         I 

m* 


216 


216  0~a* 

We  see.  then,  that  the  moments  of  inertia  relative  to  Bb,  Cc  and  Aa  are  proportional  to  —  .  -=•„ ,  -=5 . 

Ob     Oc      Oa 

Hence  if  we  inscribe  an  ellipse  in  the  triangle  tangent  at  b,  c  and  a,  it  will  be  the  central  ellipse  of  inertia, 
and  the  reciprocal  of  the  square  of  any  semi-diameter  will  be  proportional  to  the  moment  of  inertia  for  the 
coincident  axis. 

If,  then,  we  draw  the  semi-diameter  Oe  parallel  to  AC,  we  have 

_,  _  mJV  sin*  A  _  £ 
•  Oe  =          2i6/6  12' 

Now   Oe  and    Ob  are   conjugate  semi-diameters,  and    since   the    parallelogram    upon   two   conjugate 
semi-diameters  is  equal  to  the  rectangle  of  the  principal  semi-axes, 


/T     •     /i 
x  Oe  sin  6  = 


/T7i 
Ob 


.  . 
sin8  6 


Inserting  the  value  of  sin1  0  and  of  Oe*  already  found,  we  have 

JV,in", 

*  y  j;2 


CHAP.  III.] 

We  have  also 


MOMENT  OF  INERTIA— EXAMPLES. 


45 


Solving  these  two  equations  we  obtain 


—  i2^V  sin2  A, 
*A\. 


We  thus  know  the  central  principal  moments  of  inertia  I  x  = 


and  / 


If  the  principal  axis  of  X  makes  the  angle  a.  with  the  base  b  =  AC,  we  have 

Ib  =  Jx  cos2  a  +  Iy  sin2  a;     or     ^  c*  sin2  A  =  in/%  cos"  a  +  mfc2y  sin2 


Hence 


cos2  a= 


c*  sin2  A 
18 


We  thus  know  the  position  of  the  principal  axis  of  X and  the  principal  moments  of  inertia Ix,  I  ,  Iz. 
The  moment  of  inertia  for  any  axis  through  the  centre  of  mass  is  then  given  by 

/  =  /.  cos2  a  +  Iy  cos2  ft  +  Iz  cos2  y. 

(5)  Find  the  moment  of  inertia  for  a  homogeneous  circular  disc. 

ANS.  Let  b  be  the  base  AC  of  an  isosceles  triangle  ABC,  and  r,r,  the  equal  sides  and  h  the  altitude 

Then  h?  =  r9 and,  from  example  (4),  we  have  for  the  axis  OZ  through 

the  centre  of  mass  at  right  angles  to  the  plane  of  the  triangle 

For  a  parallel  axis  BZ'  through  the  vertex  B  we  have 


Now  the  circular  disc  may  be  considered  as  composed  of  an  indefi-       A  /  ^ 

nitely  large  number  of  isosceles  triangles  all  having  a  vertex  at  the  centre         ^ 

of  mass  O,  and  an  altitude  h  equal  to  r.    We  have  therefore,  for  a  circular  disc,  for  the  axis  OZ  through 
Y  the  centre  of  mass  O  at  right  angles  to  the  plane  of  the  disc 


Any  three  rectangular  planes  through  the  centre  of  mass  are 
planes  of  symmetry.  Hence  (page  36)  any  three  rectangular  axes 
with  origin  at  centre  of  mass  O  are  central  principal  axes. 

We  have  also  Jz  —  Ix+fy; 

or,  since  /*  and  Iy  are  evidently  equal, 


(5) 


For  any  axis  in  the  plane  of  the  disc  we  have,  then,  if  p  is  the  distance  from  the  centre  of  mass  to 
the  axis, 


=  —  cos2 
4 


We  can  write  this  in  the  form 


/'  =  ~(r  cos  a  +  pY  +  ^(r  cos  a  —  p)*  +  ~(r  sin  a  —  p?  +  -g(r  sin  a  +  pY  +  — 


MEASURABLE  RELATIONS  OF  MASS  AND  SPACE. 


[CHAP.  III. 


Hence  the  moment  of  inertia  for  a  homogeneous  circular  disc  is  the  same  as  for  a  particle  of  one  eighth 
the  mass  of  the  disc  at  the  extremities  of  any  two  rectangular  diameters,  and  a  particle  of  one  half  the  mass  of 
the  disc  at  the  centre. 

BY  CALCULUS. — Take  an  elementary  circular  strip  of  radius  p  and 
thickness  dp.     The  area  of  this  strip  is  ittpdp,  and  its  mass  is 

m  =  -zditpdp. 

Its  moment  of  inertia  for  the  axis  OZ  is,  then, 
mp*  =  2ditp*dp. 
For  the  disc,  then,  we  have 


But  Snr*  is  the  mass  m  of  the  disc.     Hence 


as  already  found. 

(6)  Find  the  moment  of  inertia  for  a  homogeneous  hollow  circular  disc. 

ANS.  Let  r\  be  the  outer  and  rt  the  inner  radius.    Then,  if  8  is  the  surface  density,  we  have  for  the  mass 
of  a  disc  of  radius  /-, 

m,  =  *«rr»% 
and,  from  example  (5), 

i  _   r,«  ,     r,« 

/,  =  m,  —  =  SitrS  .  — . 


For  the  mass  of  a  disc  of  radius  r»  we  have 


ma  =  8itr*, 


and 


Hence  the  polar  moment  of  inertia  for  the  hollow  disc  is 


~ 


But  Sn(ri*  —  ri1)  is  the  mass  m  of  the  hollow  disc.     Hence 


St+  r,'). 


Hence  for  any  diameter 


(r,  i 


BY  CALCULUS.— We  have,  as  in  the  preceding  example, 

7  -    f'1      »      -  — 

as  already  found. 


CHAP.  III.] 


MOMENT  OF  INERTIA— EXAMPLES. 


47 


(7)  Find  the  moment  of  inertia  for  a  homogeneous  elliptical  disc. 

ANS.  Let  the  semi-transverse  axis  Oa  be  a   and  the  semi-conjugate  axis  Ob  be  b. 
described  about  the  ellipse  so  that  its  radius  Oa  is  equal 
to  the  semi-transverse  axis  a. 

Then  we  have  for  the  ratio  of  the  mass  of  any  ele- 
ment ee  of  the  ellipse  to  that  of  the  corresponding  ele- 
ment EEoi  the  circle 

ee    _    bb  b 

~EE  ~~BB=~a' 

Hence  the  moment  of  inertia  of  the  ellipse  relative  to 
the  principal  axis  O  Y  is  —  times  that  of  the  circle.  In 
the  same  way  the  moment  of  inertia  of  the  ellipse 
relative  to  the  principal  axis  OX  is  -7-  times  that  of  the 

circle. 

We  have  then,  from  example  (5),  since  Sna*  is  the 
mass  of  the  circle  and  Snab  =  m  is  the  mass  of  the  ellipse, 


Let  a  circle    be 


and  for  the  axis  OZ 


r-. 


_  a1  + 


C 


Hence  the  moment  of  inertia  for  a  homogeneous  elliptical  disc  is  the  same  as  for  a  particle  of  one  eighth 
the  mass  of  the  ellipse  at  the  extremities  of  the  two  principal  axes,  and  a  particle  of  one  half  the  mass  of  the 
ellipse  at  its  centre  of  mass. 

(8)  Find  the  moment  of  inertia  for  a  homogeneous  right  parallelopipedon. 

ANS.  We   have  seen,  from  example  (3),  that  for  a  parallelogram  we  have  one  sixth  of  the  mass  at  the 

middle  of  each  side  and  two  sixths  the  mass  at  the  centre. 
If  this  parallelogram  moves  parallel  to  itself  it 
describes  the  parallelopipedon,  and  the  centre  and  middle 
points  describe  straight  lines,  aa',  bb' ,  dd' ,  ed ',  each  of  mass 
m\  =  £m  and  the  straight  line  cc1  of  mass  nti  =  ^m,  where 
m  is  the  mass  of  the  parallelopipedon. 

We  have,  then,  two  systems  of  three  lines  each,  viz., 
the  system  aa',  dd' ,  each  of  mass  m\  =  ^m,  and  cc1  of  mass 
mt  =  ^m,  and  the  system  bb' ,  ee' ,  each  of  mass  m\  =  ^m, 
and  cc1  of  mass  m*  =  ^m. 

From  example  (2),  page  39,  we  have,  for  the  first 
system  tn\  =  ^m  at  the  middle  points  of  aa'  and  dd' , 
and  \m\  =  y^S  at  c  and  c1 .  For  the  second  system  we  have 
#/,  =  £m  at  the  middle  points  of  bb'  and  <?/,  and  £»ii  =  TVm 
at  c  and  c1. 

The  moment  of  inertia  for  any  axis  is  then  the  same 
as  for  a  particle  of  one  sixth  the  mass  of  the  parallelopi- 


«j— - 
0 


r 


pedon  at  the  middle  point  of  each  face. 
Thus  if  AB  =  b,  BC  =  d,  and  BE  -. 
AB 


h,  we  have  for  the  axis  /*  through  the  centre  of  mass  O  parallel  to 


fb  =  "(//«  +  d*  sin'  A). 


For  the  axis  Id  through  the  centre  of  mass  O  parallel  to  BC 


Id  = 


sin9  A). 


For  the  axis  Ih  through  the  centre  of  mass  parallel  to  BE 


MEASURABLE  RELATIONS  OF  MASS  AND  SPACE. 


|C"IIA1'.    III. 


The  axis  /*  is  a  principal  axis.  The  axes  of  It  and  /^are  not  principal  axes.  The  principal  axes  in  the 
middle  section  have  the  same  position  as  for  a  parallelogram,  page  41. 

If  the  ends  are  rectangular,  we  have  A  =  90°,  and  /*,  /*  and  Id  are  the  principal  axes.  In  this  case  we 
have 


/,=™2< 

For  a  cube  6  =  d  =  h  and 


Ih  =  -  ( 


where  D  is  the  diagonal  of  a  face. 

(9)  Find  the  moment  of  inertia  for  a  homogeneous  right  cylinder. 

ANS.  We  have  seen  from  examples  (5)  and  (7)  that  for  a  circular  or  elliptical  disc  we  have  one  eighth  of 
the  mass  at  the  extremities  of  two  principal  axes  and  four  eighths  of  the 
mass  at  the  centre. 

If  the  disc  moves  parallel  to  itself,  it  generates  the  cylinder,  and  the 
centre  and  extremities  of  the  principal  axes  describe  straight  lines. 

We  have,  then,  two  systems  of  three  lines  each,  viz.,  the  system  aa',  dd' 
each  of  mass  m\  =  im,  and  c<?  of  mass  m*  =  fm,  and  the  system  6V,  ee1,  each 
of  mass  m\  =  ^m,  and  cc1  of  mass  m*  =  fm. 

From  example  (2),  page  39,  we  have  for  the  first  system  mt  =  |m  at  the 
middle  points  of  aa'  and  dd'  ,  \rn-i  =  TVm  at  c  and  c'  ',  and  \m^  =  TJ5m  at  the 
centre  of  mass  O.  For  the  second  system  we  have  also  |m  at  the  middle 
points  of  66'  and  ee1  ,  TJ5m  at  c  and  (f  and  -j^m  at  the  centre  of  mass  O. 

The  moment  of  inertia  for  any  axis  is,  then,  the  same  as  for  a  particle  of 
one  eighth  the  mass  of  the  cylinder  at  the  extremities  of  two  principal  axes  of 
the  middle  section,  and  a  particle  of  one  sixth  the  mass  at  the  centre  of  mass 
and  at  the  centre  of  mass  of  each  end. 

If  /  is  the  length  and  r  the  radius  for  circular  base,  we  have  for  the 
principal  axes  through  the  centre  of  mass 


For  a  hollow  circular  cylinder,  if  r\  is  the  outer  and  r2  the  inner  radius,  we  have,  from  example  (6), 


If  the  bases  are  ellipses  of  semi-transverse  axis  a  and  semi-coniugate  axis  6,  we  have 


4\  37  4\  3/  4 

(10)  Find  the  moment  of  inertia  of  a  homogeneous  sphere  by  Calcttlus. 
ANS.  Any  three  diameters  at  right  angles  are  central  principal 

axes. 

Let  r  be  the  radius,  and  take  a  circular  element  at  a  distance  y 

from  OX.     The  radius  x  of  this  element  is  given  by 

JT"  =  r"  -  y . 

The  area  of  the  element  is  then  itx*  =  n(r*  —  y).     Its  volume  is 
jr(r»  _ y*)dy,  and  if  5  is  the  density,  its  mass  is 


m  = 
The  moment  of  inertia  of  the  element  relative  to  O  Y  is 


CHAP.  III.]  MOMENT  OF  INERTIA—  EXAMPLES.  ,  49 

Integrating  between  the  limits  y  =  +  r  and  y  =  —  rt  we  have,  since  m  =  $ditr*  is  the  mass  of  the 
sphere,  for  the  moment  of  inertia  relative  to  any  diameter 


If  we  integrate  between  the  limits  y  —  +  r  andj  =  o,  we  have  for  the  moment  of  inertia  of  a  hemisphere 
relative  to  the  axis  O  Y  perpendicular  to  the  base  at  the  centre,  since  the  mass  of  the  hemisphere  is  m  = 

I,  =  f  in* 

The  moment  of  inertia  of  the  slice  relative  to  OX  is 


Integrating  between  the  limits  y  —  +  r  and^  =  o,  we  have  for  the  moment  of  inertia  of  a  hemisphere 
relative  to  any  line  OX  in  its  base  through  the  centre 

7*  =  |mr». 

The  expression  7  =  fmr"  evidently  holds  for  any  spheroid  of  revolution  whose  equatorial  radius  is  r. 

(i  l)  Find  the  moment  of  inertia  for  a  homogeneous  right  cone  or  pyramid,  by  Calculus. 

ANS.  Let  A  be  the  area  of  the  base,  h  the  height  or  altitude.     Take  any  slice  parallel  to  the  base  at  a 

y 

distance  y  from  the  vertex.     Then  the  area  of  this  slice  is  j#At  its 

density, 
SAy'dy 


volume  is    •*  V   ^  ",  and  if  5  is  the  density,  its  mass  is 


Let  >?vbe  the  radius  of  gyration  of  the  base  for  any  line  in  the 
plane  of  the  base.     Then  the  radius  of  gyration  of  the  slice  for  any 

parallel  line  in  its  plane  is  y^.     The  moment  of  inertia  of  the  slice 
ft 

relative  to  a  parallel  line  through  the  vertex  V  is  then 

mfkl  ,  _  SAktydy      SAy'dy 

~tf~  ~~h4        +  ~~W~' 

Integrating  between  the  limits  y  =  h  and  y  =  o,  we  have,  since  m  =  -  •  is  the  mass  of  the  cone  Or 
pyramid,  for  an  axis  through  the  apex  A  at  right  angles  to  the  axis  of  the  cone  or  pyramid 


For  a  parallel  axis  through  the  centre  of  mass  O 

h  =  I'b  -  m(f^)2  =  |m£J  +  ^gtf'  ............    (I) 

Let  kv  be  the  radius  of  gyration  of  the  base  for  the  vertical  axis  of  the  cone  or  pyramid.  Then  the 
radius  of  gyration  of  the  slice  for  this  axis  \&jkv.  The  moment  of  inertia  of  any  slice  relative  to  this  axis  is 
then 


Integrating  between  the  limits  y  =  k  and/  =  o,  we  have  for  the  moment  of  inertia  relative  to  the 
geometric  axis 

.    .    ......     ....     (2) 


Equations  (i)  and  (2)  are  general  and  hold  good  for  any  base.     We  have  only  to  substitute  for  fo  and  kv 
their  values  in  any  case. 


MEASURABLE  RELATIONS  OF  MASS  AND  SPACE. 


[CHAP.  III. 


CASE  i. — For  a  right  pyramid  with  parallelogram  base  we  have  (example  3)   for  a  line  through  the 

centre  of  mass  o  of  the  base  parallel  to  AB  =  b 
if. 

_  </'  sin*  A 

12 

Hence  for  a  parallel  line  through  the  centre  of  mass  O  we  have, 
from  equation  (i), 


In  the  same  way  for  an  axis  through  O  parallel  to  BE 


For  the  axis  VO  we  have 


Hence  from  equation  (2), 


I,  =       (*>  4-  </•). 


CASE  2.— For  a  right  cone  with  circular  base  we  have  (example  5)  for  any  line  in  the  plane  of  the  base 
through  the  centre 


Hence  for  a  parallel  line  through  the  centre  of  mass  O 
It  =  A  »'•'  + 


«-£ 


For  the  axis  VO  we  have 


Hence 


CASE  3.  —  For  a  right  cone  with  elliptic  base,  if  the  semi-axes  are  a  and  b, 
we  have  (example  7) 


Hence 


KINEMATICS  OF  A  POINT.    GENERAL  PRINCIPLES. 


CHAPTER   I. 

LINEAR  AND  ANGULAR  DISPLACEMENT. 

Kinematics. — As  we  have  seen  (page  i),  we  deal  in  mechanics  with  measurable 
relations  of  matter,  space  and  time. 

In  the  preceding  pages  we  have  given  those  measurable  relations  of  matter  and  space 
only;  of  which  we  shall  make  future  use. 

Let  us  now  consider  those  measurable  relations  of  space  and  time  only,  of  which  we 
shall  also  have  to  make  future  use. 

That  branch  of  science  which  treats  of  .the  measurable  relations  of  space  and  time  only, 
that  is,  of  pure  motion,  is  called  KINEMATICS.  It  adds  to  the  ideas  of  pure  geometry  the 
idea  of  motion. 

We  shall  first  consider  the  motion  of  a  point  and  then  the  motion  of  a  rigid  body. 
Path  of  a  Point. — Two   points  are  said   to    be    consecutive   when    they    are    so    close 
together  that  no  third  point  can  be  taken  between   them.      The  line  joining  the  successive 
consecutive  positions  of  a  point  during  its  motion  is  called  its  PATH. 

Let  «,,  a^  a,,  etc.,  represent  the  successive  consecutive  positions  of  a  point.  Then 
the  line  a^a^a^  etc.,  is  the  path. 

We    always    denote   the  length   of  the   path   or    the   distance 
described  by  the  point  by  the  letter  s. 

The  distance  between  any  two  consecutive  points,  as  al  a^ ,  atat,  etc.,  is  then  indefi- 
nitely small.  This  indefinitely  small  distance  we  denote  by  ds. 

This  portion  of  the  path  ds  between  two  consecutive  points  we  consider  as  a  straight 
line  and  call  an  element  of  the  path.  The  direction  of  any  element  is  that  of  a  tangent  to 
the  path. 

Any  path,  then,  we  consider  as  made  up  of  straight-line  elements,  each  of  length  ds, 
and  each  tangent  to  the  path,  while  the  entire  length  of  the  path  is  the  sum  of  all  the  ele- 
ments and  is  denoted  by  s  =  ^£ds. 

If  the  time  of  describing  the  path  is  denoted  by  /,  then  the  indefinitely  small  time  of 
describing  an  element  of  the  path  is  denoted  by  dt. 

Angle  described  by  a  Point. — Let  a  point  move  in  any  path 
whatever  from  the  initial  position  at  to  the  final  position  a. 

Then  the  distance  s  from  al  to  a,  measured  along  the  path, 
is  the  DISTANCE  DESCRIBED  by  the  point  relative  to  <*,.  Take 
any  point  O  as  pole,  and  draw  Oar,  Oat  .  .  .  Oa  from  O  to 
every  consecutive  point  of  the  path.  The  sum  of  all  tire  indefi- 
nitely small  elementary  angles  a  tOat-\- atOat  +  etc.  gives  the 
angle  swept  over  by  the  radius  vector  Oa  in  passing  from  the 
position  Oa,  to  the  final  position  Oa.  We  call  this  the  ANGLE 
DESCRIBED  by  the  point  relative  to  O. 

51 


KINEMATICS  OF  A  POINT.    GENERAL  PRINCIPLES. 


[CHAP.  I. 


Denote  its  magnitude   by  0.     Then  the  indefinitely  small  elementary  angles    atOat, 
atOat,  etc.,  are  denoted  by  d<f>. 

Example.— Let  a  point  move  in  a  circle  of  radius  r  ft.  through  a  half  circumference,  from  at  to  a.  Take 
the  pole  O  in  the  perpendicular  OC  through  the  centre  C  of  the  circle,  so  that  Oa\  =  Oa  =  I  ft.  The 
successive  positions  of  the  radius  vector  are  then  elements  of  a  cone.  If  we  develop  this  cone,  we  have  a 


circular  sector  atOa,  whose  radius  is  /  ft.,  length  of  arc  s  —  itr  ft.,  and  whose  developed  angle  a}Oa=  <f>  is 
the  angle  described.     Hence 


=  itr,    or 


Tfr       ,. 
—  radians. 


Linear  and  Angular  Displacement. — Let  al  and  a,  as  before,  be  the  initial  and  final 
positions  of  a  point  moving  in  any  path. 

Then,  as  already  defined,  the  distance  s  from  av  to  a, 
measured  along  the  path,  is  the  distance  described  relative  to 
flj,  and  the  angle  0  swept  over  by  the  radius  vector  is  the 
angle  described  relative  to  O. 

Now  draw  the  straight  line  a^a  from  the  initial  position 
tfj  to  the  final  position  a.  This  line  a^a  is -the  LINEAR  DIS- 
PLACEMENT of  the  moving  point.  We  denote  its  length  by 
D.  If  it  is  indefinitely  small,  the  two  points  al  and  a  be- 
come consecutive,  and  the  displacement  D  =  ds  is  then  an 
element  of  the  path. 

Denote  the  angle  a^Oa  subtended  at  O  by  the  linear  displacement  D  by  6.  This  angle 
is  the  ANGULAR  DISPLACEMENT  of  the  moving  point  relative  to  O.  If  it  is  indefinitely 
small,  we  have  dQ  =  d<t>. 

Linear  displacement  is  then  measured  in  units  of  length,  as  feet,  and  angular 
displacement  in  radians. 

Example.— In  the  preceding  example  we  have  seen  that  the  distance  described  is  5  =  itr  feet,  and  the 
angle  described  is  0  =  —  radians. 

The  linear  displacement  is,  however,  D  =  2r  ft.  in  a  direction  from  «,  to  a\  and  the  angular  displacement 
0  is  given  by 

0        r 


_  .     e 

/sin  — 

2 


•  r,     or    sin  —  =  -7" 


Line  Representative  of  Linear  and  Angular  Displacement. — We  see,  then,  that  the 
linear  displacement  a^a  —  D  is  a  straight  line.  It  has  therefore  both  magnitude  and  direc- 
tion. 

Thus  a  straight  line  a^a  represents  by  its  length  the  magnitude 
D  ft.  of  any  linear  displacement,  and  the  arrow  indicates  the  direc- 
tion. The  linear  displacement  is  thus  completely  represented  in 
both  magnitude  and  direction  by  a  straight  line  and  arrow. 


CHAP.  I.] 


VECTOR  QUANTITIES. 


53 


In  the  same  way  we  can  represent  angular  displacement.  Thus  if  al  and  a  are  the  initial 
and  final  positions  of  a  point  and  O  the  pole,  so  that  a^Oa  —  6 
is  the  angular  displacement,  then  a  straight  line  AO  through  the 
pole  at  right  angles  to  the  plane  of  a^Oa  represents  by  its 
length  the  magnitude  0  of  the  angular  displacement,  and  the 
arrow  indicates  that  if  we  look  along  this  line  in  the  direction  of 
the  arrow,  the  rotation  of  the  radius  vector  is  always  seen  clock- 
wise. In  the  figure,  for  instance,  from  Oal  to  Oa. 

The  line  representative  AO  coincides,   therefore,  with  the 
axis  about  which  the  radius  vector  turns. 

By  "  direction  "  of  angular  displacement  we  always  mean  the  direction  of  its  line  repre- 
sentative as  indicated  by  its  arrow. 

Thus  if  the  radius  vectors  Oal  and  Oa  in  the  figure  lie  in  a  vertical  east  and  west  plane, 
the  direction  of  the  angular  displacement  is  north,  because  the  axis  is  a  north  and  south  line, 
and  if  we  look  north  along  the  axis  we  see  rotation  of  the  radius  vector  clockwise. 

Example. — Let  a  point  move  in  a  circle  of  radius  r  ft.  in  an  east  and  west  plane.  The  point  starts  from 
the  top  and  moves  east-ward  through  a  quadrant.  Find  the  distance  described,  the  angle  described,  the  linear 
displacement,  the  angular  displacement,  for  pole  at  centre. 

ANS. — The  distance  described  is  s  =  —  ft.     The  angle  described  is 

<f>  =  —  radians.  The  linear  displacement  is  a^a  =  r  Vz  ft.,  making  an 
angle  of  45°  with  the  initial  radius  Oa\. 

The  angular  displacement  is  6  =—  radians  north,  because  the  line 

representative  AO  must  have  its  arrow  pointing  north  in  order  that 
rotation  of  radius  vector  may  be  seen  clockwise  when  we  look  in  the 
direction  of  the  arrow. 

Note  that  "angle  described"  0=  —  radians  has  magnitude  only, while  "angular  displacement"  has  in 
this  case  the  same  magnitude  and  also  direction,  and  is  9  =  ~  radians  north. 

Vector  Quantities. — All  quantities  which  have  magnitude  and  direction  so  that  they  can 
be  represented  by  straight  lines  and  arrows  are  called  VECTOR  quantities. 

Linear  and  angular  displacement  are  thus  vector  quantities.  In  order  that  they  may  be 
known,  therefore,  magnitude  and  direction  must  be  given.  Magnitude  alone  is  not  sufficient. 

Thus,  in  the  preceding  example,  it  is  not  sufficient  to  say  that  the  linear  displacement  is 
r  1/2  ft.  This  is  the  magnitude  only.  The  direction  must  also  be  stated. 

So,  also,  it  is  not  sufficient  to  say  that  the  angular  displacement  is  6  =  -  radians.  In 
this  case  this  is  the  angle  described,  but  it  is  only  the  magnitude  of  the  angular  displacement. 
The  direction  must  also  be  stated.  Thus  the  angular  displacement  is  0  =  —  radians  north. 

Displacement  in  General. — The  term  "displacement"  always  signifies  linear  displacement 
unless  otherwise  specified. 


CHAPTER  II. 


RESOLUTION  AND  COMPOSITION  OF  LINEAR  DISPLACEMENTS. 

Resolution  and  Composition  of  Linear  Displacements.— Suppose  a  point  to  move  by 
any  path  from  A   to  B,  so  that  AB  is  the  line  representative   of  the  linear  displacement. 
Then  let  it  move  by  any  path  from  B  to  C,  so  that  BC  is  the  line 
representative  of  the  second  displacement. 

It  is  evident  that  AC  is  the  line  representative  of  the 
RESULTANT  displacement,  when  the  two  displacements  AB  and 
BC  are  thus  successive. 

These  two  displacements  AB  and  BC  are  called  COMPONENT 

DISPLACEMENTS. 

Suppose,  however,  that  the  two  component  displacements, 
instead  of  being  successive,  are  simultaneous.  That  is,  while  the  point  goes  from  A  to  B, 
let  the  line  AB  move  parallel  to  itself  to  DC,  so  that  the  end  B  arrives  at  C  at  the  same 
instant  that  the  point  arrives  at  B.  Then  it  is  evident  that  AC  is  still  the  resultant 
displacement. 

We  see,  then,  that  if  a  point  has  two  component  displacements  AB,  BC,  either  simul- 
taneous or  successive,  the  resultant  displacement  AC  due  to  combining  them  is  easily 
found.  This  combination  is  called  COMPOSITION  of  displacements. 

Also,  if  we  have  any  given  displacement  AC,  we  can  find  the  equivalent  component 
displacements,  simultaneous  or  successive,  in  any  two  directions  we  please.  This  is  called 
RESOLUTION  of  displacement. 

We  can  RESOLVE  a  displacement  into  equivalent  components,  or  we  can  COMBINE  the 
components  into  an  equivalent  resultant. 

Triangle  and  Polygon  of  Linear  Displacements — We  can  express  this  composition  or 
resolution  as  follows: 

If  two  sides  of  a  triangle,  AB,  BC,  taken  the  same  way 
round,  as  shown  by  the  arrows,  give  the  linear  displacements, 
simultaneous  or  successive,  of  a  point,  then  the  third  side,  AC, 
taken  the  opposite  way  round,  as  shown  by  the  arrow,  gives  the 
resultant  displacement. 

Inversely,  any  displacement  AC  can  be  resolved  into  two  components  AB  and  BC, 
simultaneous  or  successive,  in  any  two  desired  directions,  by  completing  the  triangle  ABC, 
and  taking  AB  and  BC  the  opposite  way  round. 

This  is  called  the  principle  of  the  "  triangle  of  displacements." 

Again,  if  we  had  a  third  displacement  ,  CD,  the  resultant  of  AC 
already  found,  and  CD  is  AD.  But  AC  is  the  resultant  of  AB  and 
BC.  Hence  AD  is  the  resultant  of  AB,  BC  and  CD,  either  simul- 
taneous or  successive. 

Hence  if  any  number  of  displacements,  simultaneous  or  succes 
sive,  are  given  by  the  sides  of  a  polygon  A  B  CD,  etc.,  taken  tht 
same  way  round,  the  line  AD  which  closes  the  polygon,  taken  the 
other  way  round,  gives  the  resultant  displacement. 

54 


CHAP.  II.] 


LINEAR  DISPLACEMENT.— EXAMPLES. 


55 


This  is  called  "the  principle  of  the  "  polygon  of  displacements"  The  triangle  of  dis- 
placements is  evidently  only  a  special  case. 

Rectangular  Components. — When  a  displacement  is  resolved  into  two  components 
at  right  angles,  the  components  are  RECTANGULAR  COMPONENTS.  Unless  otherwise 
specified,  when  we  speak  of  the  components  of  any  displacement,  rectangular  components 
are  to  be  understood. 

Component  of  the  Resultant  equal  to  the  Algebraic  Sum 
of  the  Components  of  the  Displacements. — It  is  evident  that 
the  resultant  of  any  two  given  displacements  is  equal  to  the 
algebraic  sum  of  their  components  along  the  resultant. 

For  if  AB  and  BC  are  the  given  displacements,  the  re- 
sultant AC  is  equal  to  the  sum  of  the  components  Ad and  dC. 

So  also  for  any  number  of  displacements,  the  resultant  AE 
is  equal  to  the  algebraic  sum  of  the  components  of  the  displace- 
ments along  AE. 

The  component  in  any  direction  of  the  resultant  itself  is 
then  equal  to  the  algebraic  sum  of  the  components  of  the  dis- 
placements in  the  same  direction. 

Thus  the  projection  ae  of  the  resultant  AE  upon  any  line 
OP,  that  is,  the  component  of  AE  along  OP,  is  the  same  as  the 
algebraic  sum  of  the  components  of  AB,  BC,  CD,  DE  along 
OP. 

That  is,  the  component  of  the  resultant  in  any  direction  is 
equal  to  the  algebraic  sum  of  the  components  of  tlie  displacements 
in  that  direction. 

Examples.— (i)  A  point  has  three  displacements,  N.  60"  E.,  40  ft.; 
S.  soft.;  W.  jo"  N.,  60 ft.  Find  the  resultant  displacement. 

ANS  10  4/3  ft.  W. 

(2)  Three  component  displacements  have  magnitudes  represented  by  I,  2  and  j,  and  directions  given  by 
th^  sides  of  an  equilateral  triangle.     Find, the  magnitude  of  the  resultant.  ANS.    4/3. 

(3)  Show  that   the  resultant  of   two  equal  displacements  of  magnitude  a,  inclined  60°,  is  equal  to  the 
resultant  of  a  and  2a  inclined  120° . 

(4)  To  a  man  in  a  balloon  the  starting-point  bears  N.  20°  E.,  and  is  depressed jo°  below  the  horizontal.     A 
point  at  the  same  level  as  the  starting-point  and  10  miles  from  it  is  vertically  below  him.     Find  the  compo- 
nent displacements  of  the  balloon  relative  to  the  starting-point. 

ANS    9.39  miles  south  ;  3.42  miles  west  ;  5.77  miles  high. 

Relative  Displacement. — Since  the  displacement  of  a  point  is  change  of  position,  it 
can  only  be  determined  by  reference  to  some  chosen 
point  of  reference  (page  1 1). 

Thus  if  av  and  a  are  the  initial  and  final  positions 
of  a  moving  point  A,  and  B  is  some  chosen  point  of 
reference,  the  positions  a1  and  a  relative  to  B  are 
known,  if  we  know  the  radius  vectors  Ba^  and  Ba  and 
the  angle  a^Ba  =  0. 

To  an  observer  at  B  the  point  A  is  seen  to  move 
from  al  to  a  through  the  angle  #,  and  a^a  is  then  the 
displacement  of  A  relative  to  B. 

A  line  ab,  then,  parallel  and  equal  to  a^a  ivitJi  arrow  at  b  gives  this  displacement  a^a 
in  magnitude  and  direction,  and  it  also  indicate:  by  the  end  letters  that  the  line  ab  with 
arrow  at  b  is  the  displacement  of  A  relative  to  B. 


KINEMATICS  OF  A  POINT.     GENERAL   PRINCIPLES. 


[CHAP.  II. 


Now  to  an  observer  on  the  moving  point  A  we  see,  by  completing  "the  parallelogram, 
that  the  point  B  would  appear  to  the  observer  on  A  to  move  from  B  to  BH ,  through  the 
same  angle  0,  and  BBn  parallel,  equal  and  opposite  to  ala  gives  the  displacement  of  B  rela- 
tive to  A. 

Aline  ba,  then,  parallel  and  equal  to  BBHt  with  arrow  at  a,  gives  the  displacement  />/>„ 
in  magnitude  and  direction,  and  //  also  indicates  by  the  end  letters  that  the  line  ba  with 
arrow  at  a  is  the  displacement  of  B  relative  to  A. 

Hence  any  change  in  the  relative  position  of  any  two  points  A  and  B  may  be  regarded 
as  a  displacement  of  A  relative  to  B,  as  indicated  by  the  line  ab  with  arrow  at  b,  or  as  an 
equal  and  opposite  displacement  of  B  relative  to  A,  as  indicated  by  the  line  ba  with  arrow 
at  a. 

Notation. — The  student  should  notice  carefully  the  preceding  notation. 
A  relative  displacement  is  given  in  magnitude  and  direction  by  a  'straight 
line  and  arrow.  The  letter  at  the  arrow  end  denotes  the  point  of  reference : 
that  at  the  other  end  the  moving  point. 

Thus  the  line  ab  with  arrow  at  b  gives  the  magnitude  and  direction  of 
the  displacement  of  A  relative  to  B.  The  equal  parallel  line  with  arrow  at 
a  gives  the  displacement  of  B  relative  to  A. 

Triangle  and  Polygon  of  Relative  Displacements. — Except  for  notation  we  have 
changed  nothing,  and  the  principles  already  established  still  hold. 

Thus  let  a  moving  point  A  have  a  displacement  ab  relative  to  some  point  B,  so  that  it 
moves  from  a  to  b,  and  at  the  same  time,  or  afterwards,  let  the  point  B  have  the  displace- 
ment BBH  relative  to  some  point   C.      Then  the  line 
be  through  b,   parallel  and   equal  to  BBn ,   gives  the 
displacement  of  B  relative  to  C,  and  it  is  evident  that 
A  moves  from  a  to  c,  and  hence  ac  is   the   resultant 
displacement  of  A  relative  to  C. 

Hence  if  two  sides  ab,  be  of  a  triangle  abc,  taken 
the  same  way  round,  give  the  displacement  of  A  rela- 
tive to  B,  and  B  relative  to  C,  either  simultaneous  or 
successive,  then  the  third  side,  ac,  will  give  the  resultant 

displacement  of  A  relative  to  C  if  taken  the  opposite  way  round,  from  a  to  c. 
same  way  round,  from  c  to  a,  it  gives  the  displacement  of  C  relative  to  A. 
This  is  the  principle  of  the  triangle  of  relative  displacements. 

Again,  let  ab,  be,  cd,  etc.,  be  the  line  representa- 
tives of  the  displacements  of  A  relative  to  B,  B  rela- 
tive to  C,  and  C  relative  to  D,  etc.  Then  if  we  lay 
off  these  displacements  we  obtain  the  polygon  abed. 
As  we  have  just  seen,  ac  gives  the  displacement  of  A 
relative  to  C.  The  line  dv/ gives  then  the  displacement 
of  A  relative  to  D.  Also  any  line  in  the  polygon,  as 
bdt  gives  the  displacement  of  B  relative  to  D. 

Hence  if  any  number  of  relative  displacements, 
simultaneous  or  successive,  are  given  by  the  sides  of  a 
polygon  abed,  etc.,  the  line  which  closes  the  polygon, 
taken  the  opposite  way  round,  gives  the  displacement 
of  the  first  point  relative  to  the  last ;  taken  the  same 
way  round,  the  displacement  of  the  last  relative  to  the  first. 


If  taken  the 


CHAP.  II.] 


RELATIVE  DISPLACEMENT— EXAMPLES. 


57 


The  triangle  of  relative 


This  is  the  principle  of  the  polygon  of  relative  displacements. 
displacements  is  evidently  a  special  case. 

Examples.  —  (i)  A  circle  of  radius  r  rolls  on  a  horizontal 
plane  until  it  turns  through  a  quarter  revolution.  Find  tJie 
displacement  of  the  point  of  the  circle  initially  in  contact  with 
the  plane  relative  to  the  point  diametrically  opposite. 

ANS.  Let  O  be  the  initial  point  of  contact  on  the  plane, 
and  A  the  corresponding  point  of  the  circle,  initially  at  O, 
and  B  the  point  of  the  circle  diametrically  opposite. 

When  the  circle  rolls  through  a  quadrant,  A  has  moved 
to  An,  and  B  to  Bn.  The  displacement  of  A  relative  to  the 
fixed  point  O  of  the  plane  is  given  by  the  line  from  A  to  An. 

A  parallel  and  equal  line  aO  with  arrow  at  O  gives, 
then  the  displacement  of  A  relative  to  O. 

The  displacement  of  B  relative  to  O  is  the  line 
BBn.  A  parallel  and  equal  line  bO  with  arrow  at  O 
gives  then  the  displacement  of  B  relative  to  O.  The 
same  line  Ofrvfith  arrow  at  b  gives  the  displacement  of 
O  relative  to  B. 

If,  then,  we  lay  off  these  displacements  in  order, 
aO,  Ob,  the  closing  line  ab  gives  the  displacement  of 
A  relative  to  B.  This  displacement  we  then  easily 
find  to  be  2r  ^2,  making  an  angle  of  45"  with  the 
vertical.  The  same  line  ba  taken  in  the  opposite 
direction  gives  the  displacement  of  B  relative  to  A. 

(2)    Two  railway  trains  A  and  B  run,  one  north- 
east a  distance  d,  the  other  southeast  the  same  distance. 
Find  the  displacement  of  A  relative  to  B. 
ANS.  The  displacement  of  A  relative  to  some  fixed  point  E  on  the  earth  is  represented  by  the  line  ae 
with  arrow  at  e.     The  displacement  of  B  relative  to  the  same 
point  E  is  represented  by  the  line  be  with   arrow  at  e.      The 
same  line  with  arrow  at  b  gives,  then,  the  displacement  of  E 
relative  to  B. 

Laying  off  these  displacements  ae,  eb  in  order,  we  find  ab' 
the  displacement  of  A  relative  to  B,  to  be  d  ^2  in  a  direction 
north. 

(3)  A  point  A  moves  jo  ft.  in  a  given  direction  relative  to  a 
fixed  point  P.     Another  point,  B,  moves  relative  to  P  40  ft.  in  a 
direction  at  right  angles  to  the  direction  of  A.    Find  the  displace- 
ment of  A  relative  to  B. 

ANS.  50  ft.  in  a  direction  inclined  to  the  direction  of  A  by 
an  angle  whose  tangent  is  |-. 

(4)  The  displacement  of  a  point  A  relative  to  a  point  B  is  a 
distance  6  ft.  south,  and  relative  to  a  point  C  5  ft.  west.     If  C  is 

initially  a  distance  4ft.  south  of  B,  find  the  final  position  of  C  relative  to  B.  , 

ANS.   Distance  of  Cs  final  position  from  B  is  5  4/5  ft.,  and  the  direction  from  B  to  the  final  position  of 
C  is  east  of  south  by  an  angle  whose  tangent  is  £. 


CHAPTER  III. 

RESOLUTION  AND  COMPOSITION   OF  ANGULAR  DISPLACEMENTS. 

Resolution    and   Composition   of   Finite   Successive   Angular   Displacements   about 
Different  Axes. — Let  O  be  a  fixed  point  or  pole,  and  let  OA^  and  OB2  be  two  given  axes, 
intersecting  at  O  and  making  the  angle  A1O£2  =  ex. 

Suppose  successive  finite  angular  displacements  about  these 
axes  in  the  following  order: 

ist.    An    angular  displacement   9l   about    OAr       2d.    An 
angular  displacement  02  about  OBV 

It  is  required  to  find  the  resultant  angular  displacement. 
CONSTRUCTION. — Draw  through  O  an  axis  OBl  making  the 
angle  AlOBl  —  a  with  OAl ,  and  the  angle  Bl  -OBZ  —  ^  with  OBV 
Let   the  axis   OA t   and    O£l   be  rigidly   fixed   relative   to  each 
other,    so   that  when   the   angular  displacement    6l   takes  place 
about  OA^ ,  the  axis  OBl  will  turn  into  the  position  OBr 
Take  the  distances  OA{  =  OBl  =  OBr      Then  if  we  have  the  angular  displacement   0, 
about  OAl  ,  and  next  0.2  about  OB2 ,  the  points  Bl  and  A^  move  on  the  surface  of  a  sphere  of 
radius  OAr 

Join  AlBl  ,  B1B2  and  B2Al  by  great  circles  of  this  sphere.       Then  the  arc  BlAl  =  a,  arc 
B2Al  =  a,  arc  B^BI  =  6l ,  and  in  the  spherical  triangle  B1A1B2  the  spherical  angle  at  Ax  is  6r 
Bisect  this  angle  by  a  great  circle  A^D  meeting  B^B2  at  D.         Draw  a  great  circle  B2R 
through  B2  inclined  to  BZA^  by  the  spherical  angle  £#2  and  meeting  AXD  at  R. 
Then  the  resultant  angular  displacement  is  about  the  axis  OR. 

PROOF. — Draw  the  arc  B2R'  making  the  same  angle,  £#2,  with  B2Al  on  the  other  side,  and 

make  the  arc  B2R'  =  B2R.     Then  A^R'  will  equal  Aft,  and  the  angle  BZA^R'  =  $0,  =  B^A.R. 

Now  when  the  angular  displacement  0,  takes  place  about  OA^ ,  the  axis  OBl  moves  to 

OBi  through  the  angle  B^A^B^  =  Bv  ,  and  the  axis  OR,  if  rigidly  fixed  relative  to  OA1  ,   will 

move  to  OR'  through  the  angle  RA^R'  =  0,. 

Next,  when  the  angular  displacement  02  takes  place  about  OB2  through  the  angle 
R'B^R  —  03 ,  the  point  R'  evidently  moves  back  to  R. 

Hence  the  axis  OR  has  the  same  position  before  and  after  the  angular  displacements. 
The  resultant  angular  displacement  is  then  about  the  axis  OR.  t 

MAGNITUDE  AND  POSITION. — For  the  position  of  this  resultant  axis  and  the  magnitude  0 
of  the  resultant  angular  displacement  about  it  we  have  in  the  spherical  triangle  AXRB2  the 
side  A^B2  =  a,  the  angle  AVB^R  =  $02  and  the  angle  RA&  =  %6r  Also  the  angle 
B2RA,  =  1 80°  —  £0. 

Hence  we  have  for  the  magnitude  of  the  resultant  angular  displacement  6  the  equation 

cos  £0  =  cos  £0t  cos  i02  —  sin  £0,  sin  $0,  cos  a (I) 

58 


CHAP.  III.]  ANGULAR  DISPLACEMENTS  IN  GENERAL.  59 

Also  for  the  direction  of  OR 

sin  ROAl  _  sin  ROBZ  _  sin  a  ,,,,. 

sin  %62    '       sin  \B^    ~  sin  %V 

From  (I)  we  can  find  6,  and  from  (II)  we  can  find  the  angles  ROA^  and  ROB2  which  the  axis 
OR  makes  with  the  axes  OAl  and  OB^. 

Examples.  —  (i)    The  telescope   of  a   transit   instrument  initially   horizontal  and  pointing  north   is  first 
turned  to  an  altitude  of  60°  and  then  turned  to  the  west.     Find  the  resultant  angular  displacement. 

Axs.   We  have  6.  =  60°,  0a  =  90°,  a=  90°.      Hence,  from  (I),  cos  |0  =  £-1.  ^,    or  6  =  104°  28'  39",  or 

6  =  1.823  radians. 

Hence  sin  |Q  =  |4/|,  and  we  have,  from  (II),  for  the  position  of  the  resultant  axis 

sin  ROAi  =  ^^  sin  45°  =  2  y^,    or     ROA,  =  63°  26'  5.8"; 

sin  ROB*  =  S^^  sin  3or"  =  Vl     or     KOB*  -  39°  13'  53-4". 

(2)  In  the  preceding  example  let  the  successive  displacements  be  taken  in  reT.>erse  order. 

ANS.  We  have  then  9i  =  90°,  63  =  60",  a.  —  90°.     Hence  cos  ^9  =  \tf\  and  sin  £0  =  \^\,  just  as  before. 
But  for  the  position  of  the  resultant  axis 


i  =  4/J    and     sin  ROB*  =  2|/f 

Resolution  and  Composition  of  Angular  Displacements  in  General.  —  We  have  just  seen 
how  to  find  the  resultant  for  finite  successive  angular  displacements  about  intersecting  axes. 

For  all  other  cases  of  angular  displacement  about  intersecting  axes  we  can  combine  and 
resolve  angular  displacements  just  like  linear  displacements,  as  explained  in  tlie  preceding 
chapter, 

This  can  be  shown  as  follows  : 

All  other  cases  fall  under  the  head  of 

(a)  Finite  or  indefinitely  small  successive  or  simultaneous  angular  displacements  about 
the  same  axis; 

(b]  Indefinitely  small  successive  or  simultaneous  angular  displacements  about  different 
intersecting  axes. 

(c}   Finite  simultaneous  angular  displacements  about  different  intersecting  axes. 
Let  us  consider  these  cases  in  order. 

(a)  FINITE   OR   INDEFINITELY  SMALL  SUCCESSIVE  OR  SIMULTANEOUS  ANGULAR  DIS- 
PLACEMENTS ABOUT  THE  SAME  AXIS.  —  If  the  axis  of  all  the  angular  displacements  is  the 
same,  the  plane  of  rotation  of  the  radius  vector  does  not  change,  and  all  the,  line  repre- 
sentatives lie  in  the  same  straight  line.      The  resultant  is  then  given  by  the  algebraic  sum 
of  the  line  representatives,  and  this  evidently  holds  whether  the  angular  displacements  are 
successive  or  simultaneous,  finite  or  indefinitely  small. 

(b)  INDEFINITELY  SMALL  SUCCESSIVE  OR  SIMULTANEOUS  ANGULAR   DISPLACEMENTS 
ABOUT  DIFFERENT  INTERSECTING  AXES.  —  Let    the   angular   displacements    be    indefinitely 
small  and  successive  in  the  order  ftl,  #2.      Then  we  have  sin  £0X  =  ^6V  sin  £#2  =  £#2,  sin  £fl 
=  %0.      Hence,  from  equations  (II), 

0  :  6l  :  Oz  :  :  sin  a  :  sin  ROB2  :  sin  ROAl  .......      (i) 


6o 


KINEMATICS  OF  A  POINT.      GENERAL  PRINCIPLES. 


[CHAP.  III. 


Let  OAl  be  the  line  representative  of  0lf  OBi  the  line  representative  of  &2,  and  the 
angle  AVOBZ  =  a.  Complete  the  parallelogram  and  draw 
OR.  Then  in  the  triangle  ROB2  we  have 

OR  :  0,  :  02  :  :  sin  «  :  sin  ROB2  :  sin  ROAr    .     (2) 

Comparing  (2)  with  (i)  we  see  that  OR  =  6,  or  the  result- 
ant is  the  third  side  of  the  triangle. 

Hence    we    have    the    "  triangle   of  angular  displace- 


**, 


menfs,"  just  as  for  linear  displacements,  page  54. 

The  same  evidently  holds  for  simultaneous  angular  displacements  indefinitely  small. 

(c)  FINITE  SIMULTANEOUS  ANGULAR  DISPLACEMENTS  ABOUT  DIFFERENT  INTERSECTING 
AXES. — The  same  evidently  holds  for  finite  simultaneous  angular  displacements  about  the 
same  or  different  axes.  For  we  can  divide  each  finite  displacement  into  a  number  of 
indefinitely  small  displacements,  and  treat  each  pair  as  in  the  previous  case. 

Hence  for  all  cases  except  finite  successive  angular  displacements  about  different 
intersecting  axes,  which  case  we  have  discussed  on  page  58,  we  can  combine  and  resolve 
angular  displacements  by  means  of  their  line  representatives,  just  like  linear  displacements 
in  the  preceding  chapter. 

We  have  then  the  "triangle  and  polygon  of  angular  displacements,"  just  as  for  linear 
displacements,  page  55. 

Also,  just  as  on  page  55,  the  component  of  the  resultant  in  any  direction  is  equal  to  the 
algebraic  sum  of  the  components  of  the  displacements  in  that  direction. 

Also,  \ve  can  find  relative  angular  displacement  in  precisely  the  same  way  (page  56). 

Examples. — (i)  A  body  has  two  simultaneons  rotations  of  2  radians  and 4  radians  about  axes  inclined  60° 
to  each  other.  Find  the  resultant. 

ANS.  2  1/7  radians  about  an  axis  inclined  to  the  greater  component  at  an  angle  whose  sine  is         _  . 

(2)  A  sphere  with  one  of  its  superficial  points  fixed  has  tiuo  simultaneous  rotations,  one  of  8  radians  about  a 
tangent  line  and  one  of  75  radians  about  a  diameter.  Find  the  axis  of  the  resultant  angular  displacement,  the 
resultant  angular  displacement,  and  the  number  of  revolutions  about  the  resultant  axis. 

ANS.  Axis   inclined  to  greater  component  at  an  angle  whose  tangent  is  — .      Resultant 

angular  displacement  17  radians.    Number  of  revolutions  —  =  about  2.75  revolutions. 

(3>  A  pendulum  suspended  from  a  point  in  the  prolonged  polar  axis  of  the  earth  swings  in  a 
plane  through  the  axis.  Find  the  angular  displacement  of  this  plane  relative  to  the  earth  in 
t2  hours. 

ANS.  The  angular  displacement  of  the  earth  relative  to  the  plane  is  EP  =  it  radians  north.  That  is, 
\i  we  look  north  along  the  polar  axis  the  rotation  is  seen  clockwise.  The  angular 
displacement  of  the  plane  relative  to  the  earth  is  then  PE  =  n  radians  south. 
Chat  is,  an  observer  on  the  earth  looking  south  along  the  polar  axis  sees  the  plane 
•rotate  clockwise  through  180°  in  12  hours. 

(4)  Let  the  pendulum  be  suspended  from  a  point  in  the  prolonged  radius  of  the  earth 
at  a  place  of  latitude  A.  and  siving  in  a  meridian  plane.  Find  the  angular  displace 
ment  of  the  plane  relative  to  the  earth  in  12  hours. 

ANS.  The  line  representative  of  the  angular  displacement  of  the  earth  relative  iy 
the  plane  is  still  EP  =  it  radians  north.  The  component  of  this  along  the  radius  of 
the  place  is  Ep  =  it  sin  A  radians.  The  angular  displacement  of  the  plane  relative  to 
the  earth  is  then  pE.  That  is,  an  observer  at  A  looking  towards  the  centre  of  the 
earth  would  see  the  plane  shift  clockwise  through  Jt  sin  A  radians  in  12  hours. 


CHAPTER    IV. 

LINEAR  AND  ANGULAR  SPEED  AND  VELOCITY. 

Mean  Linear  and  Angular  Speed. — Let  a  point  move  in  any  path  from  the  .initial 
position  al  to  the  final  position  a  in  the  time  /. 

Let  s  be  the  distance  described  (page  51).     Then   -,  or  the 
distance  described  per  unit  of  time,  is  the  MEAN  LINEAR  SPEED. 
Let  <f>  be  the  angle  described  (page  51).     Then  — ,    or   the 

angle  described  per  unit  of  time,  is  the  MEAN  ANGULAR  SPEED. 

Linear    speed    is    then    measured    in  feet    per  second,  and 
angular  speed  in  radians  per  second. 

Example. — Let  a  point  move  in  a  circle  of  radius  r  ft.  through  a  half 
circumference,  from  a\  to  a,  in  /  =  2  seconds. 

Then  the  mean  linear  speed  is  —  =  —  ft.  per  sec. 
Take  the  pole  O  in  the  perpendicular  OC  through  the  centre  C  of  the  circle,  so  that  Oai  —  Oa  —  /ft. 
a,  Then  (page  52)  the  angle  described  is  <t>  —  ^  jadians,  and  the  mean  an- 
gular speed  is  f—~rr  radians  per  sec. 

Mean  Linear  and  Angular  Velocity. — In  the  preceding 
figure  a^a  =  D  is  the  linear  displacement  (page  52),  and 
the  angle  afla  =  0  is  the  angular  displacement  (page  52). 

Let  t  be  the  time  of  moving  from  a^  to  a.     Then  — ,  or  the 

a 

linear  displacement  per  unit  of  time,  is  the  MEAN  LINEAR  VELOCITY,  and  -,  or  the  angular 

displacement  per  unit  of  time,  is  the  mean  angular  velocity  relative  to  (9. 

Linear  velocity  is  measured,  then,  like  linear  speed,  in  feet  per  second,  and  angular 
velocity,  like  angular  speed,  in  radians  per  second. 

Example,— In  the  preceding  example  we  have  seen  that  the  mean  linear  speed  is  S——  —  ft.  per  sec.,  and 
the  mean  angular  speed  is  ^  =  ^  radians  per  sec. 

The  mean  linear  velocity  is,  however,  —  =  —  ft.  per  sec.  in  a  direction  from  «,  to  a,  and  the  mean 
angular  velocity  for  pole  at  O  is  —radians  per  sec.,  where  0  is  given  by  /sin  |0  =  r,  or  sin  ^0  =  -j  ,  and  has 
also  direction  as  explained  in  the  next  article. 

Line  Representatives  of  Mean  Linear  and  Angular  Velocity. — Since  mean  linear 
velocity  is  linear  displacement  per  unit  of  time,  \ve  can  represent  it  by  a  straight  line,  just 
like  linear  displacement  itself. 

61 


KINEMATICS  OF  A  POINT.    GENERAL  PRINCIPLES. 


JC'IIAP.    IV. 


Thus  the  straight  line  ao  represents  by  its  length  the  magnitude  -    of  any 

mean  linear  velocity,  and  the  arrow  represents  the  direction. 

In  the  same  way,  since  mean  angular  velocity  is  angular  displacement  per 
•0    unit  of  time,  we  can  represent  it  by  a  straight  line  like  angular  displacement 

itself. 
Thus  the  straight  line  AO  at  right  angles  to  the  plane  of  afia  represents  by  its  length 

a 

the  magnitude  —  of  any  mean  angular  velocity,  and   looking  in 

the  direction  of  the  arrow  from  A  to  O  the  rotation  of  the  radius 
vector  is  always  seen  clockwise. 

fey  direction  of  angular  velocity  we  always  mean  the  direction 
of  its  line  representative. 

Thus  if  the  initial  and  final  positions  Oal   and  Oa  of  the 
radius  vector  lie  in  a  vertical   east   and   west    plane  a^Oa,   the 
direction  of  the  mean  angular  velocity  is  north,  because  in  such 
case  the  line  representative  AO  would  point  north,  so  that  looking  in  the  direction  of  the 
arrow  from  A  to  O  we  should  see  the  rotation  of  the  radius  vector  clockwise. 

Distinction  between  Mean  Speed  and  Velocity.  —  Let  the  initial  position  of  a  point  at 
the  time  /t  be  at  a  distance  sl  measured  along  the  path  from  any  fixed  point  in  the  path, 
taken  as  an  origin  or  point  of  reference,  and  at  any  other  time  /  greater  than  tl  let  the  dis- 
tance measured  along  the  path  from  this  same  fixed  point  be  s.  Then  (s  —  sv)  is  the  dis- 
tance described  in  the  interval  of  time  (t  —  /t),  and  the  mean  linear  speed  is 


In  similar  notation  we  have  (0  —  0,  ),  the  angle  described  in  the  interval  of  time  (/  — 
and  the  mean  angular  speed  is 

0  -  0 


If  s  is  greater  than  j,,  the  distance  described  is  away  from  the  origin  and  the  mean 
linear  speed  is  positive.  If  s  is  less  than  slt  the  distance  described  is  towards  the  origin  and 
the  linear  speed  is  negative.  So,  also,  if  0  is  greater  than  0lf  the  angle  described  is  away 
from  the  initial  line  of  reference  and  the  mean  angular  speed  is  positive.  If  0  is  less  than 
0,,  the  angle  described  is  towards  the  initial  line  of  reference  and  the  mean  angular  speed  is 
negative. 

Thus  if  the  distance  from  the  origin  is  increasing  the  linear  speed  is  positive,  if  decreas- 
ing it  is  negative.  If  the  angle  described  from  the  initial  radius  is  increasing  the  angular 
speed  is  positive,  if  decreasing  it  is  negative. 

Mean  linear  speed,  then,  is  mean  time-rate  of  distance  described,  and  mean  angular  speed 
is  mean  time-rate  or  angle  described.  Both  possess  magnitude  and  sign,  according  to  whether 
the  distance  or  angle  described  increases  or  decreases  with  the  time,  but  both  are  independent 
of  direction  of  the  path. 

Such  quantities  which  have  sign  and  magnitude  but  are  independent  of  direction  are 
called  SCALAR  quantities.  They  cannot  be  represented  by  straight  lines. 

But,  as  we  have  seen,  mean  linear  velocity  —  is  mean  time-rate  of  linear  displacement, 


CHAP.  IV.]  INSTANTANEOUS  LINEAR.  AND  ANGULAR  SPEED  AND   VELOCITY. 


e 

and  mean  angular  velocity  —  is  mean  time-rate  of  angular  displacement.      Both  possess  not 

only  magnitude  but  direction.      Such  quantities  are  VECTOR  quantities.      They  can  be  repre- 
sented by  straight  lines. 

We  see,  then,  that  mean  speed  and  velocity,  although  measured  in  the  same  units,  are 
quantities  of  a  different  kind,  and  the  number  of  units  are  in  general  different  for  each. 

Example.  — Let  a  point  move  in  a  circle  of  radius  r  —  10  feet.  Let  the  plane  of  the  circle  be  an  east  and 
west  plane.  Let  the  point  move  from  the  top  a\  eastward  through  a  half  circumference,  from  a\  to  a,  in  the 
time  /  =  3  seconds.  Take  the  pole  O  in  the  perpendicular  OC  through  the  centre  C  of  the  circle,  so  that 
Oat  =  On  =  /  =  20  ft.  a 

Then  the  distance  described  is  s  =  itr  —  31.416  ft.,  and  the  mean  linear 
speed  is  —  =  -f  10.472  ft.  per  sec. 

The  angle  described  is  (page  52)  <j>  =  ~  =  1.5708  radians,  and  the  mean 

angular  speed  is  —  =  +  0.5236  radians  per  sec. 

On  the  other  hand,  the  linear  displacement  is  D  —  ir  =  20  ft.  in  the  direction  from  ai  to  a,  and  the 
mean  linear  velocity  is  —  =  6.66  ft.  per  sec.  in  the  same  direction. 

The  angular  displacement  for  pole  at  O  is  given  by  /  sin  |6  =  r,  or  sin  J0  =  —  =  £,  or  6  =  —  radians 

A          if  * 

north,  and  the  mean  angular  velocity  isy  =—  radians  per  sec.  north.     We  see,  then,  that  the  mean  linear 

speed  and  velocity,  although  measured  in  the  same  units,  are  in  this  case  not  only  different  in  magnitude,  but 
one  is  independent  of  direction,  while  the  other  has  direction.     So,  also,  for  mean  angular  speed  and  velocity. 

Instantaneous  Linear  and  Angular  Speed  and  Velocity.—  The  limiting  magnitude  of  the 
mean  linear  or  angular  speed  when  the  interval  of  time  is  indefinitely  small  is  the  INSTAN- 
TANEOUS linear  or  angular  speed. 

These  are  SCALAR  quantities  having  sign  and  magnitude  but  independent  of  direction, 
just  like  mean  linear  and  angular  speed. 

The  limiting  magnitude  and  direction  of  the  mean  linear  or  angular  velocity  when  the 
interval  of  time  is  indefinitely  small  is  the  INSTANTANEOUS  linear  or  angular  velocity. 

Instantaneous  linear  and  instantaneous  angular  velocity,  then,  are  vector  quantities, 
having  both  magnitude  and  direction,  and  can  therefore  be  represented  by  straight  lines, 
just  like  mean  linear  and  mean  angular  velocity. 

Thus,  let  the  interval  of  time  be  indefinitely  small  and  denoted  by  dt. 

Then  the  initial  and  final  positions  al  and  a  of  a  moving  point 
will  become  consecutive  points  of  the  path,  so  that  the  distance  a^a 
can  be  denoted  by  ds. 

We  see,  then,  that  in  this  case  the  distance  described  is  ds. 
That  the  linear  displacement  is  ds  in  magnitude,  and  its  direction 
is  tangent  to  the  path,  so  that  a^a  is  its  line  representative.  That  is, 
in  this  case,  distance  described  is  the  magnitude  of  the  linear  displace- 
ment. 

If,  then,  we  denote  the  instantaneous  linear  speed  by  v,  we  can 
write 

ds 


and  this  will  also  be  the  magnitude  of  the  instantaneous  linear  velocity,  the  direction  of  which 
is  always  tangent  to  the  path. 


64  KINEMATICS  OF  A  POINT.    GFNER4L  PRINCIPLES.  [CHAP.  IV. 

Similarly,  in  the  indefinitely  small  time  dt,  the  indefinitely  small  angle  described  is 
a^Oa,  which  can  be  denoted  by  d</>.  The  indefinitely  small  angular  displacement  dO  will  be 
the  same  in  magnitude,  and  its  direction  will  be  at  right  angles  to  the  plane  of  rotation  a^Oa, 
so  that  AO  is  its  line  representative.  That  is,  in  this  case,  angle  described  (*/0)  is  the 
magnitude  of  the  angular  displacement  (dff). 

If,  then,  we  denote  the  instantaneous  angular  speed  by  co,  we  can  write 

dd 
™  =  -dt> 

ana"  this  will  also  be  the  magnitude  of  the  instantaneous  angular  velocity,  the  direction  of 
which,  as  indicated  by  its  line  representative  AO,  is  always  at  right  angles  to  the  plane  of 
rotation,  so  that  looking  in  the  direction  of  its  arrow  we  see  the  rotation  of  the  radius  vector 
clock-wise.  The  line  representative  AO  coincides,  therefore,  with  the  axis  about  which  the 
radius  vector  rotates. 

Examples.— (i)  Let  the  distance  described  by  a  moving  point  be  given  by  the  equation 

s  =  y/  +  8f>, (i) 

where  s  is  the  number  of  feet  described  in  any  number  of  seconds  t. 
Then  for  any  other  number  of  seconds  /,  we  can  write 

J.  =  7A  +  8/,« (2) 

Suppose  /i  to  be  less  than  /.     Then  subtracting  (2)  from  (i)  we  have  for  the  distance  s  —  Si  described  in 
the  interval  of  time  /  —  t\ 

j-j,  =  7(t  -  /,)  +  8(/2  -  /,»)  =  ;(/  -  /,)  +  8(/  +  /!)(/•  -  /,). 
The  mean  linear  speed  is,  then,  for  this  interval  of  time 


7  +  8(/  +  /,) , (3) 


We  see  from  (3)  that  as  the  interval  of  time  /  —  t\  decreases,  t\  approaches  equality  with  /  and  the  mean 
linear  speed  given  by  (3)  approaches  the  limiting  value  7  +  i6/.  We  have  then  for  the  instantaneous  speed, 
or  the  magnitude  of  the  instantaneous  velocity, 

v  =  ~  =  7  +  i6/.  .     .  (4) 

In  order  that  the  linear  velocity  maybe  completely  known  we  must  specify,  in  addition  to  its  magnitude 
as  given  by  (4),  its  direction,  which  is  always  tangent  to  the  path  at  the  instant. 

Students  familiar  with  the  Calculus  will  note  that  (4)  is  obtained  directly  from  (i)  by  differentiating  s 
with  reference  to  /. 

If  in  (4)  we  make  /  =  2,  we  have  v  =  39  ft.  per  sec.  This  does  not  mean  that  the  point  has  described 
39  feet  in  the  preceding  second,  nor  that  it  will  describe  39  feet  in  the  next  second.  But  it  means  that  at 
the  instant  (2  seconds  from  the  start)  it  is  moving  at  such  a  rate  that,  if  that  rate  did  not  change,  the  point 
would  describe  39  feet  in  the  next  second.  It  is  the  instantaneous  speed. 

(2)  Let  a  point  move  in  a  circle  in  a  vertical  east  and  west  plane,  and  let  the  angle  described  by  the  radius 
vector  from  a  fixed  point  to  the  moving  point  be  given  by  the  equation 

0  =  7t  +  Bf, 

where  0  is  the  number  of  radians  described  in  any  number  of  seconds  t,  the  origin  being  at  the  centre.     Find,  as 
in  the  preceding  example,  the  mean  and  instantaneous  angular  speed  and  velocity, 
ANS.  The  mean  angular  speed  is 

\^ji  =  7  +  8(/  +  /,). 
The  instantaneous  angular  speed  is 


This  is  the  magnitude  of  the  instantaneous  angular  velocity,  whose  direction  is  north  if  the  point  i 
moving  eastward  in  the  plane  from  the  top  point. 


CHAPTER  V. 

LINEAR  AND  ANGULAR  VELOCITY. 

Speed  and  Velocity  in  General. — The  terms  "speed"  and  "velocity"  always  signify 
instantaneous  linear  speed  and  velocity  unless  otherwise  specified.  The  terms  linear  and 
angular  speed  and  velocity  always  signify  instantaneous  linear  and  angular  speed  and  velocity 
unless  otherwise  specified. 

Uniform  and  Variable  Linear  Velocity. — When  the  linear  velocity  has  the  same  magni- 
tude and  direction  whatever  the  interval  of  time,  it 'is  UNIFORM.  If  either  magnitude  or 
direction  change,  it  is  VARIABLE. 

If,  then,  the  velocity  is  uniform,  the  line  representative  has  always  the  same  magnitude 
and  direction ;  and  since  velocity  is  tangent  to  the  path,  we  have  uniform  speed  in  a  straight 
line.  The  velocity  in  this  case  is  the  same  as  the  mean  velocity  for  any  interval  of  time. 

If  only  the  direction  changes,  we  have  uniform  speed  in  a  curve.  In  this  case  the  mag- 
nitude of  the  velocity  is  the  same  as  the  mean  speed  for  any  interval  of  time. 

If  only  the  magnitude  changes,  we  have  variable  speed  in  a  straight  line. 

If  both  magnitude  and  direction  change,  we  have  variable  speed  in  a  curve. 

Examples.— (i)  A  point  moves  with  uniform  speed  in  a  circle  of  radius  r  =  6ft.  and  makes  a  half  revolu- 
tion in  the  time  t  =  j  sec.  Find  the  mean  speed,  the  mean  velocity,  the  instantaneous  speed  and  the  instant  an- 
eous  velocity. 

ANS.  The  mean  speed  is  —  =  lit  ft.  per  sec.,  and  the  mean  velocity  is  —  =  4  ft.  per  sec.  in  the  direction 
of  the  diameter  through  the  starting-point. 

Since  the  speed  is  uniform,  the  instantaneous  speed  must  be  the  same  as  the  mean  speed,  or  zit  ft. 
per  sec. 

The  magnitude  of  the  instantaneous  velocity  is  the  same  as  the  instantaneous  speed,  or  v  =  lit  ft.  per 
sec.,  but  the  direction  at  any  instant  is  tangent  to  the  path  at  the  position  of  the  point  at  that  instant. 

(2)   Criticise  the  statement,  "a  point  moves  in  a  circle  with  ttniform  velocity." 

ANS.  A  point  can  move  in  a  circle  or  in  any  path  with  uniform  speed.  But  if  the  velocity  is  uniform, 
it  can  only  move  in  a  straight  line  with  uniform  speed.  A  point  cannot  move  in  a  curve  with  uniform 
velocity. 

Uniform  and  Variable  Angular  Velocity. — By  direction  of  angular  velocity  we  always 
mean  the  direction  of  its  line  representative  (page  62). 

When  the  angular  velocity  has  the  same  magnitude  and  direction  whatever  the  interval 
of  time,  it  is  UNIFORM.  If  either  magnitude  or  direction  change,  it  is  VARIABLE. 

If,  then,  the  angular  velocity  is  uniform,  the  line  representative  has  always  the  same 
magnitude  and  direction,  and  we  have  uniform  angular  speed  in  an  unchanging  plane.  The 
angular  velocity  in  such  case  is  the  same  as  the  mean  angular  velocity  for  any  interval  of 
time. 

If  only  the  direction  changes,  we  have  uniform  angular  speed,  but  a  changing  plane  of 
rotation.  In  this  case  the  magnitude  of  the  angular  velocity  is  the  same  as  the  mean  angular 
speed  for  any  interval  of  time. 

65 


66  KINEMATICS  OP  A  POINT.    GENERAL  PRINCIPLES.  [CHAP.  V. 

If  only  the  magnitude  changes,  we  have  variable  angular  speed  in  an  unchanging  plane. 
If  both  magnitude  and  direction  change,  we  have  variable  angular  speed  and  a  changing 
plane  of  rotation. 

Examples. — (i)  A  point  moves  with  uniform  angular  speed  in  a  circle  of  radius  r  =  6ft.  in  a  vertical 
east  and  west  plane,  and  makes  a  half  revolution  in  the  time  t  =  2  sec.  Find  the  mean  angular  speed,  the 
mean  angular  velocity,  the  instantaneous  angular  speed  and  velocity,  origin  at  the  centre. 

ANS.  The  mean  angular  speed  is  —  —  —  radians  per  sec.    The  mean  angular  velocity,  if  the  point  moves 

eastward  from  the  top,  is  also  -  radians  per  sec.  north. 

Since  the  angular  speed  is  uniform,  the  instantaneous  angular  speed  is  the  same  as  the  mean  speed. 
The  magnitude  of  the  instantaneous  angular  velocity  is  the  same,  but  its  direction,  if  the  point  moves  east- 
ward from  the  top,  is  north. 

(2)  Criticise  the  statement,  "  a  point  moves  in  a  circle  the  plane  of  which  is  constantly  changing,  with 
uniform  angular  velocity." 

ANS.  A  point  can  move  in  a  circle  the  plane  of  which  is  changing,  with  uniform  angular  speed.  But  if 
the  angular  velocity  is  uniform,  its  line  representative  does  not  change  either  in  magnitude  or  direction,  and 
the  plane  therefore  cannot  change.  A  point  cannot  move  in  a  changing  plane  with  uniform  angular  velocity. 

Resolution  and  Composition  of  Linear  Velocity, — Since  linear  velocity  is  linear  dis- 
placement per  unit  of  time  when  the  interval  of  time  is  indefinitely  small  (page  63),  it  has 
magnitude  and  direction,  and  can  be  represented  by  a  straight  line,  just  like  linear  displace- 
ment itself.  Therefore  all  the  principles  of  Chapter  II,  page  54,  which  hold  good  for  linear 
displacements  hold  good  also  for  linear  velocities. 

We  have,  then,  the  "  triangle  and  polygon  of  velocities"  and  can  combine  and  resolve 
velocities  just  like  displacements. 

We  also  have  relative  velocity  with  similar  notation  as  for  displacements  (page  55). 

Examples. — (i)  A  ship  sails  N.  jo*  E.  with  a  speed  of  10  miles  an  hour.    Find  its  velocity  east  and  north. 
ANS.  5  miles  per  hour  east,  5l/3~miles  per  hour  north. 

(2)  Find  the  vertical  velocity  of  a  train  moving  up  a  i-per-cent 

O- *•«  gradient  at  a  speed  of  30  miles  per 'hour. 

ANS.  0.3  miles  per  hour. 

C >•  P  (3)  A  circle  rolls  on  a  horizontal  plane.     Its  centre  moves  with 

a  velocity  v  towards  the  east.      Find  the  velocity  of  the  top  and 

C-« b  bottom  points  relative  to  the  plane. 

ANS.  The  velocity  of  the  top  point  a  relative  to  the  centre  C 

— *<• —  *P     >s  given  by  the  line  ac  —  v  towards  the  east ;   of      g         G       c 

,  the  centre  C  relative  to  the  plane  by  cP  =  v  tow- 

c —  >  p  ards  the  east;   of  the  bottom  point  b  relative  to 

the  centre  C  by  be  —  v  towards  the  west. 

If,  then  ,we  lay  off  ac  and  cP  in  order,  we  have  the  velocity  of  a  relative  to  the  plane  P 
given  by  aP  —  2v  towards  the  east.  The  top  point  has,  relative  to  the  earth,  twice  the 
velocity  of  the  centre. 

If  we  lay  off  b  relative  to  C,  and  C  relative  to  P,  we  have  b  relative  to  P,  zero.     The    G 
bottom  point  is  at  rest  relative  to  the  earth.  B 

(4)  A  ball  let  fall  in  an  elevator  has  a  velocity  relative  to  the  ground  of  32  feet  per  sec., 
while  the  elevator  has  a  velocity  relative  to  the  ground  of  T2  feet  per  sec.     Find  the  velocity  of        „# 
the  ball  relative  to  the  elevator  when  it  is  rising  and  falling. 

ANS.  The  velocity  of  the  ball  relative  to  the  ground  is  given  by  BG  =  32  downwards.  If 
elevator  is  rising,  we  have  elevator  relative  to  ground  given  by  EG  =  12  upwards.  If  elevator 
is  falling,  EG  =  12  downwards. 

In  the  first  case,  laying  off  ft  relative  to  G.  and  G  relative  to  E,  we  have  B  relative  to 
£44  feet  per  sec.  down.  In  the  second  case,  laying  off  B  relative  to  G,  and  G  relative  to  E, 
we  have  B  relative  to  E  20  ft.  per  sec.  downwards. 


CHAP.  V.] 


RECTANGULAR   COMPONENTS  OF  VELOCITY. 


Rectangular  Components  of  Velocity. — Let  a  point  P  be  given  by  its  co-ordinates 
x,  y,  z,  and  let  it  have  the  velocity  v,  whose 
line  representative  makes  the  angles  <*,  ft,  y, 
with  axes  through  P  parallel  to  the  co-ordi- 
nate axes  OX,  OY,  OZ,  respectively. 

Let  the  components  of  v  parallel  to  these 
axes  be  vx,  vy,  vt,  respectively. 

Then  we  have 


dx 

vx  =•  "    —  v  cos  a> 


vy  =    -  =  v  cos  ft, 


dz 

=  —=v  cos  y. 


Analytic    Determination    of    Resultant 
Velocity.  —  If  the  point  P  has  several  simulta- 


neous velocities, 


v2,  etc.,    making  the 


angles  (arp  ft^,  yj,  (<*3,  py  y2),  (aa,  fls,  y3},  etc., 
then  we  have  for  the  resultant  components 

dx 


.  .  .  =  ^v  cos  a, 


vy  =  -j-  =  vl  cos  Pl  -j-  z/2  cos  /?2  -(-  z/3  cos  ^3  -J-  .  .  .  =  2v  cos  /?, 

dfe 

=  -^^cos  y. 


In  these  summations  we  must  take  components  in  the  directions  OX,  OY,  OZ  positive, 
in  the  opposite  directions  negative. 

We  have,  then,  for  the  magnitude  of  the  resultant  velocity  v 


V=~-  +  Vvx*+    V?  +  V? (2) 

This  resultant  passes,  of  course,  through  P,  and  its  direction  cosines  are 


cos  or  =  — ,  cos  ft  =  —  , 

z;  '  z> 


cos  Y  =  — 


Since  cos2  a  -|-  cos2  ft  -\-  cos2  y  —  I,  we  have  also,  from  (3), 
v  =  vx  cos  a-\-vy  cos  ft  -j-  v,  cos  y. 


(3) 


(4) 


In  determining  these  cosines  we  take  v  always  positive  and  measure  all  angles  a,  ft,  y 
in  such  directions  that  their  projections  on  the  co-ordinate  planes  XY,  YZ,  ZX  shall  be 
positive  from  OX  around  to  OY,  from  OY  around  to  OZ,  from  OZ  around  to  OX,  as 
indicated  by  the  arrows  in  the  figure. 

If  all  the  velocities  are  in  one  plane,  as,  for  instance,  the  plane  of  XY,  we  make  y  =  90°, 
and  hence  vz  =  o  in  (i),  (2)  and  (3). 

Examples. — (i)  A  fiotnt  has  the  component  velocities  in  the  same  plane  XY,  Vi  =J2,  Vt  =  24,  v9  =  j6,  V* 
=  48 ft.  per  sec. ,  making  angles  with  OX  of  «i  =  +  60°,  a*  =  +  fjo  °,  (*  =  +  240°,  c-t  =  Jjo°.  Find  the 
resultant  velocity. 


68  KINEMATICS  OF  A  POINT.    GENERAL  PRINCIPLES.  [CHAP.  V. 

ANS.  We  have  a—  ft  =  90°.     Hence  ft\  —  —30°,  /*»  =  +  60°,  fi»  =  +  150°,  /0«  =  +  240°.    Hence 

z/,  =  6  -  i2f/3~~  l8  +  24VT=  +  8.784  ft.  per  sec., 
vy  =  +  6  ^3  +  12  —  181/3  —  24  =  —  32.784  ft.  per  sec., 
The  resultant  is  v  =  +  33.94  ft.  per  sec.,  making  the  angle  a  with  OX  given  by 

cos  a  =  i       —  =  0.2588,     or     a  =  about  75°' 
T  33-94 

(2)  ^4  /<?/«/  has  the  component  -velocities  vt  =  40,  v*  =30,  v\  =  60  ft.  per  sec.,  making  the  angles 
a,  =  -f  60°,  ftt  =  +  60°,  YI  obtuse;  a«  =  -f-  60°  ft*  —  +  60°,  y*  acute;  ot»  =  -f-  60  °,  fit  =  -f  j>0°.  //»*/  the 
resultant. 

ANS.   We  find  the  angles  y  by.  the  equation,  page  13, 

cos"  Y  =  -  cos  (a  +  ft)  cos  (a-  ft).     Hence  yt  =  +  135°,  y*  =  +  45°,  yt  =  90°. 

.  We  have,  then,  vx  —  20  +  25  +  30  =  +  75  ft.  per  sec.,  z/y  =  20  +  25  +  30  4/3"=  -f  96.96  ft.  per  sec., 
v,  =  —  -^L  +  -5?_-r-  o  =  -f-  7.071  ft.  per  sec.  The  resultant  is  v  =  1  19.6  ft.  per  sec.,  and  its  direction  cosines 


are  cos  a  =  .'  cos  ft  =  ,  cos  y  =  ,  or  a  =  51°  ,o',  /J  =  35°  50',  y  =  86"  37'. 


Resolution  and  Composition  of  Angular  Velocity.—  Since  angular  velocity  is  angular 
displacement  per  unit  of  time  when  the  interval  of  time  is  indefinitely  small  (page  63),  it  has 
magnitude  and  direction  and  can  be  represented  by  a  straight  line,  just  like  angular  displace- 
ment itself.  But  since  the  time  is  indefinitely  small,  and  therefore  the  angular  displacement 
indefinitely  small,  we  have  the  case  (£),  page  59.  Therefore  all  the  principles  of  Chapter  II, 
page  54,  hold  good  also  for  angular  velocities  whether  simultaneous  or  successive,  and  we 
have  the  "triangle  and  polygon"  of  angular  velocities,  and  can  combine  and  resolve  them 
just  like  linear  displacements  (page  55). 

We  have  also  relative  angular  velocity  with  similar  notation  as  for  linear  displacements 
(page  56). 

Examples.  —  (i)  A  point  on  a  sphere  is  rotating  uniformly  about  a  diameter  at  the  rate  of  10  radians  per 
minute.  Find  the  component  angular  "velocity  about  another  diameter  inclined  30°  to  the  first. 

ANS.   5  4/3  radians  per  min. 

(2)  A  point  has  two  simultaneous  or  successive  angular  velocities  of  2  radians  per  sec.  and  4  radians  per  sec. 
about  axes  inclined  60°  to  each  other.  Find  the  resultant. 

ANS.  2  y^  radians  per  sec.  about  an  axis  inclined  to  the  greater  component  at  an  angle  whose  sine  is 


(3)  A  sphere  with  one  of  its  superficial  points  fixed  has  two  angular  velocities,  either  simultaneous  or 
successive,  one  of  8  radians  per  sec*  about  a  tangent  line,  and  one  of  15  radians  per  sec.  about  a  diameter.     Find 
the  resultant  angular  velocity. 

ANS.   17  radians  per  sec.  about  an  axis  inclined  to  the  greater  component  at  an  angle  whose  tangent 
is  A. 

(4)  A  pendulum  suspended  from  a  point  in  the  prolonged  polar  axis  of  the  earth  swings  in  a  plane  through 
the  axis.    Find  the  angular  velocity  of  this  plane  relative  to  the  earth.     See  example  (j),  page  60. 

ANS.  27t  radians  per  day  south.     That  is,  an  observer  looking  south  along  the  polar  axis  sees  the  plane 
rotate  clockwise  at  this  rate. 

(5)  Let  the  pendulum  be  suspended  from  a  point  in  the  prolonged  radius  of  the  earth  at  a  place  of  latitude  A. 
Find  the  angular  velocity  of  the  plane  of  s^i'iit^  relative  to  the  earth.     See  example  (4},  page  60. 


CHAP.  V.] 


RECTANGULAR   COMPONENTS   OF  ANGULAR    VELOCITY. 


69 


ANS.   -2.it  sin  A  radians  per  day  in  a  direction  towards  the  centre  of  the  earth.     That  is,  an  observer 
looking  towards  the  centre  of  the  earth  would  see  the  plane  rotate  clockwise  at  this  rate. 

- 
sm 


The  time  of  a  complete  revolution  of  the  plane  would  then  be 


=  -—  days.    If  A  is  60°  the  time  oi 

sin  A      ' 


revolution  would  be  —7-  =  i.iSS  days.     At  the  equator  sin  A  =  o,  and  the  plane  does  not  rotate  relatively  to 
the  earth.     At  the  poles  sin  A  =  i,  and  the  plane  makes  a  revolution  in  i  day. 

Rectangular  Components  of  Angular   Velocity.  —  Let  a  point  P  be  given  by  its  co- 
ordinates x,  y,  z,   and  let  it  have  the  angular 
velocity    GJ,   whose    line   representative  passes 
through  O  and  makes  the  angles  a,  ft,  y   with 
the  axes  OX,  O  Y,  OZ,  respectively. 

Then  the  components  of  GO  along  the  axes 

are   oox—  oo  cos  a,         ooy  =  GO  cos  ft,        &,  =  GO 
cos  y. 

These  components  are  positive  in  the  direc- 
tions OX,  OY,  OZ.  Since  rotation  is  always 
seen  clockwise  when  we  look  along  a  line 
representative  in  the  direction  of  its  arrow, 
we  have,  then,  positive  rotation  in  the  plane  XY 
from  OX  around  to  OY;  in  the  plane.  FZ 
from  O  Y  around  to  OZ;  in  the  plane  ZX  from 
OZ  around  to  OX,  as  shown  by  the  arrows  in 
the  figure. 

Analytic  Determination  of  Resultant  Angular  Velocity.  —  If  the  point  P  has  several 
simultaneous  angular  velocities,  calf  co2,  6?3,  etc.,  all  passing  through  O  and  making  the 
angles  (av  /?1(,  y^,  (az,  ftj,  y2),  (ar3,  /?3,  j/3),  etc.,  then  we  have  for  the  resultant  components 


GOX  =  col  cos  ctl  -\-  oo2  cos  oc2  -\-  6?3  cos 

G0y  —    wl  COS  fil  -f-  OL>2  COS  ft2  -f-  <»s  COS 

coz  =  ojl  cos  &>l  -f-  GJZ  cos  y2  -f-  ce?3  cos 


-{-•••=  2oj  cos  a) 

+     .     .     .    =  ^GO  COS  ft, 

-}-  .   .    .  =  2  on  cos  y. 


In  these  summations  we  must  take  components  in  the  directions  OX,   OY,  OZ  positive, 
in  the  opposite  direction  negative. 

We  have,  then,  for  the  magnitude  of  the  resultant 


GO  =  -f  V<*>*2  -\-  GO*  -f  w? 

This  resultant  passes,  of  course,  through  O,  and  its  direction  cosines  are 


COS  Of  =    — -  COS  ft  —    — , 

G?  OJ 


GO- 

cos  y  =  - 


(3) 


Since  cos2  a  -(-  cos2  ft  -\-  cos2  y  =  i,  we  have  also,  from  (3), 

GO  —  GJ^  cos  a  -f-  Goy  cos  ft  -\-  GJZ  cos  y. 


(4) 


In  determining  the  cosines,  we  take  GO  always  positive  and  measure  all  angles  a,  ft,  y  in 
such  directions  that  their  projections  on  the  co-ordinate  planes  XY,   YZ,  ZX  shall  be  posi- 


?o 


KINEMATICS   OF  A  POINT.     GENERAL   PRINCIPLES. 


[CHAP.  V. 


tive  from  OX  around  to  OY,  from  OY  around  to  OZ,  from  OZ  around  to  OX,  as  indicated 
by  the  arrows  in  the  figure. 

If  all  the  velocities  are  in  one  plane,  as,  for  instance,  the  plane  of  XY,  we  make  y  =  90° 
and  hence  <a,  =  o  in  (i),  (2)  and  (3). 

Example.—  (i  )  A  point  has  the  component  angular  velocities  in  the  same  plane  XY,  a>,  =  /.?,  a,a  =  24, 
o>>  =  j6.  o»4  =  48  radians  per  sec.,  making  angles  with  OX,  a,  =  -f-  60°,  a»  =  +  /jo°,  a»  =  +  240",  <*4  =  +  Jjo°. 
Find  the  resultant  velocity.  (See  example  (/),  page  68). 

ANS.  oox  =  +  8.784  radians  per  sec.;  <ay  =  —  32.784  radians  per  sec.  ;  w  —  -f-  33.99  radians  per  sec.  ; 
cos  a  =  0.2588.  or  a  =  about  75°. 

(2)  A  point  has  the  component  angular  -velocities  oo,  =  40,  oot  =  jo.  tat  =  60  radians  per  sec-,  making  the 
angles  a,  =  +  60",  fit  =  +  60°,  yl  obtuse  ;  at*  =  +  60",  ft*  =  +  60,  y*  acute  ;  o«  =  +  60°,  ft,  —  +jo°.  Find  the 
resultant. 

ANS.  See  example  (2),  page  68. 

Linear   in   Terms   of  Angular   Velocity.  —  Let    a^   and  a   represent    two   consecutive 
positions  of  a  point  moving  in  any  path.     Then  the  distance  a^a  =  ds. 

Take  any  point  O  as  a  pole,  and  draw  the  radius 
vectors  Oal  =  r^  and  Oa  =  r.  Then  the  indefinitely 
small  angle  a^Oa  is  d$. 

Drop  the  perpendicular  avn  from  a,  upon  Oa,  and 
denote  the  angle  of  inclination  a^an  between  the  radius 
vector  Oa  and  ava  by  e.  Then 

aji  =  ds  .  sin  e. 

But  if  the  points  al  and  a  are  consecutive,  we  have 
also 

a^i  —  r^dO. 
Hence  we  have 

ry/0  =  ds  .  sin  e  .....     .     .     .     .     .     .     (i) 

But  for  consecutive  points  na  =  dr  and 

rt  =  On  =  r  —  dr. 
Hence  equation  (i)  becomes 

rdO  =  dr  .  dO  -f  ds  .  sin  e.      .      ........     (2) 

As  dO  in  this  equation  decreases,  rdO  approaches  the  limiting  value  ds  .  sin  e.  We  have 
then  at  the  limit 

rdH  =  ds  .  sin  e  ............     (3) 

If  we  divide  by  the  indefinitely  small  time  dt  in  passing  from  al  to  a,  we  have 

d8      ds 


JO  . 
But  (page  64)  --    is  the  magnitude  of  the  angular  velocity  of  the  point  relative  to  O, 


which   we  have  denoted  by  &?,   and  (page  63)  -y  is   the  magnitude  of  the  linear  velocity, 


CHAP.  V.]  LINEAR.  IN   TERMS  OF  ANGULAR   VELOCITY—  EXAMPLES.  7* 

which   we  have  denoted  by  v.      Also,  v  sin   e  is  the  component  of  the  velocity  v  at  right 
angles  to  the  radius  vector  r  in  the  plane  of  v  and  r. 

Let  us  denote  this  normal  component  by  vn.      Then  we  have 

rw  =  v  sin  e  =  vn  ........     ...     (4) 

Hence  the  product  of  the  radius  vector  r  for  any  point  by  the  angular  velocity  GO  gives  the 
linear  velocity  vn  of  the  point  normal  to  the  radius  vector  in  the  plane  of  v  and  r. 

The  line  representative  of  the  angular  velocity  a?  is  a  straight  line  AO  through  the  pole 
O  at  right  angles  to  the  plane  afla  of  v  and  r,  so  that,  looking  in  the  direction  of  its  arrow, 
the  rotation  of  the  radius  vector  is  seen  clockwise., 

Equation  (4)  is  general,  whatever  the  path  or  wherever  the  pole  O  may  be  taken. 

If  the  pole  O  is  taken  at  the  centre  of  curvature  C,  so  that  Oa  is  the  radius  of  curvature 
p,  then  e  =  90°  and  we  have 


(2) 


If  the  pole  O  is  taken  anywhere  in  the  plane  through  p  perpendicular  to  v,  we  still  have 

e  =  90°,  and  in  this  case 

roo  =  v ,...,..     (3) 

Hence  in  this  case  the  product  of  the  radius  vector  by  the  angular  velocity  gives  the 
linear  velocity  itself.    ' 

Example. — Let  the  distance  described  by  a  moving  point  be  given  by  the  equation 

s  =  //  +  #2  +  2t3, 

•where  s  is  the  number  of  feet  described  in  any  number  of  seconds  t. 

Let  the  path  be  a  circle  in  an  east  and  west  plane,  and  let  the  point  start 
from  the  top  point  Z  and  move  eastward.      Let  the  radius  of  the  circle  be 


At  the  end  of  t  =  2  sec.  find  the  distance  and  angle  described;  the  linear  and 
angular  displacement;  the  mean  linear  and  angular  speed;  the  mean  linear  and 
angular  velocity  ;  the  instantaneous  linear  aad  angular  velocity.  The  origin  is 
taken  at  L  and  the  starting-point  at  Z  in  the  figure. 

ANS.   Insert  /  =  2  in  the  equation,  and  we  have  for  the  distance  described  s  =  62  ft. 

We  have  rO  —  s.     Hence  9  =  -  =   —  —  =   -radians.     The  point,  then,  has  moved  in  2  seconds  through  a 
r         1  24  "     2 

quadrant  from  Z  to  E. 

The  linear  displacement  is  then  from  Z  to  E,  or  D  =  rt/2  ft.  in  a  direction  making  an  angle  of  45°  with 
the  initial  radius  vector  CZ. 

The  magnitude  of  the  angular  displacement  is  the  same  as  the  angle  described,  or  —  radians,  and  its 
direction  is  north. 

We  find,  just  as  in  example  (r),  page  64,  the  mean  linear  speed 


For  t\  =  o  and  t  =  2  this  gives 


/  — 

We  have  for  the  mean  angular  speed 


S  —  Si  t 

-  =  31  ft.  per  sec. 


9  —  9,  f)  —  9,       v       7f 

_       -^  31,    or     =  —  ~  —  radians  per  sec. 


72  KINEMATICS  OF  A  POINT.     GENERAL   PRINCIPLES.  [CHAP.  V. 

The  mean  linear  velocity  is  the  linear  displacement  per  unit  of  time,  or  r_i_f  =  -—  ft  per  sec  in  the 
direction  from  Z  to  E. 

The  mean  angular  velocity  is  the  angular  displacement  per  unit  of  time,  or  —  =  -  radians  per  sec., 
dirtction  north. 

The  instantaneous  linear  velocity  is 

"  =  §=  7  +  16*  +  6/». 
For  /  =  2  this  gives  v  =  63  ft.  per  sec. 

The  instantaneous  angular  velocity  is  given  in  magnitude  by  ra>  =  v,  or  GO  =  -  =^^_  radians  per  sec., 
and  its  direction  is  north. 


CHAPTER   VI. 

LINEAR   AND   ANGULAR    RATE   OF   CHANGE   OF   SPEED.      LINEAR   ACCELERATION. 

Change  of  Speed.  —  When  the  speed  of  a  point  varies  in  magnitude,  the  difference  for 
any  interval  of  time  between  the  final  and  initial  instantaneous  speeds  is  the  change  of 
speed  for  that  interval  of  time. 

Mean  Rate  of  Change  of  Speed.  —  The  change  of  speed  p«r  unit  of  time  is  the  MEAN 

RATE    OF    CHANGE    OF    SPEED. 

Thus,  if  the  linear  speed  of  a  point  at  any  time  /  is  v,  and  at  an  earlier  time  tl  it  is  t\  , 
then  the  change  of  linear  speed  is  (v  —  z/J  for  the  interval  of  time  (t  —  t^,  and  the  mean 
rate  of  change  of  linear  speed  is 


This  is  positive  if  v  is  greater  than  vlt  or  if  the  linear  speed  increases  with  the  time,  and 
negative  if  v  is  less  than  vlt  or  if  the  linear  speed  decreases  with  the  time. 

Similarly,  if  the  angular  speed  of  a  point  at  any  time  t  is  &?,  and  at  an  earlier  time,  tx  ,  it 
is  colt  then  the  change  of  angular  speed  is  (GO  —  GO^  for  the  interval  of  time  (t  —  /x),  and  the 
mean  rate  of  change  of  angular  speed  is 

GO    —     Gflj 
*   —   *l    ' 

This  is  positive  if  &)  is  greater  than  <»lt  or  if  the  angular  speed  increases  with  the  time,  and 
negative  if  oj  is  less  than  o^,  or  if  the  angular  speed  decreases  with  the  time. 

Rate  of  change  of  speed,  whether  linear  or  angular,  is,  then,  like  speed  itself,  a  SCALAR 
quantity  having  magnitude  and  sign  but  independent  of  direction  (page  62). 

The  unit  of  rate  of  change  of  linear  speed  is  evidently  one  unit  of  linear  speed  per  unit 
of  time,  or  I  ft.-per-sec.  per  sec. 

The  unit  of  rate  of  change  of  angular  speed  is  one  unit  of  angular  speed  per  unit  of 
time,  or  i  radian-per-sec.  per  sec. 

Instantaneous  Rate  of  Change  of  Speed.  —  The  limiting  value  of  the  mean  rate  of 
change  of  speed  when  the  interval  of  time  is  indefinitely  small  is  the  INSTANTANEOUS  RATE 
OF  CHANGE  OF  SPEED. 

We  have,  then,  the  instantaneous  rate  of  change  of  linear  speed  given  by 

dv 


and  the  instantaneous  rate  of  change  of  angular  speed  given  by 

doa 
~dt' 


74  KINEMATICS   OF  A  POINT.    GENERAL  PRINCIPLES.  [CHAP.  VI. 

If  the  instantaneous  rate  of  change  of  speed  is  zero,  the  speed  itself  is  uniform  and 
therefore  the  same  as  the  mean  speed  for  any  interval  of  time,  which  is  also  uniform. 

If  the  instantaneous  rate  of  change  of  speed  does  not  vary  in  magnitude,  it  is  uniform 
and  therefore  the  same  as  the  mean  rate  of  change  for  any  interval  of  time,  which  is  also 
uniform. 

If  the  instantaneous  rate  of  change  ot  speed  varies  in  magnitude  with  the  time,  it  is 
variable.  In  such  case  the  mean  rate  of  change  is  also  variable. 

ExampleSi  —  (i)  Let  the  distance  described  by  a  mowing  point  be  given  by  the  equation 

s  =  jt  +  St*, 

where  s  is  the  number  of  feet  described  in  any  number  of  seconds  t. 

This  is  the  same   equation  as   in   example  (i),  page  64,  and  we  find,  as  before  in  that  example,  the 
instantaneous  linear  speed 

v  =  ~  =  7  +  i6/  ...............     (i) 

For  any  other  time,  /,  ,  we  can  write 

v,  =  7  +  i6/,  .................     (2) 

Suppose  /,  less  than  /.     Then  subtracting  (2)  from  (i),  we  have  for  the  change  of  speed  for  any  interval  of 
time  (/  —  /,) 

v  -vi  =  i6(/  -  /.)•   .................     (3) 

The  mean  rate  of  change  of  speed  is  then 

v  —  Vi 

-    =  1  6  ft.-per-sec.  per  sec. 
*  —  ii 

Since  this  does  not  change  with  the  time,  it  is  uniform  and  is  the  same  as  the  instantaneous  rate  of  change 

.    dv 
of  speed,  -jj. 

(2)  Let  the  instantaneous  linear  speed  of  a  point  be  given  by 

v  =  7  -f  ibt  +  6t*  ...............     (i) 

For  any  other  less  time,  /i  ,  we  have 

v,  =  7  -f  i6A  -|-  6/1. 

Subtracting  (2)  from  (i).  we  have  for  the  change  of  speed  for  any  interval  of  time  (/  —  A) 

v  —  vi-  i6(/  -  /,)  -|-  6(/»  -  /')  =  i6(/  -  /,)  -f  6(/  -f  /,)  (/  -  /,). 
The  mean  rate  of  change  of  speed  is  then 


This,  we  see,  is  variable  and  changes  with  the  time.  As  the  interval  of  time  (/  —  /,)  decreases, 
/i  approaches  equality  with  t  and  the  mean  rate  of  change  of  speed  given  by  (3)  approaches  the  limiting 
value  16  +  i2/.  We  have,  then,  for  the  instantaneous  rate  of  change  of  speed 

'  £-,6+,*. 

This,  we  see,  changes  with  the  time  and  is  therefore  variable. 
(3)  Let  the  angular  speed  of  a  point  be  g'iven  by 

o>  =  7  +  i6/  +  6A 
•where  oo  is  the  angular  speed  in  radians  per  sec.  for  any  instant  t. 


CHAP.  VI.]  MEAN  LINEAR  ACCELERATION  OF  A  POINT. 

Then  we  have,  just  as  before,  the  change  of  angular  speed  for  any  interval  of  time  (/  -  /,) 

<»  -  «i  =  i6(/  -  A)  +  6(/2  -  t\). 
The  mean  rate  of  change  of  angular  speed  is,  then, 

00    —     ft}, 

~-/7  =  l6  +  6(/  +  /,), 
and  the  instantaneous  rate  of  change  of  angular  speed  is 

da) 


Mean  Linear  Acceleration  of  a  Point.  —  Let  av  and  a  be  two  positions  of  a  point  moving 
in  any  path  and  passing  from  al  to  a  in  the  time  interval  t  —  /\.  Let  vl  and  v  be  the  corre^ 
spending  instantaneous  linear  velocities.  FlG  ,j  FlG  ,„ 

Each  velocity,  Fig.  (a),  is  tangent  to 
the  path  at  the  corresponding  point,  and 
is  equal  in  magnitude  to  the  speed  at  that 
point,  so  that  (v  —  vv  )  is  the  change  of  speed, 

v  —  v, 
and  —  -  —  is  the  mean  rate  of  change  of 

speed  (page  73). 

In  Fig.  (b)  take  any  point  B  as  a  pole  and  lay  off  Bb^  and  Bb,  the  line  representatives 
of  7,'j  and  v,  and  join  bl  and  b  by  a  straight  line. 

Then    the  straight   line  bvb    is  the   line   representative    of    the  change  of  velocity,   and 

J_      is  the  0*Az#  rate  of  change  of  velocity  for  the  interval  of  time  /  —  /r 
/  —  tl 

This  is  called  the  MEAN  LINEAR  ACCELERATION  of  the  point  for  the  interval  of  time. 
It   is  therefore   a  vector   quantity  given  in  magnitude  and  direction  by  a  straight  line 
parallel   to  b^b,  having  the  direction   from  bl  to  b  as  shown  by  the  arrow  in   Fig.  (£),   and 

b.b 

equal  in   magnitude  to  —  •  —  . 
/  —  tl 

Mean  linear  acceleration  can  be  defined,  then,  as  mean  time-rate  of  change  of  velocity, 
whether  that  change  takes  place  in  the  direction  of  the  velocity  or  not. 

The  unit  of  mean  linear  acceleration  is  evidently  one  unit  of  linear  speed  per  unit  of 
time,  or  I  ft.-per-sec.  per  sec.,  just  as  for  rate  of  change  of  linear  speed  (page  73). 

Instantaneous  Linear  Acceleration.  —  The  limiting  magnitude  and  direction  of  the  mean 
linear  acceleration  when  the  interval  of  time  is  indefinitely  small  is  the  INSTANTANEOUS 

LINEAR  ACCELERATION. 

Thus  if  the  interval  of  time  dt  is  indefinitely 
small,  the  two  positions  al  and  a  of  the  mov- 
ing point,  Fig.  (a),  are  consecutive. 


FlG-  («) 


FIG.  (6). 


In   Fig. 


,  then,—  Vr  is  the 

fit 


instantaneous 


linear  acceleration. 

It  is   therefore  a  vector  quantity  given    in  magnitude  and  direction  by  a  straight  line 
parallel  to  £/,  having  the  direction  from  b^  to   b  as  shown   by  the   arrow   in   Fig.  (b),  and 

b,b 

equal  in  magnitude  to    ~-r-  . 


76  KINEMATICS  OF  A  POINT.     GENERAL   PRINCIPLES.  [CHAP.  VI. 

Instantaneous  linear  acceleration  can  then  be  defined  as  limiting  time-rate  of  change  of 
velocity,  whether  that  change  (~y— )  take  place  in  the  direction  of  the  velocity  or  not. 

Its  unit  is  evidently  one  unit  of  speed  per  unit  of  time,  or  i  ft.-per-sec.  per  sec.,  just 
as  for  mean  linear  acceleration. 

We  shall  always  denote  instantaneous  linear  acceleration  by  the  letter  f. 

The  term  "acceleration"  always  signifies  instantaneous  linear  acceleration  unless 
otherwise  specified. 

We  shall  see  later  how  to  determine  it  in  magnitude  and  direction  (page  78) ;  it  is 
sufficient  to  note  here  that  if  the  path  is  a  straight  line,  bfi  coincides  with  v,  and  we  have 

/  =  -r .     That  is,  the  magnitude  of  the  acceleration  in  this  case  is  the  instantaneous  rate  of 

change  of  speed  (page  73). 

Resolution  and  Composition  of  Linear  Acceleration. — Since  acceleration  is  thus  time- 
rate  of  change  of  velocity  when  the  interval  of  time  is  indefinitely  small,  and  can  be 
represented  by  a  straight  line,  it  follows  that  all  the  principles  of  Chap.  II,  page  55,  for 
displacements  hold  good  also  for  accelerations.  We  can  then  combine  and  resolve  linear 
accelerations  just  like  displacements  (page  56),  and  we  have  the  triangle  and  polygon  of 
accelerations  just  as  for  displacements. 

We  have  also  relative  acceleration  with  similar  notation  as  for  displacement  (page  56). 

Let  Bb^  and  Bb  be  the  initial  and  final  velocities  z>,  and 

v  of  a  point  in  an   indefinitely  small  time  dt,   so  that  -—  is 

a     the  acceleration  /.      Draw  BA  and  ba  at  right  angles  to  any 
line  Aa  through  bl  in  any  given  direction. 

Then  -^-  is  the  component/,  in  the  direction  Aa  of  the 
acceleration  /.     But 

b,a      Aa  —  Ab. 


~dt 


and  Aa  and  Ab^  are  the  components  in  the  direction  Aa  of  the  velocities  v  and  », 

Hence  the  component  acceleration  in  any  direction  is  equal  to  the  time-rate  of  change  of 
velocity  in  that  direction. 

Examples.—  (i)  A  ball  let  fall  in  an  elevator  has  an  acceleration  downwards  relative  to  the  ground  of  32 
ft.-per-sec.  per  sec.  ,  while  the  elevator  has  an  acceleration  relative  to  the  ground  of  12  ft.-per-sec.  per  sec.  Find 
the  acceleration  of  the  ball  relative  to  the  elevator  when  the  acceleration  of  the  elevator  is  upwards  and  down- 
wards. 

ANS.  Solution  precisely  the  same  as  for  example  (4),  page  66.  When  acceleration  of  elevator  is 
upwards,  speed  of  the  elevator  increases  if  it  is  going  up,  decreases  if  it  is  going  down,  and  acceleration 
of  ball  relative  to  elevator  is  44  ft.-per-sec.  per  sec.  downwards. 

When  acceleration  of  elevator  is  downwards,  speed  of  the  elevator  increases  if  it  is  going  down, 
decreases  if  it  is  going  up,  and  acceleration  of  ball  relative  to  elevator  is  20  ft.-per-sec.  per  sec.  downwards. 

(2)  Two  trains  move  in  straight  lines  making  an  angle  of  bo".  The  one.  A,  is  increasing  its  speed  at  the 
rate  of  4  ft.-per-min.  per  min.  The  other,  B,  has  the  brakes  on  and  is  losing  speed  at  the  rate  of  8  ft.-per-min. 
per  min.  Find  the  relative  accelerations. 

ANS.  A  relative  to  B,  4^7  ft.-per-min.  per  min.  inclined  to  the  direction  of  A  at  an  angle  whose  sine 
is  V 


CHAP.  VI.] 


RECTANGULAR  COMPONENTS  OF  ACCELERATION. 


77 


(3)  The  initial  and  final  "velocities  of  a  moving  point  during  an  interval  of  two  hottrs  are  8  miles  per  hour 
E.  30°  A7.,  and  4  miles  per  hour  N.     Find  the  change  of  velocity  and  the  mean  acceleration. 

ANS.  41/3  miles  per  hour  W.  ;   2^/3  miles-per-hour  per  hour  W. 

(4)  A  point  has  the  simultaneous  accelerations,  bo  ft.-per-sec.  per  sec.  N.;  88  ft.-per-sec.  per  sec.  W.  30°  S.; 
60  ft.-per-sec.  per  sec.  E.  30°  S.     Find  the  magnitude  and  direction  of  the  resultant, 

ANS.  28  ft.-per-sec.  per  sec.  W.  30°  S. 

Rectangular  Components  of  Acceleration. — The  same  equations  and  conventions  hold 
for  the  components  of  an  acceleration  in  three  rectangular  directions  as  for  velocity  (page 
66).  We  have  therefore  only  to  replace  v  in  the  equations  and  figure  of  page  67  by /"with 
the  proper  subscripts. 

Analytic  Determination  of  Resultant  Acceleration. — So,  also,  the  same  equations  and 
conventions  hold  for  finding  the  resultant  acceleration  as  for  finding  the  resultant  velocity, 
page  67.  We  have  therefore  only  to  replace  v  in  equations  (i),  (2),  (3),  (4),  page  67,  by/ 
with  the  proper  subscripts. 

Examples. — Students  should  solve  examples  (/)  and  (2),  page  68,  for  accelerations  instead  of  velocities. 

Tangential  and  Central  Acceleration. — We  have  seen  (page  75)  that  if  vl  and  v  are  the 
velocities  at  al  and  a,  Fig.  (a),  and  if  the  interval  of  time  dt  is  indefinitely  small,  so  that  al 
and  a  are  consecutive  points  of  the  path,  then, 

in    Fig.    (£),    -~-    is    the    instantaneous    linear    a, 

acceleration /at  a. 

This  has  magnitude  and  direction,  can  be 
represented  by  a  straight  line,  and  combined     p 
and  resolved  just  like  displacement. 

We  can  therefore  resolve  the  acceleration 

bfi  cb 

f  =  -j-   at  a    into  a  component  -=-  along  the 


FIG.  (a). 


dt 


dt 


velocity  v,  and  -j-  at  right  angles  to  the  velocity  v.     Since  v  is  tangent  to  the  path  at  a,  we 


call  the  first  component,  -y-,  the  TANGENTIAL  ACCELERATION  and  denote  it  by  ft  ,  the  subscript 
/  denoting  that  the  acceleration  is  along  v  and  therefore  tangent  to  the  path.      The  second 


component,  -^-,  is  perpendicular  to  v  and  must  evidently  always  act  towards  the  centre  of 

curvature  C  of  the  path  along  the  radius  of  curvature  aC  —  p.  We  therefore  call  it  the 
CENTRAL  ACCELERATION  and  denote  it  by/p,  the  subscrpit  p  denoting  that  the  acceleration 
is  along  the  radius  of  curvature  p  and  therefore  towards  the  centre  of  curvature. 

Now,  since  the  points  al  and  a  are  consecutive,  we  have  cb  =  v  —  vl  =  dv,  and  hence 


cb      dv 

~dt =  ~di' 


dv 


But  -j-  is  the  instantaneous  rate  of  change  of  linear  speed  (page  73). 

Hence    the    magnitude   of  the    tangential  linear   acceleration  ft  at   any  instant  is  the 

dv 
instantaneous  rate  of  change  of  linear  speed  -j-  at  that  instant. 

We  see  also  from  Fig.  (b)  that,  since  vl  and  v  are  perpendicular  to  the  radius  of  curvature 


7 8  KINEMATICS  OF  A  POINT.    GENERAL  PRINCIPLES.  [CHA1-.  VI. 

Cav  and  Ca  in  Fig.  (a),  the  indefinitely  small  angle  a^Ca  =  dO  is  equal  to  b^Bb.      We  have, 
then,  when  £,  and  b  are  consecutive,  bvc  —  zy/#  —  vdti,  and  hence 

V  _       <*0 
•/'~  dt  ~V~dt' 

jn 

But  -i-  is  the  angular  velocity  <*>  for  the  centre  of  curvature  C,  and   if  p  is  the  radius  of 

curvature,  we  have  (page  71)  pw  —  v.     Therefore 

2, 

fp  =  VGO   =  pa?  =   — '.        .        .        .        .        .        (2) 

Hence  the  magnitude  of  the  central  acceleration  ff  at  any  instant  is  given  by  the  product  voo 
of  the  linear  velocity  v  and  the  angular  velocity  a?  for  the  centre  of  curvature  at  that  instant ; 
or  by  the  product  pco3  of  the  radius  of  curvature  p  and  the  square  of  the  angular  velocity  a?  for 

the  centre  of  curvature  at  that    instant ;    or  by  the  quotient  —  of  the  square  of  the  linear 

velocity  v3  divided  by  the  radius  of  curvature  p  at  that  instant. 

The  direction  of  the  central  acceleration  is  always  towards  the  centre  of  curvature. 
Since  the  acceleration /is  the  resultant  of/p  and/,,  we  have  for  the  magnitude  of/ 


/=  Vft+S!  ............     (3) 

and  for  its  direction  cosine  with  ff 


(4) 


Example.—  A  point  moves  in  a  vertical  circle  of  radius  r  =  21  ft.  in  an  east  and  -west  plane  with  a  velocity 
given  at  any  instant  by 

v  —  7  +  i6t  +  6P, 

where  t  is  the  interval  of  time  from  the  start.     At  the  end  of  t  =  2  sec.  the  point  is  at  the  top  of  the  circle 
moving  eastward.     Find  o»,  ft  and  fp  at  this  instant. 
ANS.  We  find,  just  as  in  example  (2),  page  74, 


or  for  /  =  2  sec.  ft  =  40  ft.-per-sec.  per  sec.  towards  the  east.     Also,  for  /  =  2  sec.  v  =  63  ft.  per  sec.  east, 

v*  f 

and  hence  at  this  instant  /p  =  —  =  189  ft.-per-sec.  per  sec.  towards  the  centre,  and  o>  =   —  =  3  radians  per 

sec.  north. 

The  acceleration  /at  the  instant  is,  then, 

/  =  ^ft  +  fl  =  63  i/«o"ft.-per-sec.  per  sec., 

making  with  the  radius  through  the  top  point  an  angle  whose  cosine  is  —  =  —  ^— 

/        Vio 

Uniform  and  Variable  Acceleration.  —  If  the  acceleration  /has  the  same  magnitude  and 
direction,  whatever  the  time,  it  is  UNIFORM.  If  either  magnitude  or  direction  change,  it  is 
VARIABLE. 

If  /  is  zero,  there  is  no  change  of  velocity  and  we  have  uniform  speed  in  a  straight 
line,  or  uniform  velocity. 

If/p  is  zero  but  /,  is  not,  we  have  variable  speed  in  a  straight  line.  The  velocity  is 
variable  (page  65). 


CHAP.  VI.] 


THE  HODOGR.APH. 


79 


lift  is  zero  but/p  is  not,  we  have  uniform  speed  in  some  curved  path.     The  velocity 
is  variable  (page  65).      If  at  the  same  time/p  is  constant  in  magnitude,  the  curved  path  is  a 

circle,     For/~p  =  — ,  and  if  fp  is  constant  and  v  is  constant,  p  must  also  be  constant. 

If  ft  and  ff  are   both  variable,  we  have   variable    speed  in  some  curved    path.     The 

velocity  is  variable  (page  65). 

Example. — Criticise  the  statement,  "  a  point  moves  in  a  circle  with  uniform  acceleration." 
ANS. — If  the  acceleration  is  uniform,  it  does  not  change  either  in  magnitude  or  direction.  But  if  a  point 
moves  in  a  circle,  we  must  have  a  radial  acceleration  fp  at  every  instant.  The  acceleration  is  therefore  not 
uniform.  If  the  magnitude  of  fp  is  constant,  then  we  have  uniform  speed  in  the  circle.  If  not,  we  have 
varying  speed  in  the  circle.  A  point  can  move  in  a  circle  with  constant  central  acceleration,  but  not  with 
uniform  acceleration. 

The  Hodograph.  —Let    a    point    moving  in  any  path  have    the  consecutive  positions 
al ,  #2,  <?3,  etc.,  and  let  the  corresponding  velocities  be  vlt  vv    v&,  etc.,    Fig.  (a).     These 


FlG.  (a). 


FIG. 


velocities  are  tangent  to  the 
path  at  0J,  a2,  as,  etc.,  and 
are  equal  in  magnitude  to  the 
speed  at  these  points. 

Now  from  any  point  B, 
Fig.  (b),  draw  the  line  repre- 
sentatives Bblt  Bb2,  Bb^  of 
v\*  V2  »  ^3>  e*c-  The  extremi- 
ties of  these  lines  will  form  a 
polygon  b^b^b^,  and  if  the 
points  al,  a2,  az  are  consecutive,  the  points  blt  #2,  bz  will  also  be  consecutive,  and  the 
polygon  ^/2£3  becomes  a  curve  also.  Thus  as  the  point  a  describes  the  path  a^fa,  Fig.  (a), 
we  can  conceive  a  point  b  to  describe  a  curve  b^b2bz ,  Fig.  (b). 

This  curve  is  called  the  HODOGRAPH.     The  point  B  is  the  POLE  of  the  hodograph ;   the 
points  blt  b2,  b3  of  the  hodograph  are  the  points  corresponding  to  alt  a2,  a3  of  the  path. 

When  the  point  a  moves  from  a^  to  a2  in  the  indefinitely  small  time  dt,  the  corre- 


sponding point  b  moves  in  the  same  time  from 


to  bz. 


Now  in  such  case 


dt 


velocity  in  the  path,  and  -~~  is,  then,  the  corresponding  velocity  in  the  hodograph. 


the 


But  we  have  seen  (page  75)  that    -^  is  the  acceleration  f  at  ar 

Hence  any  radius  vector  Bb  in  the  hodograph  is  the  line  representative  of  the  velocity  v 
at  the  corresponding  point  of  the  path,  and  the  velocity  at  any  point  of  the  hodograph  is 
the  acceleration  f  at  the  corresponding  point  of  the  path. 

Let  Ca  =  p  be  the  radius  of  curvature,  and  GO  the  angular  velocity  of  the  point  relative 
to  C,  so  that  the  radius  of  curvature  p  turns  about  C  with  the  angular  velocity  oa. 

Then,  since  v  is  perpendicular  to  p,  the  angular  velocity  of  Bb  in  Fig.  (b)  is  also  GO.  and 
hence  VGO  =  /p,  where  fp  is  the  acceleration  perpendicular  to  v  and  therefore  along  the  radius 
of  curvature  p  towards  the  centre  of  curvature  C. 

We  have  then 


=  vao, 


just  as  already  found  (page  78). 


8o 


KINEMATICS  OF  A   POINT.    GENERAL  PRINCIPLES. 


[CHAP.  VI. 


Examples. — (i)  A  point  moves  with  uniform  velocity.   What  is  the  hodograph? 
ANS.  A  point. 

(2)  A  point  moves  with  uniform  acceleration.      What  is  the  hodograph  ? 

ANS.  A  straight  line. 

(3)  A  point  moves  with  unfiorm  speed  v  in  any  plane  path.      What  is  the  hodograph  f 

ANS.  A  circle  of  radius  v. 

Axial  and  Radial  or  Deflecting  and  Deviating  Acceleration. — Let  a  point  P  have  the 
angular  velocity  oj  about  an  axis  CA  through  the  centre  of 
curvature  C,  so  that  PC  =  p  is  the  radius  of  curvature. 
Let  v  =  pco  be  the  velocity  of  P.  Then  ff  =  VGO. 

We  can  resolve  co  at  any  point  O  of  the  axis  CA  into 
a  component  car  about  an  axis  OP  through  the  point  P,  and 
a  component  a>a  about  an  axis  Oa  at  right  angles  to  OP. 
We  can  also  resolve  the  central  acceleration  fp  into  a  com- 
ponent fr  along  OP,  and  a  component  fa  parallel  to  Oa. 

The  component  oor  does  not  affect  the  velocity  of  P, 
since  it  passes  through  P.  Hence,  if  r  is  the  distance  OPt 
we  have 


v  =  rooa, 

and  the  plane  of  rotation  is  the  plane  POX.  The  component  cor  causes  this  plane  to  rotate 
about  the  axis  OP. 

We  see,  then,  that  if  a  point  rotates  with  angular  velocity  &>  about  an  axis  through  the 
centre  of  curvature,  we  can  resolve  the  motion  at  any  point  of  that  axis  into  rotation  /;/  a 
rotating  plane. 

If  the  radius  of  curvature  in  this  plane  is  r,  we  have  v  =  rcoa.  The  central  acceleration 
in  this  plane  is  then 

/r=T>«.  =  ro£ (i) 

We  call  this  the  radial  or  deflecting  acceleration  because  it  is  in  the  direction  of  the  radius 
and  causes  change  of  direction  of  i>  in  the  plane  of  rotation  POX. 

But  the  rotation  oar  about  OP  causes  the  plane  of  rotation  POX  to  change  direction. 
We  have,  then,  the  acceleration  at  right  angles  to  the  plane  of  rotation  POX 


fa  =  voor  = 


(2) 


We  call  this  the  axial  or  deviating  acceleration  because  it  is  in  the  direction  of  the  axis  of 
rotation  Oa  and  causes  deviation  of  the  plane  of  rotation  POX.  The  radial  or  deflecting 
acceleration  fr  then  causes  change  of  direction  of  v  in  the  plane  of  rotation,  that  is  P  moves 
in  a  curve  in  this  plane.  The  axial  or  deviating  acceleration  /,  causes  change  of  this  plane 
of  rotation. 

If  fr  is  zero,  the  path  is  a  straight  line  in  a  rotating  plane. 

If/,  is  zero,  the  path  is  a  curve  in  a  constant  plane. 

If  /,  and  /.  are  zero,  the  path  is  a  straight  line.  In  all  cases,  since /,  and  /.  are  com- 
ponents of  the  central  acceleration/,,  we  have 


(3) 


CHAP.  VI.] 


DEFLECTING  AND  DEVIATING  ACCELERATION— EXAMPLES. 


81 


Example.—  A  point  moves  in  a  vertical  circle  of  radius  r  =  21  ft.  in  an  east  and  west  plane  with  a  velocity 
given  at  any  instant  by 


where  t  is  the  interval  of  time  from  the  start.  At  the  end  of  t  -  2  sec.  the  point  is  at  the  top  of  the  circle 
moving  east.  At  the  same  instant  the  plane  of  the  circle  rotates  about  the  vertical  diameter  with  an  angular 
velocity  of  2  radians  per  sec.  upwards.  Find  oo,ft,  p,  andfp  andf. 

ANS.  We  have,  just  as  in  example,  page  78,  for  rotation  in  the  plane  the  tangential  acceleration 


or  for  /  =  2  sec.  ft  =  40  ft.-per-sec.  per  sec.  towards  the  east.     Also  for  /  =  2  sec.  v  =  63  ft.  per  sec.  east,  and 
hence  at  this  instant  the  deflecting  or  radial  acceleration  is  fr  =  —  =  189  ft.-per-sec.   per  sec.  towards  the 

centre  and  &>a  =  -£-  =  3  radians  per  sec.  north. 

We  have  also  due  to  rotation  of  the  plane,  since  oar  =  2  radians  per  sec.,  the  axial  or  deviating  acceleration 
fa  =  voor  =  126  ft.  per  sec.  north. 
Hence 

oo  =  \/ GO,? -\-  a»r2  =   4/J3  radians  per  sec., 

making  an  angle  with  r  in  the  plane  of  r  and  K>a  whose  cosine  is 
<or  2 

The  radius  of  curvature  p  is  given   by  poo  —  v,  or  p  =  —  =  —-=  **•' 
The  central  acceleration  is 

fp  =  \/f<?  +fr*  =  227.14  ft.-per-sec.  per  sec. 

The  total  acceleration  is 

/  =  \/ff  +/•  =  230.64  ft.-per-sec.  per  sec. 
The  direction  cosines  with_/i,/a  and/r  are 

cosa+4  =  -l-  ±>        cos/S  =  4=+  — -       cosr=-^  =  -  ^ 

f  230 


CHAPTER  VII. 

ANGULAR    ACCELERATION. 

Mean  Angular  Acceleration  of  a  Point. — Let  ^  and  &  be  the  initial  and  final  angular 

5  velocities  of  a  point  for  any  interval  of  time  t  —  tr 

v     ,       >v  Take  any  point  O  as  a  pole  and   lay  off  Ob^  and   Ob,  the  line 

representatives  of  GJ,  and  GO,  and  join  b\b  by  a  straight  line. 

Then  the  straight  line  bj>  is  the  line  representative  of  the  change 

of  angular  velocity,  and  — - —  is  the  mean  rate  of  change  of  angular  velocity  for  the  interval 

*  ~~  *i 
of  time  /  — /,. 

This  is  called  the  MEAN  ANGULAR  ACCELERATION  of  the  point  for  the  interval  of  time. 

It  is  therefore  a  vector  quantity  given  in  magnitude  and  direction  by  a  straight  line  parallel 
to  b^b,  having  the  direction  from  b^  to  b  as  shown  by  the  arrow  in  the  figure,  and  equal  in 

bJ> 

magnitude  to  — - — . 

Mean  angul-ir  acceleration  can  be  defined,  then,  as  mean  time-rate  of  change  of  angular 
velocity,  whether  that  change  takes  place  in  the  direction  of  the  angular  velocity  or  not. 

The  unit  of  mean  angular  acceleration  is  evidently  one  unit  of  angular  speed  per  unit 
of  time,  or  I  radian-per-sec.  per  sec.,  just  as  for  rate  of  change  of  angular  speed  (page  73). 

Instantaneous  Angular  Acceleration. — The  limiting  magnitude  and  direction  of  the 
mean  angular  acceleration  when  the  interval  of  time  is  indefinitely  small  is  the  INSTAN- 
TANEOUS angular  acceleration. 

Thus  if  the  interval  of  time  is  dt  or  indefinitely  small, 

then  •  V    is  the  instantaneous  angular  acceleration. 

o- 
It  is  therefore  a  vector  quantity  given  in  magnitude  and 

direction  by  a  straight  line  parallel  to  bj),  having  the  direction  from  bl  to  b  as  shown  by  the 

bb 
figure  and  equal  in  magnitude  to -4-- 

Instantaneous  angular  acceleration  can  then  be  defined  as  limiting  time-rate  of  change  of 
angular  velocity,  whether  that  change  take  place  in  the  direction  of  the  velocity  or  not. 

Its  unit  is  evidently  one  unit  of  angular  speed  per  unit  of  time,  or  i  radian-per-sec. 
per  sec.,  just  as  for  mean  angular  acceleration. 

We  shall  always  denote  instantaneous  angular  acceleration  by  the  letter  a. 

The  term  "angular  acceleration"  always  signifies  instantaneous  angular  acceleration 
unless  otherwise  specified. 

We  shall  see  later  how  to  determine  it  in  magnitude  and  direction  (page  84);  it  is 
sufficient  to  note  here  that  if  the  point  moves  in  an  unchanging  plane,  bj>  coincides  with 

•  82 


CHAP.  VII.]  RESOLUTION  AND  COMPOSITION  OF  ANGULAR  ACCELERATION.  83 

a?,  and  we  have  OL  —  -y-.     That  is,  the  magnitude  of  the  angular  acceleration  in  this  case 

is  the  instantaneous  rate  of  change  of  angular  speed  (page  73). 

Resolution  and  Composition  of  Angular  Acceleration. — Since  angular  acceleration  is 
time-rate  of  change  of  angular  velocity  when  the  interval  of  time  is  indefinitely  small,  it  has 
magnitude  and  direction  and  can  be  represented  by  a  straight  line,  just  like  angular  velocity 
(page  64).  Therefore  all  the  principles  of  Chapter  II,  page  54,  hold  good  also  for  angular 
accelerations  whether  simultaneous  or  successive,  and  we  have  the  "triangle  and  polygon" 
of  angular  accelerations,  and  can  combine  and  resolve  them  just  like  linear  displacement. 

We  have  also  relative  angular  acceleration  with  similar  notation  as  for  linear  displace- 
ments (page  55). 

Example, — (r)  A  sphere  has  angular  acceleration  about  a  diameter  at  the  rate  of  10  radians-per-sec.  per 
sec.  Find  tJie  component  angular  acceleration  about  another  diameter  inclined  30°  to  the  first. 

ANS. — 51/3  radians-per-sec.  per  sec. 

(2)  A  body  has  two  sitmdtaneous  or  successive  angular  accelerations  of  2  radians-per-sec.  per  sec.  and  4 
radians-per-sec.  per  sec.  about  axes  inclined  60°  to  each  other.  Find  the  resultant. 

ANS.  2  Vj  radians-per-sec.  per  sec.  about  an  axis  inclined  to  the  greater  component  at  an  angle  whose 


(3)  A  sphere  with  one  of  its  superficial  points  fixed  has  two  angular  accelerations,  either  simultaneous  or 
succe-ssive,  one  of  8  radians-per-sec.  per  sec.  about  a  tangent  line,  and  one  of  15  radians-per-sec.  per  sec.  about  a 
diameter.  Find  the  resultant. 

ANS.  17  radians-per-sec.  per  sec.  about  an  axis  inclined  to  the  greater  component  at  an  angle  whose 
tangent  is  T8r. 

Rectangular  Components  of  Angular  Acceleration. — The  same  equations  and  conven- 
tions hold  for  the  components  of  an  angular  acceleration  in  three  rectangular  directions  as 
for  angular  velocity,  page  69.  We  have  therefore  only  to  replace  &>  in  the  equations  and 
figure  of  page  69  by  a  with  the  proper  subscripts. 

Analytic  Determination  of  Resultant  Angular  Acceleration. — So,  also,  the  same  equa- 
tions and  conventions  hold  for  finding  the  resultant  angular  acceleration  as  for  finding  the 
resultant  angular  velocity,  page  69.  We  have  therefore  only  to  replace  oa  in  equations  (i), 
(2),  (3)  and  (4),  page  69,  by  a  with  the  proper  subscripts. 

Examples.—  Students  should  solve  examples  (/)  and  (2),  page  70,  for  angular  accelerations  instead  of 
velocities. 

Axial  and  Normal  Angular  Acceleration. — We  have  seen  that  if  ool  and  oaa  are  the 
initial  and  final  angular  velocities  of  a  point  during  an  indefinitely  small  interval  of  time  dt, 

then  ~-  is  the  instantaneous  angular  acceleration  a. 

bb 
We   can   resolve   this   angular  acceleration  a  =  — 

into   a   component  -j-  along  <*>a,  and  -~   normal   to    coa.     O 

Since  &?a  coincides  with  the  axis  of  rotat!on  of  the  radius 

vector,  we  call  the  first  component  —  the  AXIAL  angular 

acceleration  and  denote  it  by  aa,  the  subscript  a  denoting  that  the  acceleration  is  along  the 

axis.     The  second  component,   ~-  normal  to  &7a,  we  denote  by  <*„,  and  call  it  the  NORMAL 
angular  acceleration. 


KINEMATICS  OF  A  POINT.    GENERAL  PRINCIPLES. 


[CHAP.  Vll. 


Now  let  <»r  be  the  angular  velocity  of  the  line  representative   Ob  =  o?fl  about  an  axis 
perpendicular  to  the  plane  of  bflb. 

Then,  just  as  on  page  80,  we  have 


CO 
(2) 


But  -—•  is  the  instantaneous  rate  of  change  of  angular  speed. 

Hence  the  magnitude  of  the  axial  angular  acceleration  aa  is  the  instantaneous  rate  of 
c/tange  of  angular  speed. 

The  magnitude  of  the  normal  angular  acceleration  an  is  the  product  coawr  of  the  angular 
velocity  va  about  the  axis  and  the  angular  velocity  oor  of  the  axis  itself. 

Since  the  acceleration  a  is  the  resultant  of  aa  and  an,  we  have  for  the  magnitude  of  a 


a  = 
and  for  its  direction  cosine  with  the  axis 


(3) 


cos  a  =  ^ 
a 


(4) 


Uniform  and  Variable  Angular  Acceleration. — If  the  acceleration  a  has  the  same 
magnitude  and  direction  whatever  the  time,  it  is  UNIFORM.  If  either  magnitude  or 
direction  change,  it  is  VARIABLE. 

If  a  is  zero,  there  is  no  change  of  angular  velocity,  and  we  have  uniform  angular  speed 
in  an  unchanging  plane. 

If  «„  is  zero  but  ota  is  not,  we  have  variable  angular  speed  in  an  unchanging  plane. 

If  aa  is  zero  but  an  is  not,  we  have  uniform  angular  speed  in  a  changing  plane. 

If  aa  and  an  are  both  variable,  we  have  variable  angular  speed  in  a  changing  plane. 

Linear  in  Terms  of  Angular  Acceleration. — -Let  O  be  any  pole,  and  Oa  —  r  the  radius 

vector  to  any  point  a  moving  in  any  path. 
We  have  proved,  on  page  71,  that 

rco  =  sin  e, 

where  a?  is  the  angular  velocity  and  v  is  the  linear 
velocity  making  the  angle  e  with  Oa.  We  can 
therefore  write 

rdco  =  dv  sin  e, 


and  dividing  by  dt, 


dao      dv  . 
—r-  =  —  sin  e. 
dt       dt 


But        is  the  magnitude  of  the  axial  angular  acceleration  aa,  and^  is  the  magnitude  of  the 

tangential  acceleration  ft. 
Hence  we  have 

raa  =//sinc=/l,     ..........     (i) 

where  /«  is  the  component  of  f  normal  to  r. 


CHAP.  VII.] 


LINEAR  IN   TERMS  OF  ANGULAR  ACCELERATION. 


Hence  the  product  of  the  radius  vector  r  by  the  axial  angular  acceleration  a a  gives  the 
linear  acceleration  fn  normal  to  the  radius  vector  in  the  plane  of  ft  (or  v)  and  r. 

If  the  pole  O  is  taken  at  the  centre  of  curvature  C  so  that  Ba  is  the  radius  of  curvature 
p,  then  e  =  90°,  and  we  have 

Pa.=ft (2) 

If  the  pole  O  is  taken  anywhere  in  the  plane  through  p  perpendicular  to/,  we  still  have 
e  —  90°.  and  in  this  case 

'*.=/«.  .  V (3) 

Hence,  in  this  case,  the  product  of  the  radius  vector  by  the  axial  angular  acceleration  gives 
the  tangential  acceleration. 

If,  then,  a  point  P  has  the  angular  acceleration 
aa  about  an  axis  AO,  so  that  the  radius  vector  OP 
=  r  is  in  a  plane  perpendicular  to  /  through  the 
radius  of  curvature,  we  have 


If  at  the  same  time  the  plane  of  rotation  has  the 
angular  acceleration  ocn  ,  we  have 


(4) 


But,  as  we  have  seen,  page  84,  if  aoa  is  the  angular  velocity  about  the  axis  and  <*>r  is  the 
angular  velocity  of  the  axis  about  OP,  we  have 

an    —    ^a^r' 

Hence  we  have 

/.  =  r«v»r, 
just  as  already  found  on  page  80. 

We  have  already  found  (page  80)  the  radial  acceleration 

/»•=  real. 

The  acceleration  /  is,   then,    the  resultant  of  fr,fa  and/,,   or  of/p  and  ft,  since  /p  is 
the  resultant  of  fr  and/,. 

Example.—^  point  moves  in  a  vertical  circle  of  radius  r  —  2  ft.  in  an  east  and  west  plane  with  an  angular 
velocity  given  at  any  instant  by 

&)a  =  2t  +  4?. 

At  the  end  of  t  =  %  sec.  the  point  is  at  the  top  of  the  circle  moving  east,  and  at  the  same  instant  the  plane  of  the 
circle  is  rotating  about  the  radius  through  the  point  with  the  angtilar  velocity  oor  =  j  radians  per  sec. 

Find  aa,  a«,  a,  v,  fr,  fa.,  fp,  ft  and  f. 

ANS.   We  find,  just  as  in  example  (2),  page  74,  the  axial  angular  acceleration 


For  /  =  J  sec.  we  have,  then,  o.t  =  6  radians-per-sec.  per  sec. 
north. 

The  angular  velocity  for  /  =  \  sec.  is  ooa  =  2  radians  per  sec. 
north' 

The  normal  angular  acceleration  is 

a,,  =  ooaaar  =  6  radians-per-sec.  per  sec.  west 
if  oor  has  the  direction  given  in  the  figure. 


86  KINEMATICS  OF  A  POINT.    GENERAL   PRINCIPLES.  LC"A1'-  V11' 

The  angular  acceleration  a  is  then  given  by 

a  =  v'aj  +  a*  =  6  vTradians-per-sec.  per  sec., 

and  its  direction  cosine  with  the  axis  Is 

cos  ft  =  —  =  — =. 

a        A/2 

The  linear  velocity  is  v  =  ra>a  =  4  ft.  per  sec.  east. 
The  tangential  acceleration  is/<=  r**=  12  ft.-per-sec.  per  sec.  east. 

The  radial  acceleration  is/r  =  -  =  roal  -  voaa  =  8  ft.-per-sec.  per  sec.  towards  the  centre  C. 
The  axial  acceleration  is/,  =  rooAoar  =  12  ft.-per-sec.  per  sec.  north. 

The  central  acceleration  is/,  =  Vfr  +/I  =  4  V~i3  ft.-per-sec.  per  sec.,  and  its  direction  cosine  with  the 
radius  is 

/»       4/13 
The  resultant  acceleration /is  given  by 


/=  t//7  +/r  +/-  =  1/22  ft.-per-sec.  per  sec., 
and  its  direction  cosines  with/,,/,  and/-  are 

4=    ;L,     cos0=£  =  -/L,    cosr=4  =  -^. 

/        4/22  /         i/22  /        i/22 

Homogeneous  Equations. — We  have  already  called  attention  (page  4)  to  the  fact  that 
the  units  in  all  numeric  equations  are  always  understood,  and  when  these  units  are  inserted 
the  equation  must  be  HOMOGENEOUS,  th=it  is,  every  term  must  express  a  quantity  of  the 
same  kind.  When  this  is  not  the  case  the  equation  is  impossible,  and  some  error  must  have 
been  made  in  its  derivation. 

Thus  suppose  that  the  result  of  some  investigation  is  expressed  by 

s  -j-  at  =  bv, 

where  a  and  b  are  numbers  only,  s  is  a  number  of  feet,  t  a  number  of  seconds,  and  v  a 
number  of  feet  per  second.  Without  reference  to  the  various  steps  of  the  investigation  by 
which  this  result  has  been  reached,  we  can  say  at  once,  from  inspection,  that  it  is  incorrect 
and  some  error  has  been  committed.  For  we  cannot  have  a  number  of  feet  plus  a  number 
of  seconds  equal  a  number  of  feet  per  second. 

If,  however,  a  stands  for  a  number  of  feet  per  second,  and  b  stands  for  a  number  of 
seconds,  the  equation  is  homogeneous  and  possible.  For  if  we  insert  the  units,  we  now 
have 

s  (fU  +  a  (J^-  )  X  /  (sec.)  =  b  (sec.)  v  (^™). 
or 

s  (ft.)  +<t/(ft.)=  fc»(ft.). 

Here  all  the  terms  are  quantities  of  the  same  kind  and  the  equation  is  homogeneous.  The 
relations  expressed  by  it  are  possible.  It  does  not  follow  that  the  equation  is  necessarily 
correct.  It  may  still  have  been  incorrectly  derived.  But  it  is  not  impossible  and  we 
cannot  reject  it  upon  simple  inspection. 

The  student  should  form  the  habit  of  thus  testing  all  equations.     It  will  often  prevent 


CHAP.  VII.]  HOMOGENEOUS  EQUATIONS—EXAMPLES.  87 

him  from  making  errors  in  a  long  investigation,  and  save  him  the  trouble  of  going  over  every 
step  in  order  to  locate  some  error  which  may  have  been  otherwise  committed. 
Examples. — (i)  In  the  equation 

B  =  ?t  +  $f  +  2t\ 

what  must  be  the  units  of  J,  8  and  2  f 

radians   0  rad.       rad. 
ANS.  7 ,  8 5,  2  — -j,  or  7  radians  per  sec.,  8  rad.  per  sec.2,  2  rad.  per  sec.3 

(2)  ///  the  example,  page  71,  we  have  given  the  equation 

What  must  be  the  units  7,  8  and  2  f 

ANS.  7  — L,  8  — '-j,  2  — 4  or  7  ft.  per  sec.,  8  ft.  per  sec.4,  2  ft.  per  sec.3, 
sec.      sec.        sec. 

The  student  should  go  over  the  solution  of  this  example  on  page  71  and  test  thus  every  equation  in  it. 


CHAPTER   VIII. 


MOMENTS.      MOMENT  OF  DISPLACEMENT,  VELOCITY  AND  ACCELERATION. 


Moment  of  a  Vector  Quantity. — The  product  of  a  vector  quantity  by  the  length  of  the 
perpendicular  let  fall  from  any  point  upon  the  direction  of  the  line  representative  of  the 
vector  is  called  the  MOMENT  of  the  vector  relative  to  that  point. 

The  perpendicular  is  called  the  LEVER-ARM,  and  the  point  is  the  CENTRE  OF  MOMENTS. 
Thus  let  AB  be  the  line  representative  of  any  vector  whose  magnitude  is  m.     Take  any 
w  B    point  O  as  a  centre  of  moments  and  draw  the  lever-arm,  or 

perpendicular    Oa  upon   the  direction   of  AB    produced  if 
necessary.      Let   the  length  of  this   perpendicular  or  lever- 

,  M  arm  be  /.      Then  the  product  ml  gives  the  magnitude  of 

the  moment  of  the  vector  AB  relative  to  O. 

This  moment  is  itself  a  vector  quantity,  that  is  it  has 
both  magnitude  and  direction  and  can   therefore  be  itself 
represented  by  a  straight  line. 

Thus  the  line  representative  of  the  moment  is  a  straight  line  <9J/at  right  angles  to  the 
plane  of  BAO,  whose  magnitude  is  ;///,  and  so  directed  by  its  arrow  that  when  we  look 
along  it  in  the  direction  of  its  arrow,  the  rotation  indicated  by  the  direction  of  AB  relative 
to  O  is  seen  clockwise. 

Resolution  and  Composition  of  Moments. — We  can  then  combine  and  resolve  moments 
and  we  have  component  and  resultant  moments  and 
the  triangle  and  polygon  of  moments,  just  as  for  dis- 
placements (page  55).  Also,  the  component  of  the 
resultant  in  any  direction  is  equal  to  the  agebraic  sum 
of  the  components  in  that  direction. 

Moment  about  an  Axis. — Let  OZ  be  a  given 
axis  and  AB  a  vector.  Take  any  plane  XY  per- 
pendicular to  the  axis  OZ  intersecting  this  axis 
at  O.  Project  AB  upon  this  plane  in  A'B'.  Now 
AB  can  be  resolved  at  A  into  a  component  paral- 
lel to  A' B'  and  a  component  parallel  to  the  axis  OZ. 

This  latter  component   gives  no   rotation   about       X 
OZ.     The  moment  of  AB  about  the  axis  OZ  is,  then,  the  moment  of  A' B'  about  O. 

Hence  the  moment  of  a  vector  relative  to  any  axis  is  tlic  same  as  the  moment  of  the  com- 
ponent in  a  plant  perpendicular  to  the  axis,  relative  to  the  point  of  intersection  of  that  axis  and 
plane. 


1 


CHAP.  VIII.] 


MOMENT  OF  RESULTANT. 


89 


Completing   the  paral- 


Moment  of  Resultant. — Let  AB  and  AC  be  two  components, 
lelogram,  we  have  the  resultant  AR. 

Choose  any  point  O  in  the  plane  of  BAC.      Then  the  moment  of  AB  relative  to   O  is 
the  length  of  AB  multiplied  by  the  perpendicular  -from 
O  on  AB,  or  twice  the  area  of  the  triangle  ABO. 

In  the  same  way,  the  moment  of  AC  relative  to  O 
is  twice  the  area  of  the  triangle  ACO,  and  the  moment 
of  the  resultant  AR  relative  to  O  is  twice  the  area  of 
the  triangle  ARO. 

Draw  a  line  through  O  and  A,  and  drop  the  per- 
pendiculars Bhlt  Ch.2,  Rhy  Also  draw  Bh  parallel  to 
OA.  Then  C/i2  =  Rh,  and  Ch2  +  Bh^  =  Rhy 

The  area  of  the  triangle  ACO  =  AO  X  $C/i2,  the 
area  of  the  triangle  ABO  =  AO  X  \BJiv,  the  area  of  the 
triangle  ARO  =  AO  X  $R/iy 

We  have,  then,  the  area  of  ARO  equal  to  the  sum  of  the  areas  of  ACO  and  ABO. 

If  we  had  a  third  component,  the  same  would  hold  for  the  resultant  of  this  and  R,  and 
so  on. 

If  we  had  taken  O  inside  the  angle  CAB,  the  moments  of  AC  and  AB  would  have  been 
opposite  in  direction  and  we  should  have  found  the  area  of  ARO  equal  to  the  difference  of 
the  areas  ACO  and  ABO. 

Hence  for  any  number  of  components  the  moment  of  the  resultant  is  equal  to  the  alge- 
braic sum  of  the  moments  of  the  components. 

We  have  dealt  thus  far  with  linear  and  angular  displacements,  velocities  and  accelera- 
tions, and  have  seen  that  they  are  all  vector  quantities.  Each  and  all  of  them,  then,  can 
have  a  moment  relative  to  a  point  or  axis,  and  the  preceding  principles  apply  to  all  alike. 

Moment  of  Displacement. — Let   AB   be   a   linear  displacement  D.      Then  the  moment 
Dl  is  twice  the  area  of  the  triangle  AOB. 

Hence  the  moment  of  a  linear  displacement  relative  to  any 
point  O  is  twice  the  area  swept  through  by  the  radius  vector  OA 
in  moving  from  OA  to,  OB  in  the  plane  AOB. 

The  unit  of  moment  of  linear  displacement  is  therefore  one 
square  unit  of  length,  or  one  square  foot. 

The  line  representative  is  OM  at  right  angles  to  the  plane  of  AOB,  so  that  looking  in 
its  direction,  as  indicated  by  the  arrow,  the  rotation  of  the  radius  vector  is  seen  clockwise. 

Moment  of  Velocity. — It  AB,  in  the  preceding  figure,  represents  a  linear  velocity  v, 
then  the  moment  vl  relative  to  O  is  twice  the  areal  velocity  of  the  radius  vector  OA. 

The  unit  of  moment  of  a  linear  velocity  is  then  the  square 
of  the  unit  of  length  per  unit  of  time,  or  one  square  foot  per 
second. 

The  line  representative  OM  is  as  before.  If  r  is  the  radius 
vector  and  a?  the  angular  velocity  relative  to  O,  we  have  (page 
71)  rco  =  v  sin  e  —  vn  ,  or  vn  =  r&),  and  the  lever-arm  /  = 
r  sin  e.  Therefore  the  moment 

vl  =  r*oo  =  vnr. (0 

Moment  of  Acceleration. — If  AB,  represents  a  linear  acceleration/,  then  the  moment 
fl  relative  to  O  is  twice  the  areal  acceleration  of  the  radius  vector  OA . 


9 


KIN EM A TICS  OF  A  POINT.     GENERAL   PRINCIPLES. 


[CHAP.  VIII. 


The  unit  of  moment  of  a  linear  acceleration  is  then  the  square  of  the  unit  of  length  per 
unit  of  time  squared,  or  one  square  ft.-per-sec.  per  sec. 

The  line  representative  OM  is  as  before.  ,  If  r  is  the  radius 
vector  and  ora  the  axial  angular  acceleration,  we  have  (page  84) 
r«,  =  /,  sin  e  =  /„.  The  lever  arm  /=  r  sin  e.  Therefore 
the  moment  //  of  the  tangential  linear  acceleration  is 


(2) 


V«rfne-V, 


Moment  of  Angular  Velocity. — If  OA  represents  an  angular 
velocity  <«>,  then,  as  we  have  seen  (page  71),  the  moment  roo 
relative  to  any  point  a  gives  the  linear  velocity  VH  =  v  sin  e  of 
that  point  at  right  angles  to  the  plane  of  the  axis  OA  and  the 
radius  vector  Oa,  in  such  a  direction  that,  looking  along  OA  in  its 
direction,  the  rotation  of  the  radius  vector  is  seen  clockwise. 

The  moment  of  an  angular  velocity  is,  then,  a  linear  velocity, 
and  its  unit  is  one  foot  per  second.  O 

Moment  of  Angular  Acceleration. — If  OA  represents  an  angular  acceleration,  a  then,  as 
we  have  seen,  page  84,  the  moment  roc  relative  to  any  point  a  gives 
the  linear  acceleration  fn  of  that  point  at  right  angles  to  the  plane  of 
the  axis  OA  and  the  radius  vector  Oa,  in  such  a  direction  that,  look- 
ing along  OA  in  its  direction,  the  rotation  of  the  radius  vector  is  seen 
clockwise. 

The  moment  of  an  angular  acceleration  is,  then,  a  linear  accelera- 
tion, and  its  unit  is  one  foot-per-sec.  per  sec. 

Example.—  The  distance  in  feet  s  described  by  a  point  in  any  time  t  sec.  is  given 
by  s  =  a  +  b?. 

The  path  is  a  circle  of  radius  r  feet  in  a  vertical  east  and  west  plane.  The  point  when  t  =  o  is  at  (he  top 
«nd  it  morses  (wards  the  east. 

(.1)  State  the  units  of  a  and  b.  (b)  Find  the  linear  •velocity,  the  tangential,  radial  and  resultant  accelera- 
tion at  any  instant,  (c)  The  angular  velocity  and  acceleration  at  any  instant.  (d)  The  area/  velocity  and 
acceleration  of  the  radius  at  any  instant. 

At  the  end  of  j  sees,  find  (e)  the  distance  described,  the  mean  speed,  the  angular  and  linear  displacement. 

ANS.  (a)  The  unit  of  a  must  be  i  ft.,  of  b  \  ft.-per-sec.  per  sec. 

(b)  v  =  -r  =  2bt  ft.  per  sec.  tangent  to  the  path  at  the  instant ; 

/rc=~  =  2b  ft.-per-sec.  per  sec.  tangent  to  the  path  at  the  instant ; 

/p  =  —  =  -^-  ft.-per-sec.  per  sec.  towards  the  centre; 

V* 

ft.-per-sec.  per  sec.,  making  an  angle  with  the  radius  whose  tangent 


2bt 


(c)  «  =  p  =  ^f  radians  per  sec.  north;  aa  =  —  =  —  radians-per-sec.  per  sec.  north. 

(d)  —  =  brt  sq.  ft.  per  sec.  north;  ^ft-=br  sq. -ft.-per-sec.  per  sec.  north. 

(c)  When  /  =  o,  j,  =  a.    When  /  =  3,  s  =  a  +  9*.     Hence  the  distance  described  is 

s  —  si  =  gb  ft. 
The  mean  speed  =  —  =  3^  ft.  per  sec. 

Angular  displacement  8  —  0,  —  -  =  —  radians.      Linear  displacement  =  2r  sin 


in  a  direction  from  the  initial  to  the  final  position. 


KINEMATICS     OF    A     POINT.        APPLICATION     OF 

PRINCIPLES. 


CHAPTER    I. 

MOTION   OF   A    POINT.      CONSTANT    AND    VARIABLE    RATE    OF    CHANGE    OF   SPEED. 

Rate  of  Change  of  Speed  Zero. — As  we  have  seen,  page  79.  if  the  tangential  acceleration 
/(  is  zero,  the  resultant  acceleration  /  is  the  same  as  the  central  acceleration  fp  and  therefore 
always  at  right  angles  to  the  velocity  v.  In  such  case  we  have  in  general  uniform  speed 

7,2 

in  some  curve,  and  f  =  fp  =  —  (page  78).      If  the  central  acceleration  ff  =  f  is  constant   in 

magnitude,  then,  since  v  is  also  constant  in  magnitude,  we  must  have  p  constant  in  magnitude 
and  the  path  is  a  circle.  If  /p  is  zero,  then  p  is  infinity  or  the  path  is  a  straight  line. 

Whatever  the  path  may  be,  let  sl  be  the  initial  distance  of  the  moving  point  measured 
along  the  path  from  any  given  fixed  point  of  the  path  taken  as  origin,  and  s  the  final  distance 
of  the  moving  point  from  that  origin,  measured  in  the  same  way,  at  the  end  of  any  interval 
of  time  t,  so  that  the  distance  described  in  this  interval  of  time  /  is  s  —  sr 

Now  if  the  tangential  acceleration  f,  is  zero,  the  speed  in  the  path  does  not  change 
and  hence  the  instantaneous  speed  at  any  instant  is  equal  to  the  mean  speed  for  any  interval 
of  time.  We  have,  then, 


This  equation  is  general  whatever  the  path,  provided  ft  is  zero.  If  s  is  greater  than  slt  v  is 
positive.  If  s  is  less  than  sl  ,  v  is  negative.  Hence  a  positive  (-{-)  value  for  v  denotes  that 
the  distance  from  the  origin  increases  as  the  interval  of  time  increases,  a  negative  (  —  )  value 
for  v  denotes  that  the  distance  from  the  origin  decreases  as  the  interval  of  time  increases, 
without  regard  to  direction  of  motion. 

Rate  of  Change  of  Speed  Constant.  —  If  the  tangential  acceleration  ft  is  constant  in 
magnitude,  we  have  in  general  motion  in  a  curved  path  with  uniform  rate  of  change  of  speed. 
If  in  this  case  the  central  acceleration  fp  =  o,  the  path  is  a  straight  line. 

Whatever  the  path  may  be,  let  v^  and  v  be  the  initial  and  final  speeds  for  any  interval  of 
time  /.  Then,  since  for  constant/,  the  instantaneous  rate  of  change  of  speed  at  any  instant 
is  equal  to  the  mean  rate  of  change  of  speed  for  any  interval  of  time,  we  have 


91 


92  KINEMATICS  OF  A  POINT.     APPLICATIONS.  [CHAP.  1. 

The  value  of/)  is  positive  (-}-)  when  v  is  greater  than  ?',,  that  is  when  the  speed   increases 
as  the  interval  of  time  increases,  and  negative  (  —  )  when  the  speed  decreases  as  the  interval 
of  time  increases,  without  regard  to  direction  of  motion. 
From  (2)  we  have 

(3) 


The  average  speed  during  the  interval  /  is 

(4) 


The  distance  described  (s  —  j,)  between  the  initial  and  final  positions,  measured  along  the 
path,  is  the  mean  speed  multiplied  by  the  time,  or 


Inserting  the  value  of  /  from  (2)  we  have  also 


(5) 


Hence 

V*  =  ^2  +  2/t  (S  -  S,}  ..........       (7) 

These  equations  are  general  whatever  the  path,  provided  ft  is  constant.  In  applying 
them/,  is  positive  (+)  when  the  speed  increases,  and  negative  (—  )  when  the  speed  decreases, 
during  the  interval  of  time  /,  without  regard  to  direction  of  motion. 

So,  also,  i>  is  positive  (-[-)  when  the  distance  from  the  origin  increases,  and  negative  (  —  ) 
when  the  distance  from  the  origin  decreases,  during  the  interval  of  time  /,  without  regard  to 
direction  of  motion. 

Rate  of  Change  of  Speed  Variable.  —  If  the  rate  of  change  of  speed  is  variable,  we  have  from  (i),  in 
Calculus  notation,  for  the  velocity  at  any  instant 


ds 


and  from  (2)  for  the  tangential  acceleration  at  any  instant 

"=«  ................  « 

and  from  (8)  for  the  distance  described 

s  —  st  =  J  vdt.      ...............  (10) 

We  have  also,  from  (9), 

fids  =  vdvt 


and  hence 


/• 


These  equations  are  general,  and  the  preceding  equations  can  be  deduced  from  them. 
Thus  if//  is  zero,  we  have  uniform  speed  and 


CHAP.  L]  GRAPHIC  REPRESENTATION  OF  RATE  OF  CHANGE  OF  SPEED. 

lift  is  constant,  we  have,  by  integrating  (9). 

v=ftt+C, 
where  C  is  the  constant  of  integration.     When  /  =  o,  v  equals  fi  and  hence  C  =  v\.     Hence 

v  =  vl  +ftt. 
Inserting  this  value  of  v  in  (8)  and  integrating,  we  have 

s**vj+  Iff  +  C. 
But  when  /  =  o  we  have  s  =  s\,  hence  C  ;=  s\.     Therefore 


93 


1 

V 

\          b          c           d    N 

V 

£,     B 


Graphic  Representation  of  Rate  of  Change  of  Speed. — If  we  represent  intervals  of  time 
by  distances  laid  off  horizontally  along  the  axis  of  x,  and  the  corresponding  speeds  by 
ordinates  parallel  to  the  axis  of  y,  we  shall  have  in  general  a  curve  for  which  the  change  of 
y  with  x  will  show  the  law  of  change  of  speed  with  the  time. 

(a)  RATE  OF  CHANGE  OF  SPEED  ZERO.— Lay  off  from  A  along  AB  equal  distances,  so 
that  the  distances  from  A  to  I,  i  to  2,  2  to  3,  etc., 
are  all  equal  and  represent  each  one  second  of  time, 
and  let  AB  represent  the  entire  time  /. 

Then  at  A,  i,  2,  3,  and  B  erect  the  perpen- 
diculars AM,  \b,  2c,  -$d,  BN,  and  let  the  length  of 
each  represent  the  speed  at  the  corresponding 
instant. 

Since  there  is  no  change  of  speed,  these  per- 
pendiculars will  all  be  of  equal  length,  we  shall 
have  AM  =  i  b  =  2c  =  ^d  =  BN  =  v,  and  the  speed 

at  any  interval  of  time  will  be  given  by  the  ordinate  at  that  instant  to  the  line  MN  parallel 
to  AB. 

The  space  described  in  any  time  is  given  by  s  —  sl=  vt.     This  is  evidently  given  by 
N  the  area  AMNB  in   the  diagram.      Therefore  the 

area    corresponding   to   any    time    gives  the   space 
described  in  that  time. 

(b)  RATE  OF  CHANGE  OF  SPEED  CONSTANT. — 
Lay  off  as  before  the  time  along  AB,  and  at  A,  i, 
2,  3,  B,  the  corresponding  speeds,  so  that  AMis  the 
initial  speed  vlt  and  BN  the  final  speed  v.  Draw 
MC,  be' ,  cd',  parallel  to  AB. 

Then  bb'  will  be  the  change  of  speed  in  the 
first  sec.,  cc'  the  change  of  speed  in  the  next  sec., 
and  so  on.  Since  these  are  to  be  constant,  NM  is  a  straight  line,  the  ordinate  to  which  at 
any  instant  will  give  the  speed  at  that  instant. 

The  rate  of  change  of  speed  is  then =. =  =  ft.      But  = -1 

I  sec.       i  sec.        I  sec.  i  sec.  t 

=  ft.      Hence  the  rate  of  change  of  speed  is  the  tangent  'of  the  angle  NMC  which  the  line  MN 
Hence  ft  —  —  ,     or     NC  —  ftt. 


Y 

*^ 

\^ 

* 

< 

^^ 

^_. 

d, 

I 

c 

M 

^^ 

v 

b< 

)                i 

^        * 

Ic            & 

A 

-     t 

A 

t 

makes  with  the  horizontal. 

The  distance  described  in  the  time  t  is,  from  equation  (5),  given  by  — 


But  this 


9-1 


KINEMATICS  OF  A  POINT.    APPLICATIONS. 


[CHAP.  I. 


is    the    area    of   AMNB.     Therefore    the   area  corresponding  to  any  time  gives   the  space 
described  in  that  time. 

We  have  then  directly  from  the  figure,  since  NC  =  ftt, 


X  /  = 


If   v,    is   greater   than   7%    a   will    be    negative,    and  the    line  MN  is  inclined  below 
the  horizontal  MC. 

[(r)  Rate  of  Change  of  Speed  Variable].—  If  the  rate  of  change  of  speed  is  not  constant,  we  shall  have  in 
„  general  a  curve  MNn.     The  tangent  to  this  curve  at  any 

point  N  makes  an  angle  with  the  axis  of  X,  whose  tan- 

gent is  -—  •  =//,  equation  (9),  or  the  rate  of  change  of 
speed.  The  elementary  area  BNnb  =  vdt  =  ds,  equation 
(8),  and  the  total  area  AMNB  =  /  vdt  =  s  —  Si,  equa- 

€//=0 

tion  (to). 

dV 


ftt 


Bl, 


V  {S 

When  -j7—°>     or    ~j-t  =  °.     or    ft  =  o,  the  tangent 


to  the  curve  is  horizontal  at  the  corresponding  point,  and  we  have  the  speed  at  that  point  a  maximum  or 
minimum  according  as  the  curve  is  concave  or  convex  to  the  axis  of  X. 

Examples.  —  (i)  The  speed  of  a  point  changes  from  50  to  30  ft.  per  sec.  in  passing  over  80  ft.    Find  the  con- 
stant rate  of  change  of  speed  and  the  time  of  motion. 

—  °°  =  —  10  ft.-per-sec.  per  sec.    The  minus  sign  indicates  decreasing  speed. 

2(j  —  Si)  2   X   oO 


ANS.    // 


(2)  Draw  a  figure  representing  the  motion  in  the  preceding  example,  and  deduce  the  results  directly  from  it 
Ans. — Average  speed  =  —         —  =  40  ft.  per  sec.    Hence  40*  =  80,  or  /  =  2  sec. 

Also/  = ^—  =  —  10  ft.-per-sec.  per  sec. 

(3)  A  point  starts  from  rest  and  moves  with  a  constant  rate  of  change  of  speed.     Show  that  this  rate  is 
numerically  equal  to  twice  the  number  of  units  of  distance  described  in  the  first,  second. 


ANS.     We   have  /  =  i   and  z/i  =  o  ;  hence,   from    eq.  (5),  —      \  =-/'.  or//  =  — 


-   ,  , 
i  sec. 


hich  is  numer- 


i  sec.        2 
ically  equal  to  z(s  —  st). 

(4)  In  an  air-brake  trial,  a  train  running  at  40  miles  an  hour  was  stopped  in  62  5.6  ft.  If  the  rate  of 
change  of  spted  was  constant  during  stoppage,  what  was  it  ? 

ANS.     From   eq.  (6)   we     have     for  v  =  o,   s  -  s,  =  625^6,  and    vt 


~  ~  T     o      6 
~~  (6ox6o)f  X2X  625.6  =  ~  2<75  ft<-Per-sec-  Pcr  sec-     The  (-)  s'8n  shows  retardation. 

(5)  4  point  starts  with  a  speed  v\  and  has  a  constant  rate  of  change  of  speed  —  //.      When  will  it  come  to 
rest,  and  what  distance  does  it  describe? 

ANS.     From  eq.  (3),  when  v  =  o,  we  have  r/,  -/,/  =  o.  or  /  =  !£•     From  eq.  (5),  s  -  s,  =  -/  =  — 

ft  2  2ft. 

(6)  A  point  describes  /jo  ft.  in  the  first  three  seconds  of  its  motion  and  50  ft.  in  the  next  two  seconds.    If 
the  rate  of  change  of  speed  is  constant,  when  will  it  come  to  rest  f    I  Vhen  will  it  have  a  speed  of  30  ft.  per  sec.  f 

ANS.   From  cq.  (5)  we  have  for  s  -  s,  •=-.  \  50  and  /  =  3,  1  50  =3^,  +  |  /,;  and  for  s  -  s,  =  200  and  /  =  5, 
too  =  $Vl  +  "l/i.     Combine  thes<?  two  equations  and  we  have/  =  -  10  ft.-per-sec.  per  sec.,  and  v,  =  65  ft. 

per  sec.     From  eq.  (3).  if  v  =  o.  we  have  65  -  io/  =  o,  or  /  =  6.  5  sec.     From  eq.  (3)  we  also  have,  if  v  =  30, 
30  =  65-  io/.  or/  =  3.; 


CHAP.  L] 


CONSTANT  RATE  OF  CHANGE  OF  SPEED— EXAMPLES. 


(7)  A  point  whose  speed  is  initially  30  meters  per  sec.  and  is  decreasing  at  the  rate  of  40  centimeters-per- 
sec.  per  sec.,  moves  in  its  path  until  its  speed  is  24.0  meters  per  minute.     Find  the  distance  traversed  and  the 
time. 

ANS.  We   have  Vi=  30  and  v  =  4  meters  per  sec.,  and ft  — — 0,4  meter-per-sec.  per  sec.     From  eq. 

(6),  s  —  Si  =        ~  9°°  =  1 105  meters.    From  (3)  we  have  4  =  30  —  o.4/,  or  /  =  65  sec. 
—  0.8 

(8)  A  point  has  an  initial  speed  of  v\  and  a  variable  rate  of  change  of  speed  given  by  -\-  kt,  where  k  is  a 
constant.      What  is  the  speed  and  distance  described  at  the  end  of  a  time  t  ? 

ANS.  From   eq.   (9)  we  have/i  =  —  =  kt,  and  integrating,  v  = [-  C.     If,  when  /  =  o,  we  have  v  =  v\ , 

fcfi 
we  obtain  C  —  TA  ,  and  hence  v  =  v\  -j-  — . 


From  eq.  (8),  ds  =  vdt  =  Vidt  +  3 .     Integrating,  s  =  Vit  +  -^-  -(-  C.    If,  when  /  =  o,  we  have  ^  =  o,  we 

obtain  C  =  o,  and  hence  .s  =  v\t  +  -?- 

o  . 

(9)  ,4  /<?/«/  /&<*£  an  initial  speed  of  60  ft.  per  sec.  and  a  rate  of  change  of  speed  of  +  4O.ft.-per-sec.  per  sec. 
Find  the  speed  after  8  sec.;  the  time  required  to  traverse  joo  ft.  ;  the  change  of  speed  in  traversing  that 
distance  ;  the  final  speed. 

ANS.   From  eq.  (3)  we  have  7/  =  6o  +  4ox8  =  380  ft.  per  sec.     From  eq.  (5)  we  have  300  =  6o/  +  2O/2, 


or  /  =  ±  y  —      --=  +  2.65  sec.  or  —  5.65  sec.     The  first  value  only  applies. 

From  eq.  (3)  we  have  v  —  v\  =  \ot  =  2o(±  1/69  —  3)  =  +  106  ft.  per  sec.  or  —  226  ft.  per  sec.  The  first 
value  only  applies. 

WTe  have  for  the  final  speed  v  =  +  166  ft.  per  sec.  or  —  166  ft.  per  sec.     The  first  value  only  applies. 

That  is,  the  point  starts  from  A  with  the  speed  v\  •=  60  ft.  per  sec. ___^ 

and  describes  the  path  AB  =  300  ft.  in  /  =  2.65  sec.,  the  speed  at  B 

being  v  —   166  ft.  per  sec.  _^- 300ft        ~--\ 

In  order  to  interpret  the  negative  values  obtained,  we  observe     A^" 

that  v  —  —  166   ft.  per  sec.  means  that  the  distance  from  the  origin  t=2.G5  B 

decreases.     Take  A  as  origin  and  let  the  point  then  start  from  B  towards  A  with  the  speed  z/i  =  —  166  ft. 

per  sec.      Then  from  eq.  (3)  we  have  v  =  —  166  +  4o/.     We 
see  that  when  /  =  4.15  sec.,  v  =  o,  and  the  point  has  passed  to 
some  point  P,  where  the  speed  is  zero.     This  point  is  the  turn- 
ing-point.     For  /  greater  than  4.15  sec.  v  becomes   positive; 
that  is,  the  point  moves  back  towards  B,  and  arrives  at  a  point 
A,  where  the  speed  is  v  —  +  60  in  the  time  given  by  60  =  4O/, 
or  /  =  1.5  sec.     The  entire  time  from  B  to  P  and  back  to  A  is 
/  =  5.65  sec.    This  is  the  time  given  by  the  negative  value  of  / 
in  the  example;  that  is,  it  is  the  time  before  the  start,  during  which  the  point  moves  from  B  to  P  and  back 
to  A.     The  change  of  speed  v  —  v\  is  +  60  +  166  =  +  226,  which  is  the  negative  value  in  the  example. 
For  the  space  BA  described  between  the  initial  and  final  positions  we  have 


v  +  v\ 


/,     or 


+60  —  166 


..  , 

5.65  =  —  300  ft., 


the  (— y*sign  showing  that  the  distance  is  on  the  other  side  of  the  origin  from  the  case  of  the  example. 

We  see,  then,  that  our  equations  are  general  if  we  have  regard  to  the  signs  of  v,  v\,  s,  s\,  andy/. 

( i  o)  If  the  motion  in  example  (9)  is  retarded, find  (a}  the  distance  described  from  the  starting-  to  the  turning- 
point  ;  (b)  the  distance  described  from  the  starting-point  after  10  sec.,  the  speed  acquired  and  the  distance  between 
the  final  and  initial  positions  ;  (c)  the  distance  described  during  the  time  in  which  the  speed  changes  to  —  go  ft. 
per  sec.,  and  this  time  ;  (d)  the  time  required  by  the  moving  point 
to  return  to  the  starting-point. 

ANS.  The  initial  speed  is  v\  =  +  60  ft.  per  sec.,  the  rate  of 
change  of  speed  is/<  =  —  40  ft.-per-sec.  per  sec.  Let  the  time 
count  from  the  start  at  A,  so  that  Si  =  o,  when  /  =  o. 

(a)  We  have  from  eq.  (6)  for  the  distance  from  A  to  the 
turning-point  P,  where  v  =  o, 

_  ~"  Vl*  _  ~  36o° 
S  ~  ~IfT  ~  ~^8b~ 


96  KINEMATICS  OF  A  POMT- APPLICATIONS.  [CHAP.  I. 

(£)  From  eq.  (5)  we  have  for  the  distance  between  the  initial  and  final  positions  after  10  sec. 

AB  =  s  =  vtt  +  -fit*  =  60  x  10  —  20  x  100  =  -  1400  ft. 

The  minus  sign  shows  distance  on  left  of  A.    The  total  distance  described  is  then  1490  ft.    The  speed 
acquired  is  given  by  eq.  (5)  -  1400  =  ^j-—  x  10,  or  w  =  -  340  ft.  per  sec.  The  minus  sign  shows  motion 

from  A  towards  B.  > 

(c)  From  eq.(6)  the  distance  between  the  initial  and  final  positions,  when  the  speed  is  —  90  ft.  per  sec.,  is 

AC-  s  =  gLn^L  =  8|0°  ~  36o°  =  -  56.25  ft.     The  minus  sign  shows  that  C  is  on  the  left  of  A.     The 

a/i  —  80 

total  distance  described  from  the  start  is  then  56.25  +  90  =  146.25  ft.    The  time,  from  eq.  (5),  is 

oo  +  60 

_  56.25  =  — ^~- — /,    or    /  =  3.75  sec. 

(</)  The  time  to  reach  the  turning-point,  as  we  have  seen,  is  1.5  sec.     The  time  to  return  is,  from  eq.  (5)' 
45  =  k?/  =  1.5  sec.     The  entire  time  to  go  and  return  is  then  3  sec. 

Rate  of  Change  of  Angular  Speed  Zero. — As  we  have  seen,  page  84,  if  the  angular 
acceleration  a  is  zero,  there  is  no  change  of  angular  velocity.  The  angular  velocity  oo  is  then 
uniform,  its  line  representative  has  always  the  same  magnitude  and  direction,  and  we  have 
uniform  angular  speed  in  an  unchanging  plane, 

In  this  case,  if  Bl  is  the  initial  angle  of  the  radius  vector  from  a  fixed  line  in  the  plane, 
and  0  is  the  final  angle,  we  have 

e  -  e 

co  =  — — l-,     or     oof  =  6  —Ol (i) 

[Compare  equation  (i),  page  91.] 

This  equation  is  general.  If  0  is  greater  than  0lt  co  is  positive.  If  0  is  less  than  0l , 
co  is  negative.  Hence  a  positive  (-{-)  value  for  co  denotes  that  the  angle  from  the  initial 
radius  vector  increases  as  the  interval  of  time  increases,  a  negative  (— )  value  for  GO  denotes 
that  the  angle  from  the  initial  position  of  the  radius  vector  decreases  as  the  interval  of  time 
increases,  without  regard  to  direction  of  rotation. 

Rate  of  Change  of  Angular  Speed  Uniform — Motion  in  a  Plane. — As  we  have  seen,  page 
84,  if  aH  is  zero,  we  have  a  =  aa  coinciding  with  <»,  and  hence  rotation  in  a  plane  with 
varying  angular  speed.  If,  in  this  case,  aa  is  uniforn,  we  have  uniformly  varying  angular 
speed  in  an  unchanging  plane.  We  have  then  (compare  with  equations  (2)  to  (7),  page  92) 


(2) 


where  oj,  and  oo  are  the  initial  and  final  angular  velocities.     The  value  of  a  is  (-J-)  when 
the  angular  velocity  increases,  and  (-<•)  when  it  decreases  during  the  time. 
From  (2)  we  have,  just  as  on  page  92, 

«  =  «»1  +  a/ (3) 

The  average  angular  speed  is 

The  angle  described  in  the  time  /  is 

0  —  8.  =  —        — /=  co.t  4-  la/*.  (t\ 

•  •      •      •      •      Y.  j  / 


CHAP.  I.J  VARIABLE  RATE  OF  CHANGE  OF  ANGULAR  SPEED.  97 

Inserting  the  value  of  t  from  (2),  we  have 


Hence 

a?2  =  <**  +  2a(Q  _  ey  ..........     (7) 

It  will  be  noted  that  all  these  equations  are  precisely  similar  to  equations  (2)  to  (7),  page 
92.  We  have  only  to  replace  s,  v,  ft  by  8,  a?,  a. 

In  applying  these  equations  a  is  positive  when  the  angular  speed  increases,  and  negative 
when  it  decreases.  So,  also,  a?  is  positive  (-)-)  when  the  angle  from  the  initial  position  of  the 
radius  vector  increases  as  the  interval  of  time  increases,  and  negative  (—  )  when  the  angle 
from  the  initial  position  of  the  radius  vector  decreases  as  the  interval  of  time  increases, 
•without  regard  to  direction  of  rotation. 

Rate  of  Change  of  Angular  Speed  Variable.  —  If  the  rate  of  change  of  angular  speed  is  variable,  we  have 
from  (i),  in  Calculus  notation  for  the  angular  velocity  at  any  instant, 


<8> 


and  from  (2),  for  the  axial  angular  acceleration  at  any  instant, 


d<*> 

Ota  = 

and  from  (8),  for  the  angle  described, 


a"  =  -di  = 


Q  -0!  =    r*<odt (10) 

t/e 

We  have  also,  from  (9), 

aadQ  =  Godoo, 
and  hence 


These  equations  are  general,  and  the  preceding  equations  can  be  deduced  from  them. 
Thus,  if  cta  is  zero,  we  have  uniform  angular  speed  and 


If  oca  is  constant,  we  have,  by  integrating  (9), 

where  C  is  the  constant  of  integration.     When  /  =  o,  co  =  oa, ,  and  hence  C  =  <»i.     Hence 

00   =   GOi   +  Ctat. 

Inserting  this  value  of  GO  in  (8)  and  integrating,  we  have 

6  =  coit  +  \a.a?  +  C. 
But  when  /  =  o  we  have  0  =  61 ,  hence  C  =  si.    Therefore 

0  -  0!  =  wit  +  \aaf. 

Graphic  Representation. — The  graphic  representation  is  precisely  the  same  as  on  page 
93.  Thus  we  can  represent  intervals  of  time  by  distances  laid  off  horizontally,  and  the  cor- 
responding angular  speeds  by  distances  laid  off  vertically,  and  thus  obtain  the  same  diagrams 
as  for  linear  speed  given  on  page  93. 


98  KINEMATICS  OF  A  POINT-APPLICATIONS.  [£HAP.  I. 


Examples.—  (0    The  angular  speed  of  a,  point  moving  in  a  plane   about  some  assumed  point 
from  50  to  jo  radians  per  sec.  in  passing  through  So  radians.     Find  the  constant  rate  of  change  of  angular 
speed  ami  the  time  of  motion. 

The    minus   sign    denotes   decreasing    speed. 


-.  _  I0  radians-per-sec.  per  sec 


(2)  Draw  a  figure  representing  ike  motion  in  the  preceding  example,  and  deduce  the  results  directly  from  it. 

ANS.  Average  speed  =   -         —  =  40   radians   per   sec.      Hence 

\a>-30  40/  =  80,  or  /  =  2  sec.     Also  a  =  —  —  to  radians-per-sec.  per  sec. 

(3)  A  point  moving  in  a  plane  has  an  initial  speed  of  60  radians  per  sec.  about  an  assumed  point  and  a 
rate  of  change  of  speed  of  +  40  radians-per-sec.  per  sec.     Find  the  speed  after  8  sec.;  the  time  required  to 
describe  300  radians  ;  the  change  of  speed  while  describing  that  angle  ;  the  final  speed. 

ANS.  Sec  example  (9),  page  95. 

(4)  If  the  motion  in  the  last  example  is  retarded,  find  (a)  the  angular  revolution  from  the  start  to  the  turn- 
ing-point;  (/*)  the  angle  described  from  the  start  after  10  sec.;  the  speed  acquired  and  the  angle  between  the 
final  and  initial  positions  ;  (c)  the  angle  described  during  the  time  in  which  the  speed  changes  to  —go  radians 
per  sec.,  and  this  time  ;  (d)  the  time  required  by  the  moving  point  to  return  to  the  initial  position. 

ANS.  See  example  (ro),  page  95. 

(5)  A  point  moving  in  a  plane  describes  about  a  fixed  point  angles  of  120  radians,  228  radians  and  336 
radiant  in  successive  tenths  of  a  second.     Show  that  this  is  consistent  with  uniform  rate  of  change  of  angular 
speed,  and  fin. t  this  rate. 

ANS.  a  —  10800  radians-per-sec.  per  sec. 

(6)  Two  fioints  A  and  B  mor>e  in  the  circumference  of  a  circle  with  uniform  angular  speeds  GO  and  a/. 
The  angle  between  them  at  the  start  is  a.     Find  the  time  of  the  nth  meeting,  the  angles  described  by  A  and  R, 
and  the  interval  of  time  between  two  successive  meetings. 

ANS.  Time  of  the  »th  meeting,  /*  =  ±  a  +  <"—^2*. 

oo  ±  ao' 

Angle  described  by  A  is  0  =  <»/„. 

"  "  B  is  T  a. 

Interval  of  time  between  two  successive  conjunctions  is 

/,_/,  =  /,_/.,  = 


oo  ±  oaf 

where  we  take  the  (+)  or  (— )  sign  for  a  according  as  B  is  in  front  of  or  behind  A  at  start,  and  (+)  or  (-) 
sign  for  oj'  according  as  the  points  move  in  opposite  or  the  same  directions. 

(7)  What  is  the  angular  speed  of  a  fly-wheel 5  ft.  in  diameter  which  makes  thirty  revolutions  per  minute, 
and  what  />  the  linear  velocity  of  a  point  on  its  circumference  ?     Also  find  its  linear  central  acceleration. 

ANS.  it  radians  per  sec. ;   2.5*  ft.  per  sec.,  tangent  to  circ.;  2.5^"  ft.-per-sec.  per  sec. 

(8)  Find  the  linear  and  angular  speed  of  a  point  on  the  earth's  equator,  taking  radius  4.000  miles  ;  also  the 
linear  centr.il  acceleration. 

ANS.  1 535.9  ft.  per  sec. ;  —^  radians,  or  15°  per  hour;  o.i  12  ft.-per-sec.  per  sec. 

(9)  The  angular  speed  of  a  wheel  is  -it  radians  per  sec.     Find  the  linear  speed  of  points  at  a  distance  of 
3  ft.,  4ft.  and  ro  ft.  from  the  centre,  also  tht  linear  central  acceleration. 

ANS.   ^w.  3*.  7-5*  ft-  per  sec- 

|je\-  **.  ^ir»  ft.-per-sec.  per  sec. 

(10)  Tf  thf  linear  speed  of  a  point  at  the  equator  is  v,  find  the  speed  linear  and  angular  at  any  latitude  \ 
ANS.  v  cos  A;  —radians  per  hour,  or  15°  per  hour. 


CHAP.  I.]  RATE   GF  CHANGE   OF  ANGULAR  SPEED— EXAMPLES.  gg 

(11)  A  point  moves  with  uniform  velocity  v.     Find  at  any  instant  its  angular  speed  about  a  fixed  point 
whose  distance  from  the  path  is  a. 

ANS.   -^  radians  per  sec.,  where  r  is  the  radius  vector.     Uniform  velocity  means  uniform  speed  in  a 

straight  line  (page  65). 

Hence  the  angular  speed  of  a  point  moving  with  uniform  speed  in  a  straight  line  is  inversely  proportional 
to  the  square  of  the  distance  of  the  point  from  a  fixed  point  not  in  the  line. 

(12)  The  speed  of  the  periphery  of  a  mill-wheel  12 feet  in  diameter  is  6  feet  per  stc.     How  many  revolu- 
tions does  the  wheel  make  per  sec.  ? 

ANS.  —  revolutions. 

27T 

(13)  Deduce  the  equivalent  of  longitude  for  one  minute  of  time  and  for  one  second  of  time. 
ANS.   15'  to  i  min.,  15"  to  i  sec. 

(14)  77/i?  diameter  of  the  earth  is  nearly  8000  miles.     Required  the  circumference  at  the  equator  and  the 
linear  speed  at  latitude  60°. 

ANS.   25000  miles;  521  miles  per  hour. 

(15)  The  wheel  of  a  bicycle  is  52  inches  in  diameter  and  performs  3040  revolutions  in  a  journey  of  63 
minutes'.     Find  the  speed  in  miles  per  hour  ;  the  angular  speed  of  any  point  about  the  axle  ;  the  areal  velocity 
of  a  spoke  ;  the  relative  velocity  of  the  highest  point  with  respect  to  the  centre. 

ANS.  12  miles  per  hour;  8.12  radians  per  sec.  ;   19.06  sq.  ft.  per  sec.;   12  miles  per  hour. 

(16)  In  going  120  yards  the  front  wheel  of  a  carriage  makes  six  revolutions  more  than  the  hind  wheel.     If 
each  circumference  were  a  yard  longer,  it  would  make  only  four  revolutions  more.     Find  the  cireumference  of 
each  wheel. 

ANS.  4  yards  and  5  yards. 

(17)  If  the  velocity  of  a  point  is  resolved  into  several  components  in  one  plane,  show  that  its  angular  speed 
about  any  fixed  point  in  the  plane  is  the  sum  of  the  angular  speeds  due  to  the  several  components. 

(18)  A  point  moves  with  uniform  speed  v  in  a  circle  of  radius  r.     Show  that  its  angular  speed  about  any 

point  in  the  circumference  is  — . 

(19)  A  point  starting  from  rest  moves  in  a  circle  with  a  constant  rate  of  change  of  angular  speed  of  2 
radians-per-sec.  per  sec.     Find  the  angular  speed  at  the  end  of  20  sec.  and  the  angular  displacement ;  also  the 
linear  speed  and  distance  described  and  the  mimber  of  revolutions  ;  also  the  linear  tangential  acceleration  and 
the  central  linear  acceleration  at  the  end  of  20  sec. 

ANS.  40  rad.  per  sec.;  400  radians;  40^  ft.  per  sec.  ;  4oor  ft.;  - —  revolutions  ;    2r   ft.-per-sec.   per  sec. 

27t 

tangential  acceleration;  i6oor  ft.-per-sec.  per  sec.  central  acceleration. 

(20)  A  point  moving  with  uniform  rate  of  change  of  angular  speed  in  a  circle  is  found  to  revolve  at  the 
rate  of  8|  revolutions  in  the  eighth  second  after  starting  and  "j\  revolutions  in  the  thirteenth  second  after 
starting.     Find  its  initial  angular  speed  and  its  uniform  rate  of  change  of  angular  speed ;  also  the  initial 
linear  speed  and  rate  of  change  of  speed ;  also  the  initial  central  acceleration. 

ANS.  20.27T  radians  per  sec.;  —  o.^it  radians-per-sec.  per  sec.;  2o.2?rr  ft.  per  sec.  ;  —  o.^itr  ft.-per-sec. 
per  sec.  ;  4o8.o4^V  ft.-per-sec.  per  sec. 

(21)  A  point  starts  from  rest  and  moves  in  a  circle  with  a  uniform  rate  of  change  of  angular  speed  of  18 
radians-per-sec.  per  sec.     Find  the  time  in  which  it  makes  the  first,  second  and  third  revolutions. 

ANS.  ^,  1^*-*™,  fa-  *rt  Secs. 


CHAPTER   II. 

UNIFORM   ACCELERATION. 

Uniform  Acceleration — Motion  in  a  Straight  Line. — If  the  central  acceleration/,  of  a 
moving  point  is  zero,  the  path  must  be  a  straight  line  and  the  resultant  acceleration  /  is 
equal  to  the  tangential  acceleration  /,.  If,  then,  the  acceleration  /  of  a  moving  point  is 
uniform,  so  that  it  does  not  change  in  magnitude  or  direction,  and  if  its  direction  coincides 
with  that  of  the  velocity,  we  have  motion  in  a  straight  line  in  a  given  direction  with  uniform 
rate  of  change  of  speed,  and  equations  (2)  to  (6),  page  92,  apply. 

In  applying  these  equations  we  should  note  that  sign  now  indicates  direction.  If  v  or/ 
are  positive  (-J-),  they  act  away  from  the  origin;  if  negative  (92),  they  act  towards  the 
origin. 

The  most  common  instance  of  such  motion  is  that  of  a  body  falling  near  the  earth's 
surface.  All  points  of  such  a  body  move  in  parallel  straight  lines  with  the  same  speed  at 
any  instant,  so  that  the  body  has  motion  of  translation  only,  and  may  be  considered  as  a 
point. 

The  acceleration  due  to  gravity  is  known  to  be  practically  uniform  and  is  always 
denoted  by  the  letter^. 

Value  of  g. — The  value  of  g  is  usually  given  in  feet-per-sec.  per  sec.  or  in  centimeters- 
per-scc.  per  sec. 

It  has  been  determined  by  much  careful  experiment  and  found  to  vary  with  the 
latitude  \  and  the  height  //  above  sea-level. 

The  general  value  is  given  by 

g=  32. -173  —  0.0821  cos  2*  —  o. 0000037*, 
where  //  is  the  height  above  sea-level  in  feet,  and  g  is  given  in  feet-per-sec.  per  sec. ,  or 

g  =  980.6056  —  2.  5028  COS  2\  —  0.000003/*, 

where  //  is  the  height  above  sea-level  in  centimeters,  and  g  is  given  in  centimeters-per-sec. 
per  sec. 

It  will  be  seen  that  the  value  of  g  increases  with  the  latitude,  and  is  greatest  at  the  poles 
and  least  at  the  equator.  It  also  decreases  as  the  height  above  sea- level  increases. 

The  following  table  gives  the  value  of  g  at  sea-level  in  a  few  localities : 

g  g 

Latitude.  F.  S.  Units.  C.  S.  Units. 

Equator o°  o'  32.091  978.10 

New  Haven 4118  32.162  980.284 

Latitude  45° 45    o  32.173  980.61 

Paris 4850  32.183  980.94 

London 5140  32.182  980.889 

Greenwich 5129  32.191  981.17 

Berlin   5230  32.194  981.25 

Edinburgh 5557  32.203  981.54 

Pole 90    o  32.255  983.11 


CHAP.  II.]  BODY  PROJECTED   VERTICALLY   UP  OR  DOWN.  101 

For  calculations  where  great  accuracy  is  not  required  it  is  customary  to  take^-=  32  ft.- 
per-sec.  per  sec.  or  g  —  981  cm.-per-sec.  per  sec. 

For  the  United  States  g  =  32^  is  a  good  average  value  and  is  therefore  very  often  used. 

In  exact  calculations  the  value  of  g  for  the  place  must  be  used. 

Body  Projected  Vertically  Up  or  Down.  —  For  a  body  profected  vertically  upwards  in 
vacua,  taking  the  origin  at  the  starting-point,  we  have  then  vl  positive,  since  it  is  away  from 
the  origin,  and  g  negative,  since  it  is  towards  the  origin.  If  the  body  is  projected  vertically 
downwards,  taking  still  the  origin  at  the  starting-point,  we  have  vl  positive  and  g  also 
positive,  since  both  act  away  from  the  origin.  We  have  then  simply  to  replace  ft  in 
equations  (2)  to  (7),  page  92,  by  ±.g,  taking  the  plus  sign  for  the  falling  body  and  the 
minus  sign  for  the  rising  body. 

When  the  final  velocity  for  a  body  projected  upwards  is  zero,  we  have  from  (3),  page  92, 
for  the  time  of  rising  to  the  highest  or  turning  point,  by  making  v  =  o  and  ft  =  —  g, 


g 

For  the  time  of  rising  to  the  highest  or  turning  point  and  then  returning  to  the  starting- 
point  we  make  s  —  sl  =  o  in  (5),  page  92,  and/,  =  —  g,  and  obtain 


g 

Hence  the  times  of  rising  and  returning  are  equal  for  motion  in  vacua. 

The  distance  from  the  starting-point  to  the  turning-point  is  found  from   (6),  page  92, 

v% 
by  making  v  =  o  and  ft  —  —  g,  to  be   —  . 

v2          v* 
The  distance    —    or    —  is   called   the  height  due  to  the  velocity  vl  or  v;    that  is,  the 

distance  a  body  must  fall  from  rest  in  order  to  acquire  the  velocity  i\  or  v. 

vi       rf 
When  the   distances  in   rising  and  falling    are   equal  we   have  s—sl  =  o  or  —  =  —  or 

v^  =  v.     That  is,  for   motion   in  vacua  the  velocity  of  return   is   equal  to  the   velocity  of 
projection. 

If  the  time  of  rising  is  less  than  T  =  —  ,  the  displacement  s  —  sl  is  equal  to  the  distance 
described.  But  if  the  time  is  greater  than  T  —  —  ,  the  body  reaches  the  turning-point  and 

c> 

then  falls  from  rest,  and  the  entire  distance  described  is 

V  2  VZ  Z'2     \     Z? 

distance  described  =  -^-  +  \g(t  —  7)'  —     l  ---  s=  —~  —  . 

The  student  will  note  that  the  motion  is  supposed  to  take  place  in  a  vacuum.  That 
is,  the  effect  of  the  resistance  of  the  air  is  neglected.  As  a  matter  of  fact  this  resistance  has 
a  great  influence,  and  hence  the  following  examples  have  little  practical  value  except  as  illus- 
trating the  application  of  the  equations, 


102  KINEMATICS  OF  A  POINT-APPLICATIONS.  [^HAP.  II. 

Examples.—  (Unless  otherwise  specified  g  =  32.2-ft.-per-sec.  per  sec.  or  981  cm.-per-sec.  per  sec.     All 
bodies  supposed  to  move  in  vacuum.) 

(1)  A  point  moves  with  a  uniform  velocity  of  a  ft.  per  sec.     Find  the  distance  from  the  starting-point  at 
tht  end  of  one  hour. 

ANS.  7200  ft.     Motion  in  a  straight  line. 

(2)  Two  trains  have  equal  and  opposite  uniform   velocities  and  each  consists    of  12  cars  of  50  ft. 
They  are  observed  to  take  18  sec,  to  pass.    Find  their  velocities. 

ANS.  22.73  miles  per  hour. 

(3)  Two  points  move  with  uniform  velocities  of  8  and  15  ft.  per  sec.  in  directions  inclined  90°.   At  a  given 
instant  their  distance  is  toft,  ami  their  relative  velocity  is  inclined  30"  to  the  line  joining  them.     Find  (a} 
their  distance  when  nearest  ;  (A)  the  time  after  the  given  instant  at  which  their  distance  is  least. 

ANS.  (a)  5  ft.  ;  (b)  ^-  ^sec. 

(4)  A  body  is  projected  vertically  upwards  with  a  velocity  of  300  ft.  per  sec.     Find  (a)  its  velocity  after  2 
stc.  ;  (b)  its  velocity  <ifter  15  sec.  ;  (c)  the   time  required  for  it  to  reach  its  greatest  height  ;  (d)  the  greatest 
height  reached  ;  (f)  its  displacement  at  the  end  of  15  sec.  ;  (f)  the  space  traversed  by  it  in  the  first  13  sec.  ; 
(g)  its  displacement  when   its  velocity  is  200  ft.  per  sec.  upwards;    (//)  the  time  required  for  it  to  attain  a 
displacement  of  320  ft. 

ANS.  (<i)  235.6    ft.    per  sec.;  (b)  183  ft.  per  sec.  downwards;  (c)  9.3  sec.  ;    (d)  1397.5  ft.;  (<?)  877.5    ft. 
upwards;   (/)  1917-5  ft-  I  (g)  7?6-3  ft.  upwards;  (h)  1.13  sec.  in  ascending.  17.5  sec.  in  descending. 

(5)  A  ball  is  projected  upwards  from  a  window  half  way  up  a  toiver  f  17.72  meters  high,  with  a  velocity  of 
39.24  m.  per  sec.  Find  the  time  and  speed  (a)  with  which  it  passes  the  top  of  the  tower  ascending  ;  (d)  the  same 
point  descending  ;  (c)  reaches  the  foot  of  the  firmer. 

ANS   (a)  2  sec.  ;  19.62  m.  per  sec.;  (b)  6  sec.;  19.62  m.  per.  sec.;  (c)  (4  +  2|77)  sec.;  1962  V?  m.  per  sec. 

(6)  A  stone  is  dropped  into  a  well  and  the  splash  is  heard  in  3.  13  sec.     If  sound  travels  in  air  with  a  uni- 
form velocity  of  332  meters  per  sec.,  find  the  depth  of  the  well. 

ANS.  44.1  meters. 

(7)  If  in  the  preceding  example  the  time  until  the  splash  is  heard  is  T  and  the  velocity  of  sound  in  air  is 
I',  find  the  depth. 


ANS.  Depth  =         (7*  +  V)  - 

(S)  Show  that  a  body  projected  vertically  upwards  requires  twice  as  long  a  time  to  return  to  its  initial 
Position  as  to  reach  the  highest  point  of  its  path,  and  has  on  returning  to  its  initial  position  a  speed  equal 
to  its  initial  speed. 

(9)  A  stone  projected  vertically  upwards  returns  to  its  initial  position  in  6  sec.     Find  (a)  its  height  at  the 
end  of  the  first  second,  and  (b)  what  additional  speed  would  have  kept  it  I  sec.  longer  in  the  nir. 

ANS.  (a)  80  5  ft.  ;  (fi)   16.  i  ft.  per  sec. 

(10)  A  body  let  fall  near  the  surface  of  a  small  planet  is  found  to  traverse  204  ft.  between  the  fifth  and 
sixth  seconds.     Find  the  acceleration. 

ANS.  20.4  ft.-per-sec.  per  sec. 

(11)  A  particle  describes  in  the  nth  second  of  its  fall  from  rest  a  space  equal  to  p  times  the  apace  described 
in  the  («  —  /)///  second.     Find  the  whole  space  described. 

ANS  *(l  "3^- 
8<r-»-" 

(12)  A  body  uniformly  accelerated,  and  starting  without  initial  velocity,  passes  over  b  feet  in  the  first  p 
secondf.     Find  the  time  of  passing  over  the  next  bft. 

ANS.  p(  i/2~—  i)  sec. 

(13)  A  ball  is  dropped  from  the  top  of  an  elevator  4.905  meters  high.     Acceleration  of  gravity  i*  9.  Si 
meter  s-per-sec.  per  sec.     Find  the  times  in  which  it  will  reach  the  floor  (<i)  when  the  elevator  is  at  rest  ;  (/>) 
when  it  is  moving  with  a  uniform  downward  acceleration  of  9.81  >n.-per-sec.  per  sec.  ;  (c)   wlnn  moving  with 
a   uniform  downward  acceleration  of  4.905  m.-per-sec.  per  sec.  ;  (a)  when  moving  with   a   uniform  upward 
acceleration  of  4.905  m.-per-sec.  per  sec. 

ANS.  (a)  I  sec.;  (b)  co  ;  (c)  f'Jsec.  ;   (d)   -J/isec. 


CHAP.  II.] 


UNIFORM  ACCELERATION -MOTION  IN  A   CURVE. 


103 


FIG.  (a). 


FIG.  (b). 


Uniform  Acceleration — Motion  in  a  Curve. — If  the  acceleration  /  is  uniform,  so  that  it 
d  )es  not  change  either  in  magnitude  or  direction,  and  if~its  direction  doe-s  not  coincide  with 
that  of  the  velocity,  we  have  motion  in  a  curve  with  uniform  acceleration. 

A  common  case  of  such  motion  is  that  of  a  body  projected  with  any  given  velocity  in 
any  given  direction  at  the  surface  of 
the  earth,  neglecting  the  resistance  of 
the  air.  In  such  case  the  accelera- 
tion due  to  gravity  is  practically  uni- 
form, acts  downward  and  is  equal  to 
g  ft.-per-sec.  per  sec.  The  curve  or 
path  in  such  case  is  called  the  TRA- 
JECTORY. 

EQUATION  OF  THE  PATH. — Let  the  uniform  acceleration  y  be  vertical  and  act  downwards. 
(In  the  case  of  gravity  f  =  g.}  Take  the  origin  O  at  the  initial  position  of  the  point  as 
shown  in  Fig.  (a),  and  let  the  initial  velocity  v.^  make  the  angle  al  with  the  horizontal. 

Let  the  co-ordinates  of  any  point  P  of  the  path  or  trajectory  be  OB  —  x  and  BP  =  y. 

Let  the  angle  POB  -  9. 

If  in  Fig.  (b}  we  lay  off  from  Q  the  line  representative  Qbl  =  v^  of  vl ,  the  line  repre- 
sentative Qb  =  v  of  v,  etc.,  we  see  that  the  hodograph  b^,  etc.,  is  a  straight  line.  (See 
example  (2),  page  80.) 

It  is  at  once  evident  that  the  horizontal  component  vx  of  the  velocity  at  every  point  is 
constant  and  equal  to 

,  vx  =  v^  cos  a^ (i) 

In  any  time  /,  then,  the  horizontal  distance  described  is 

x  =  i\t  cos  al (2) 

The  vertical  component  of  the  initial  velocity  7\  is  7'xsin  ael  upwards.  But  the  uniform 
acceleration  f  is  downwards.  Hence  the  vertical  velocity  at  the  end  of  the  time  /  is,  from 
equation  (3),  page  92, 

vy  =vlsinal—/f (3) 

The  mean  vertical  velocity  during  the  time  /  is,  then, 
z\  sin  al  -J-  vy       2vl  sin  crl  —  ft 


=  v,  sin  a, 


The  vertical  distance  passed  over  in  the  time  t  is,  then, 

y  =  v^t  sin  atl  —  \ffj 

If  we  combine  (2)  and  (4)  by  eliminating  /,  we  have  for  the  equation  of  the  path 


(4) 


y  —  x  tan  ofl  — 


2V?  cos"  a, 


This  is  the  equation  of  a.  parabola. 

The  time  of  reaching  the  highest  point  C  is  the  time  of  describing  the  vertical  distance 
DC.  Denote  this  time  by  Tv.  Since  at  this  point  the  vertical  velocity  is  zero,  we  have, 
by  making  vy  =  o  in  (3), 


_ 

— 


r 


I04  KINEMATICS  OF  A  POINT    APPLICATIONS.  [CHAP.  II. 

If  we  substitute  this  for  /  in  (2)  and  (4),  we  have  for  the  co-ordinates  of  the  vertex  C  of 
the  parabola 


ec^fjgA...        - (8) 


The  parameter  of  the  parabola  is  then 


The  directrix  is  parallel  to  OD  at  a  distance  above  the  vertex  C  equal  to  one  half  the 
parameter,  or 

v*  cos2  al 

7'* 

or  at  a  distance  of  —,  above  O.     That  is,  the  distance  of  the  directrix  above  O  is  the  height 

due  to  tht  velocity  vr 

If  we  transfer  the  origin  to  the  vertex  C,  then  the  horizontal  velocity  at  C  is  vl  cos  <*}  , 
the  horizontal  distance  described  in  any  time  /  is  x  =  i\t  cos  <rr  The  mean  vertical  velocity 
is  \  ft.,  and  the  vertical  distance  is  y  =  £  ft.2.  Eliminating  /,  we  have 


which  is  the  equation  of  the  path  for  origin  at  the  vertex  C. 

VKLOCITV  AT   ANY   POINT  OF  THE  PATH.  —  The  magnitude   of   the   velocity  at  any 
point  Pis  the  resultant  of  the  vertical  and  horizontal  components,  or,  from  (i)  and  (3), 

V*  =  v\  +  v*  =  v*  -  2vjt  sin  at  -f/8/2  .......      (10) 

Inserting  the  value  of  y  from  (4), 

*  =  v*-*fy  ...........    (ii) 

If  the  acceleration  is  due  to  gravity,  we  replace  f  by  g  and  have 


V* 

But  we  have  just  seen  that  -       is  the  distance   of   the   directrix  above  O.     Therefore 

2S 

v*  v> 

—  —  ^  is  the  distance  of  the  directrix  above  any  point  P,  and    —  is  the  height  due  to  the 

velocity  v  (page  ioiV      ftcnce  for  a  body  acted  upon  by  gravity  the  speed  at  any  point  is 
the  same  as  that  acquired  by  a  body  falling  from  the  directrix  to  that  point. 

To  find  the  direction  of  the  velocity  v  at  any  point  P,  the  magnitude  of  which  is  given 
by  (10)  or  (11),  let  a  be  the  angle  which  it  makes  with  the  horizont.il.  Then  we  have 
directly  from  the  hodograph  (Fig.  (£),  page  103).  and  from  equations  (3)  and  (i), 

v  sin  a  —  Vy  -  ^  sin  <r,  —  //, 
V  COS  «  —  VM  —  v^  cos  atv  . 


CHAP.  II.]  UNIFORM  ACCELERATION— TIME  OF  FLIGHT  AND  RANGE.  105 

Hence;  from  equation  (2), 

ft  ft* 

tan  a  =  tan  al -  =  tan  av ,         (12) 

vl  cos  al  x 

or 

tan  a  =  tan  a,  —  — ^ — ...  .     (13) 

v*  cos''  a  l 

.   TIME   OF  FLIGHT   AND   RANGE. — If  in  (4)   we  make  y  =  o,  we  have  for  the  time  Th 
in  which  the  body  reaches  the  line  OX,  or  the  time  of  flight  in  a  horizontal  direction, 


Inserting  this  value  of  t  in  (2),  we  have  for  the  horizontal  range  OA  =  Rh 

2v*  sin  <r,  cos  «,       v*  sin  2 a, 


04) 


(15) 


This,  we  see  from  (7),  is  twice  the  distance  OD. 

We  see  from  (15)  that  the  range  is  greatest  when  sin  2txl  is  greatest,  or  when  20^  —  90° 
or  «t  =  45°.  Therefore  the  range,  neglecting  resistance  of  the  air,  is  greatest  for  an  angle 

v* 

of  elevation  of  45°,  and  is  equal  to  -~. 

DISPLACEMENT  IN  ANY  GIVEN  DIRECTION  AND  THE  CORRESPONDING  TIME.  —  Let 
6  be  the  angle  which  any  displacement  OP  =  R  makes  with  the  horizontal.  Then  we  have 
BP  —  y  =  x  tan  6.  Substituting  this  value  of  y  in  (5),  we  have  for  the  abscissa  of  the 
point  P 

_  2v*  cos  al  sin  (a^  —  6}  _  v*[sm  (2al  —  0)  —  sin  ff] 
*~~  ~Tc^~e~  /cos0  ~~'      '  (lb) 

and  therefore,  since  R  = 


--  7,, 
cos  0' 

_  2z/j2  cos  o^  sin  (flfj  —  0)  _  ^'[sin  (20^  —  0)  —  sin  ff] 

f  cos2  e  "  ,  ~         .     .    .    .    (17) 


If  in  (17)  we  make  0  =  o,  we  have  the  horizontal  range  given  by  (15). 
If  we  divide  (16)  by  the  horizontal  component  of  the  velocity,  vl  cos  av  we  have  for 
the  time  of  flight 

_  2v,  sin  (ael   -  6) 

T^s~0  ..........      (I8) 

This  reduces  to  (14)  for  6  =  o. 

ANGLE  OF  ELEVATION  FOR  GREATEST  RANGE  IN  ANY  DIRECTION.  —  The  range  R 
given  (by  (17)  is  greatest  when  sin  (2av  —  0}  is  greatest,  or  when  2^  —  &  =  90°,  or 
ai  =  i(9°°  H~  ^)-  The  direction  for  greatest  range  makes,  therefore,  with  the  vertical  an 
angle  of  90°  —  at^  =  £(90°  —  6),  that  is,  it  bisects  the  angle  between  the  vertical  and  the  range.} 

ELEVATION  NECESSARY  TO  HIT  A  GIVEN  POINT.  —  To  determine  the  direction  of  the 
velocity  vl  in  order  that  the  path  may  pass  through  a  given  point  given  by  x  and  /,  we 


io6 


KINEMATICS  OF  A  POINT-APPLICATIONS. 


substitute  for — g in  equation  (5)  the  equivalent  expression 


once 


\Vc  have  also,  from  (16), 
or,  since  R  cos  6  =  xt 


0      i         Jfx  cos  0 
= 


fx  co 

V      ?'.a 


[CHAP.  II. 
tan2  al  and  obtain  at 

(19) 

(20) 


We  see  from  (19)  that  nt  has  two  values.  If  «(  is  an  angle  such  that  sin  (2a\  —  0)  = 
sin  (2<r,  —  0),  then  2a\  —  6  =  180°  —  (2crl  —  0),  or  a(  =  90°  —  (tfj  —  8),  and  either  a|  or 
a,  will  satisfy  equation  (16). 

With  a  given  acceleration  and  initial  velocity  there  are,  then,  two  directions  of  the 
initial  velocity,  or,  and  90°  —(or,  —0),  and  therefore  two  paths  by  which  the  projectile  may 
hit  the  same  point. 


7.  2Z'V 

If,  in  (19),  we  put—-    =  I  -f-  -yV  »  we  nave 


~    tfx2  -f-  j2)     and     tan  al  = 

Smaller  values  of  i\  make  tan  al  imaginary.  Larger  values  of  vl  give  two  values  for 
tan  ",.  In  the  first  case  the  point  cannot  be  hit.  In  the  second  case  it  can  be  hit  during 
cither  the  rise  or  fall  of  the  projectile. 

ENVELOPE  OF  ALL  TRAJECTORIES. — Equation  (5)  gives  the  equation  of  the  trajectory 

corresponding    to    the    angle    of   elevation    av      If   we    substitute    i  -j-  tan2  al    for  — ^ — , 
equation  (5)  becomes 


y  ==•  x  tan  al  — 


(22) 


where  x  and  y  are  the  co-ordinates  of  any  point  of  the  path. 

For  another  angle  of  elevation,  «J,  and  the  same  initial  speed,  vlt  we  have 


, 
=  x,  tan  a,  - 


where  -i',  and  y\  are  the  co-ordinates  of  any  point  of  the  new  trajectory. 

If  we  make  x  =  xl  and  y  =  y, 
and  equate  these  two  equations.  \vi 
have  for  the  point  of  intersection  of 
the  two  trajectories 

—  (tana   +  tan  a'}  -  I 

As  the  angles  a(  and  «,  approach 
equality,  this  expression  approaches  the  limit 


•*  v, 

-,  tan  al  =  i ,     or     tan  tf,  =  -7-  . 


V* 


(23) 


CHAP.  II.]  UNIFORM  ACCELERATION—  EXAMPLES.  107 

Equation  (23)  then  gives  the  value  of  tan  afl  when  the  two  trajectories  starting  from  the 
same  point  O  with  the  same  speed  i\  have  angles  of  elevation  at  O  whose  difference 
ij  indefinitely  small. 

Substituting  this  value  of  tan  a^  in  (22)  we  obtain 


(24) 


Equation  (24)  -is  then  the  equation  of  a  curve  which  passes  through  all  the  points  in 
whicn  every  two  trajectories,  starting  from  O  at  angles  of  elevation  whose  difference  is 
indefinitely  small,  cut  each  other.  It  is  therefore  the  equation  of  the  envelope  or  curve 
which  touches  all  the  trajectories  or  parabolas  described  from  O  with  the  same  speed  vr 

Equation  (24)  is  the  equation  of  a  parabola  AC  A  whose  axis  OC  is  vertical,  whose  focus 
is  at  O,  and  whose  vertex  C  is  in  the  common  directrix  of  the  trajectories. 

With  the  given  initial  speed  •vl  the  projectile  can  reach  any  point  inside  this  envelope 
by  two  angles  of  elevation  and  two  trajectories,  as  already  proved.  It  can  reach  any  point 
in  the  envelope  by  only  one  elevation  and  path.  It  cannot  reach  any  point  outside  this 
envelope  with  any  elevation  and  the  given  speed  vr 

The  point,  therefore,  where  this  envelope  cuts  any  range  gives  the  maximum  range  in 
that  direction  for  given  vr 

v  2 
Thus  the  maximum  horizontal  range  is  found  from  (24)  by  makingjj/  =  o  to  be  -i.     This 

is  the  same  as  given  by  (15)  when  we  make  <yx  =  45°,  or  the  same  as  given  by  (17)  when  we 
make  al  —  45°  and  #  =  o. 

Questions  of  maximum  range  may  thus  be  readily  solved  by  the  equation  of  the  envelope. 

From  (2)  we  have     cos  al  =  -  —  ,      and  from  (4)     sin  a\  =  •—        —  . 

i\t  i\t 

Since  cos2  al  -+-  sin2  al  =  I,  we  have 

v?fi  ..........    (25) 


This  is  the  equation  of  a  circle  whose  radius  is  i\t  and  whose  centre  is  situated  vertically 
below  O  at  a  distance  OD  =  %ft~. 

The  circumference  of  this  circle  is  reached  in  the  same  time  by  a  projectile  starting 
from  O  with  the  velocity  vl  in  any  direction. 

In  all  equations  the  resistance  of  the  air  is  neglected.  Hence  the  following  examples 
have  little  practical  value  except  as  illustrating  the  application  of  the  equations. 

Examples,  —  (g  =  3?..i6  ft.-per-sec.  per  sec.) 

(1)  If  the  angle  of  elevation  is  30"  ,  find  the  velocity  of  projection  in  order  to  hit  a  point  at  a  distance  of 
25l  °  ft-  on  an  ascent  of  i  in  40. 

ANS.  311.5  ft.  per  «ec. 

(2)  Find  the  direction  and  magnitude  of  the  velocity  of  projection  in  order  that  a  projectile  may  reach  its 
maximum  height  at  a  point  whose  horizontal  and  vertical  distances  from  the  starting-point  are  JTO  and  yo. 


ANS.  From  equations  (7)  and  (8),     tan  <r,  =  —  °,     v,  =      l-4-J—  -t— )^. 

-TO  \  2y0 

(3")  A  gun  is  fired  horizontally  at  a  height  of  144.72  ft.  above  the  surface  of  a  lake,  and  the  initial  speed  of 
the  ball  is  1000  ft.  per  sec.  Find  after  -what  time  and  at  uihat  horizontal  distance  the  ball  strikes  the  lake  and 
with  what  velocity. 

ANS.  Time  3  sec. ;  distance  3000  ft. ;  velocity  1004.64  ft.  per  sec.  at  an  angle  below  the  horizontal  whose 
tangent  is  0.09648. 


io8  KINEMATICS  OF  A  POINT— APPLICATIONS.  [CHAP.  II. 

(4)  A  bait  is  projected  with  a  velocity  of  too  ft.  per  sec.  at  an  angle  of  75"  to  the  horizon.     Find  the  hori- 
zontal range  ;  the  range  on  a  line  jo'  to  the  horizon;  and  what  other  directions  of.  the  initial  velocity  would 
give  the  same  ranges. 

ANS.  Horizontal  range  155.5  ft. ;  range  on  30°  line  207.3  (i^3  —  0  '•  '5°  a"d  45°. 

(5)  Find  the  angle  of  elevation  in  oriier  to  hit  a  point  at  a  distance  of  1000  ft.  and  an  elevation  of  500  ft. 
with  a  velocity  of  projection  of  jio.8ft.per  sec. 

ANS.  39°  17'  or  80*  43'. 

(6)  A  body  is  projected  with  a  velocity  of  30  ft.  per  sec.  inclined  60'  to  the  horizon.    Find  the  velocity  after 
to  seconds. 

ANS.  617.3  ft-  P61"  acc'  inclined  148°  36'. 6  to  the  direction  of  the  initial  velocity. 

(7)  A  projectile  is  fired  at  an  angle  of  jo'    at  a  target  distant  1200  meters  horizontally,    (g  =  Q.&I 
meter  s-per-sec.  per  sec.)     Find  (a)  the  initial  velocity  ;  (b)  the  time  of  flight  ;    (c)  the  highest  point  of  the  tra- 
jtctory  ;   (d)  the  velocity  of  striking. 

ANS.  (a)  116.6  meters  per  sec.;  (b)  about  12  seconds;  (c)  173.21  meters;  (d)  same  as  the  initial  velocity 
making  angle  30°  below  horizontal. 

(8)  A  projectile  is  fired  with  an  initial  velocity  of  130  meters  per  sec.  from  a  point  too  meters  below  a 
target  which  is  distant  horizontally  1525  meters,    (g  =  9.8 1  meter s-per-sec.  per  sec.     Find  (a)  the  angle  of  ele- 
vation ;  (b)  the  velocity  of  striking  ;   (c)  the  time  of  flight. 

ANS.  (a)  68°  28'  50"  or  25°  16';  (b)  143.3  meters  per  sec.;  (c)  27.715  sec.  or  11.24  sec. 


CHAPTER    III. 

MOTION  UNDER  VARIABLE  ACCELERATION   IN   GENERAL.     CENTRAL  ACCELERATION. 
CENTRAL  ACCELERATION  INVERSELY  AS  THE  SQUARE  OF  THE  DISTANCE. 

PLANETARY  MOTION. 

Motion  under  Variable  Acceleration  in  General. — Let  P  be  a  point  moving  in  any  path, 
and  /its  acceleration.  The  acceleration  /can  be  resolved  into  the  tangential  acceleration 
/  and  the  central  acceleration  />. 

Let  a  be  the  angle  between  /  and  /.     Then  we  have 

?ds 
f  cos  a  =  ff.  ^^— 

Let  ds  be  the  element  of  the  path  at  P,  and  dp  the      / 
projection  of  this  element  upon  /.     Then  we  have  ^     . 

dp  =  ds  .  cos  a. 
Multiplying  these  two  equations  we  have 

fdp=ftds. 

That  is,  the  acceleration  /  multiplied  by  the  elementary  displacement  in  the  direction 
of  /is  equal  to  the  tangential  acceleration  /  multiplied  by  the  elementary  displacement  in 
the  direction  of/. 

Since  this  holds  at  every  point  of  the  path,  we  have  for  the  entire  path 

CO 


Let  vl  and  v2  be  the  velocities  at  two  consecutive  points  Pl  and  Pt  of  the  path.      Then 
for  the  indefinitely  small  time  r 


The  average  velocity  for  this  time  is  — —  — -,  and  the  distance  ds  described  is 


We  have  then 


IOQ 


1 10  KINEMATICS  OF  A  POINT— APPLICATIONS.  [CHAP.  111. 

For  the  next  two  consective  points,  Pa  and  P3 ,  we  have  then 

/*-*f*     •      '  .y' 

For  the  next  two,  P3  and  P4, 

*-*?*.  •      :'   I 

Adding  all  these,  we  have,  if  vl  is  the  initial  and  v  the  final  velocity, 


Equation  (i)  then  becomes 

?£  .........    (I) 


Equation  (I)  is  general,  as  is  seen  by  its  derivation,  whatever  the  path  and  whatever  the 
acceleration  f  in  magnitude,  and  however  it  may  change  in  direction.  It  will  therefore  in 
all  cases  give  us  the  relation  between  v  and  f  or  ft.  In  performing  the  summations  in  (I) 
we  should  take  f  or  ft  positive  when  acting  away  from  the  origin,  and  negative  when  acting 
towards  the  origin. 

ILLUSTRATIONS.  —  Thus    for   a    point   moving    in  any  path  with 

0  ___  uniform  rate  of  change  of  speed  we  have./)  constant  in  magnitude,  and 

positive  if  away  from  the  origin  O,  as  shown  in  the  figure.      Hence 

2ftds  =  ft2ds  =  ft(s  -  sj, 


where  s  is  the  final  and  s{  the  initial  distance  of  the  point  measured  along  the  path  from  any 
point  O  of  the  path  assumed  as  origin.      We  have  then,  from  (I), 


(2) 


which  holds  no  matter  whether  the  distance  from  O  is  increasing  or  decreasing.      If  increas- 
ing, s  is  greater  than  sv      If  decreasing,  s  is  less  than  sr     This  is  equation  (7),  page  92. 
•  lift  acts  towards  the  origin,  we  should  take/  negative  and  obtain 

*  =   V*  -   2ft(s   -   5l), 


and  this  again  holds  whether  the  distance  from  O  is  increasing  or  decreasing. 

Again,   for  a  point  moving  with  uniform  acceleration  in  a  straight  line,  we 
have  for  acceleration  towards  the  origin/  =  ft  negative.      Hence 


=  -f(2ds  =  -  ft(s  - 
and,  from  (I), 


If  //  —g^  this  is  the  case  of  a  body  under  the  action  of  the  earth's  attraction,  either 
projected  upwards  or  falling,  origin  at  surface  of  earth  (page  100).  If  projected  upwards,  s  is 
greater  than  sv  If  falling,  s  is  less  than  sr 


CHAP.  III.]  CENTRAL  ACCELERATION.  til 

Again,  if  the  acceleration/" is  uniform  and  does  not  coincide  with  the  velocity,  we  have 
for   origin   at    O,  as   shown   in  the  figure, /"negative  if  p 

downwards.      Hence 


—  2f .  dp  —  —  fSdp  =  —  fy, 
and,  from  (I), 

which  is  equation  (i  i)  (page  104).     We  see,  then,  that  equation  (I)  is  general  and  applies  to 
all  cases. 

Central  Acceleration. — If  the  acceleration  of  a  moving  point  is  always  directed  towards 
or  away  from  a  fixed  point  or  CENTRE  OF  ACCELERATION,  the  acceleration  is  said  to  be 
CENTRAL. 

The  velocity  v  =  Bb  of  the  moving  point  at  any  instant  is  the  resultant  of  the  velocity 
^  —  Bbl  at  the  preceding  instant,  and  of  the    change  of  velocity 
(page  75). 

But  if  b^b  is  always  directed  towards  or  away  from  a  fixed 
point,  its  moment  relative  to  that  point  is  zero.  Since  the  mo- 
ment of  the  resultant  is  equal  to  the  algebraic  sum  of  the  moments 
of  the  components  (page  89),  and  since  in  this  case  the  moment  of  one  of  the  two  compo- 
nents b^b  is  always  zero,  it  follows  that  the  moment  of  any  velocity  v  relative  to  the  centre  of 
acceleration  is  for  central  acceleration  always  constant. 

Conversely,  if  the  moment  of  the  velocity  of  a  moving  point  relative  to  any  fixed  point 
is  constant,  the  acceleration  must  be  central. 

But  the  moment  of  the  velocity  is  twice  the  areal  velocity  of  the  radius  vector  (page 
89).  Hence  in  all  cases  of  central  acceleration  the  radius  vector  describes  equal  areas  in 
equal  times. 

If  vl  is  the  known  initial  velocity  at  any  given  instant  and  /x  is  its  lever-arm,  and  if  at 
any  other  instant  the  velocity  is  v  and  lever-arm  /,  we  have  for  constant  moment 


If  rl  is  the  initial  radius  vector  and  e1  is  the 
angle  of  i\  with  rl  ,  we  have  /x  =  rl  sin  er  In  the 
same  way  /  =  r  sin  e.  Hence 

vl  —  vr  sin  e  =  vlrl  sin  et.       .      .      (3) 

If  a?!  is  the  initial  angular  velocity,  r^oo^  = 
Vi  sin  er  In  the  same  way  r&)  —  v  sin  e.  Hence 

GO       r2  f  N 

f*ca  =  r^  ,     or    —  =  - (4) 

That  is,  in  all  cases  of  central  acceleration  the  angular  velocity  is  inversely  as  the  square 
of  the  radius  vector. 

Cases  of  Central  Acceleration. — Two  of  the  most  important  cases  of  central  accelera- 
tion are : 

ist.  Central  acceleration  varying  inversely  as  the  square  of  the  distance  from  the  centre 
of  acceleration. 

2d.   Central  acceleration  varying  directly  as  the  distance  from  the  centre  of  acceleration. 

We  shall  discuss  in  this  chapter  the  first  case. 


lia  KINEMATICS  OF  A  POINT— APPLICATIONS.  [CHAP.  III. 

Central  Acceleration  Inversely  as  the  Square  of  the  Distance.— Let  rt  be  the  initial 
distance  from  the  centre  of  acceleration  O,  the  velocity  at  the  point  Pl  being  vl ,  and  r  the 
distance  to  any  point  P  at  which  the  velocity  is  v. 

Let  /„  be  the  known  acceleration  at  a  given  distance 
r0,  and  /  the  acceleration  at  any  distance  r.     Then 


This  gives  the  magnitude  of  the  central  acceleration.      If  / 
Po    is  towards  the  centre  O,  it  is  negative  and  we  have 

f-    liL 

j-       +  • 

If  /is  away  from  the  centre  O,  it  is  positive. 

(i)  ACCELERATION  TOWARDS  THE  CENTRE.  —  From  (I),  page  no,  we  have 

v2  —  v*  frz 

2f.dp=V—^-,    where    /  =  -  &£-. 

Let  />,  and  Pt  be  two  consecutive  points  at  distances  rv  and  ra  from  O.     Then,  if  the 
points  are  consecutive,  dp  =  r2  —  rl  ,  and  r2  =  r1r2.      Hence 


>*--•('.  -'•)-/*•  (7  "7). 

r,r2  \  r2        rt/ 

For  the  next  two  consecutive  points,  P2  and  P3  ,  we  have,  in  the  same  way, 


For  the  next  two  consecutive  points,  P3  and  P4  , 

>*-/*•     -• 


and  so  on. 

Summing  up,  if  rl  is  the  initial  and  r  the  final  distance,  we  have 


and  hence,  from  (I), 


If  the  distance   is  increasing,  r  is  greater  than  rr      If  decreasing,  r  is  less  than  rr 
Equation  (i)  holds  in  both  cases  if  the  acceleration  is  towards  the  centre. 
(2)  ACCELERATION  AWAY  FROM  CENTRE. — In  this  case  we  have 

/  =  +•&£, 

and  proceeding  just  as  before,  we  obtain 

/ 1        t\ 

(2) 


CHAP.  III.]        CENTRAL  ACCELERATION  INVERSELY   AS  SQUARE  OF  DISTANCE— VELOCITY.        1 13 

This  also  holds  whether  r  is  increasing  or  decreasing.  We  see  that  in  order  to  obtain 
(2)  we  have  simply  to  change  the  sign  of  fQ  in  (i). 

COR.  It  is  evident  that  the  same  equations  hold  for  motion  in  any  path,  if  we  take  the 
pole  O  in  the  path  and  measure  all  distances  along  the  patk,  if  the  tangential  acceleration  ft 
is  inversely  as  the  square  of  the  distance. 

We  have,  in  such  case, 


for  ft  towards 


=  v*  —  2frf<?\  ---  j  ; 


for/,  away  from  O,  i?  =  v?  +  2/0s0\j  -  i), (4) 

where  /0  is  the  tangential  acceleration  at  a  distance  s0,  sl  is  the  initial  distance 
from  O  to  Pr  s  the  final  distance  from  O  to  P,  all  distances  measured  along 
the  path.  O 

Central  Acceleration  Inversely  as  the  Square  of  the  Distance — Motion  Rectilinear. — 
If  the  central  acceleration  coincides  with  the  direction  of  the  velocity,  the  motion  is 
rectilinear.  In  this  case  equations  (3)  and  (4)  hold. 

This  is  the  case  of  a  body  falling  freely  under  the  action  of  gravity  at  a  very  great 
distance  from  the  earth. 

In  this  case  let  s0  =  r0  be  the  radius  of  the  earth.  The  known  acceleration  at  this 
distance  is/0  =  g.  We  have  then,  from  (3), 


(5) 
If  the  body  falls  from  rest  from  a  distance  sv  we  have  from  (5),  making  vl  =  o, 


..... 

If  the  distance  sl  is  infinite,  we  have 


for  the  velocity  acquired  by  a  body  falling  from  an  infinite  distance. 

If  the  body  is  projected  upwards,  we  have  from  (5),  making  v  =  O, 


If  the  distance  s  is  infinite,  we  have 


for  the  velocity  of  projection  which  would  carry  the  body  to  an  infinite  distance. 

If  we  take£-  =  32^  ft.-per-sec.  per  sec.,  the  mean  radius  of  the  earth  rQ  =  3960  miles, 
we  have,  making  s  —  r0  and  sl  =  r0 ,  for  the  velocity  acquired  in  falling  to  the  earth  from  an 
infinite  distance,  or  for  the  velocity  necessary  to  project  a  body  from  the  earth's  surface  to 
an  infinite  distance,  in  vacuo, 

v  =  Vt  =  ^2^r0  =  6.95  miles  per  sec. 


KINEMATICS  OF  A  POINT -APPLICATIONS. 
We  can  put  equation  (5)  in  the  form 


[CHAP.  III. 


If  the  fall  takes  place  near  the  earth's  surface,  ssl  will  be  practically  equal  to  rca  and 
we  have 

This  is  the  same  as  equation  (7),  page  92,  when  applied  to  a  falling  body  (page  101). 

Examples.— (l)  A  body  falls  to  the  earth  from  a  point  1000  miles  above  the  surface.  Find  the  speed  on 
reaching  the  surface,  neglecting  air-resistance,  and  taking  the  earth's  radius  4000  miles  andg  =  32.10. 

ANS.  v  —  3. 12  miles  per  sec. 

(2)  At  what  point  on  a  line  joining  the  centres  of  the  earth  and  moon  would  the  acceleration  of  a  body  be 
terot  ( Take  f  at  the  surface  of  the  moon  5.5  ft.-per-sec.  per  sec.;  radius  of  moon  1080  miles;  distance  between 
centres  of  earth  and  moon  240000  miles;  g  =  32.10  ft.-per-sec.  per  sec.;  radius  of  earth  4000  miles.) 

ANS.  Let  x\  =  distance  from  earth's  centre,  xt  from  moon's  centre,  A*  radius  of  earth,  r  radius  of  moon, 
/acceleration  at  moon's  surface,  f\  acceleration  at  xt  towards  earth,  /,  acceleration  at  x,  towards  moon. 

Then  £  =  —,     or      /,  =  ^.        Als 

If/,  and/,  are  equal 


Also  x,  +  xt  =  240000.     Hence  substituting  numerical  values  Xi  =  215893  miles. 

Central  Acceleration  Inversely  as  the  Square  of  the  Distance —Motion  in  a  Curve.— 
If  the  central  acceleration  does  not  coincide  with  the  velocity,  we  have  motion  in  a  curve. 

(a)  HODOGRAPH. — Since  the  acceleration  is  central  we  have,  from  equation  (4), 
page  in, 

2  _  r\GO\ 


where  r  is  the  radius  vector  for  any  point  whose  angular  velocity  is  GJ,  and  rl ,  GO^ 
known  radius  vector  and  angular  velocity  at  some  given  point. 

We  have  also,  by  assumption  (page  112), 
v. 
P 


are  the 


FIG.  (a). 


d< 


FIG.  (*). 


where  /is  the  central  acceleration  at  any  point  whose 
radius  vector  is  r,  and  /0  is  the  known  central  accelera- 
tion at  a  given  point  whose  radius  vector  is  r0. 

Let  P,  Fig.  (a),  be  a  point  which  has  the  velocity 
v  and  central  acceleration  directed  always  to  the  point 
O,  the  radius  vector  being  OP  =  r. 

Take  O',  Fig.  (£),  as  the  pole  of  the  hodograph 
(page  79)  and  draw  O'Q  parallel  and  equal  to  i>. 
Then  the  tangent  to  the  hodograph  at  Q  is  the  direc- 
tion of  the  acceleration  /at  Pand  is  parallel  to  OP 
=  r  (page  79). 

Since  the   angular  velocity  of  r,   Fig.  (a),  is  GO, 
the  angular  velocity  of  the  tangent  at  Q,  Fig.  (£),  is 
also  co. 
Let  C,   Fig.  (£),  be  the  centfe  of  curvature  of  the  hodograph,  so  that  CQ  is  perpen- 


CHAP.  III.]  CENTRAL  ACCELERATION  INVERSELY  AS    SQUARE  OF  DISTANCE— PATH.  115 

dicular  to  the  tangent  at  Q,  and  CQ  is  the  radius  of  the  hodograph.     Then,  since  the  accel- 


vj\_ii._y   \JL    5^,    x-  ig.    \t/J)    we   Hdvcy    —    ix  «^    .    u/   Ul    L' J^ 

But     G?^11-^,     and    /=*  " 


eration/of  P,  Fig.  (a),  is  the  velocity  of  Q,  Fig.  (&),  we  have/=  CQ  .  GO  or  CQ  =  ~. 


/r2 

<7<2  =  ^f-2-  =  a  constant  quantity. 
rfco, 

The  radius  of  curvature  of  the  hodograph  is  therefore  constant,  and  the  hodograph  is  a 
circle. 

(6)  PATH.  —  Draw  O'R,  Fig.  (£),  at  right  angles  to  CQ  and  therefore  parallel  to  r.  O'R 
is  the  component  of  the  velocity  v  along  the  radius  vector.  Draw  QN  perpendicular  to  O'C. 
Then  QN  is  the  component  of  v  at  right  angles  to  the  line  O'C,  or  the  diameter  A'B'  of 
the  hodograph. 

But  by  similar  triangles 

OR  _QN  O'R  _  O'C 

0'C~  CQ'  QN  ~  CQ' 

But  O'C  is  a  constant  quantity   by  construction,    and   CQ,   as    we    have    just    proved,  is 
also  constant.      Hence 

O'R 

-       =  a  constant  quantity  =  c. 


That  is,  the  ratio  of  the  velocity  along  the  radius  vector  to  the  velocity  at  right  angles  to  the 
diameter  A'B'  of  the  hodograph  is  a  constant  ratio  c. 

If,  then,  r^  and  r2  are  the  initial  and  final  values  of  r  for  an  indefinitely  short  time  t,  and 
d^  d2  are  the  corresponding  distances  of  P  from  any  line  AB,  Fig.  (a),  parallel  to  A'B'  ,  we 
have  for  the  velocity  along  the  radius  vector 


and  for  the  velocity  at  right  angles  to  AB 


Hence 

O'R 


Since  this   holds  wherever    we    take    the    line    AB,   let    us   take   the   initial  distance 


=     l  or  c 

c 


=  -r.     Then  we  have 


That    is,    the    ratio     ^  =  c  of  the   distance  r  of  the  point  P  from  a  fixed  point    O  to  its 
d 

distance  d  from  a  fixed  line  AB  is  constant. 


n6 


KlNEM/tTlCS  OF  A  POINT— APPLICATIONS. 


[CHAP.  Hi. 


This  is  the  property  of  a  conic  section  for  a  focus  at  O  and  directrix  AB. 

When,  therefore,  a  point  has  a  central  acceleration  inversely  as  the  square  of  the  radius 
vector,  it  must  move  in  a  conic  section  with  the  centre  of  acceleration  at  a  focus,  and,  as  we 
have  seen  (page  1 1 1)  in  all  cases  of  central  acceleration,  the  radius  vector  describes  equal 
areas  in  equal  times. 

Conversely,  if  the  radius  vector  describes  equal  areas  in  equal  times,  the  acceleration 
must  be  central ;  and  if  the  path  is  a  conic  section  and  the  centre  of  acceleration  is  at  a.  focus, 
the  acceleration  is  inversely  as  the  square  of  the  radius  vector. 

If  c  =  i  or  r  =  d,  then  the  path  is  a  parabola,  O'R  —  QN,  and  the  pole  O'  is  on  the 
circumference  of  the  hodograph.  The  parabola  becomes  a  straight  line  when  the  directrix 
passes  through  the  focus. 

If  c  is  less  than  unity,  or  r  is  less  than  d,  then  the  path  is  an  ellipse,  O'R  is  less  than 
QN,  and  the  pole  O'  is  inside  the  hodograph.  If  d  is  infinity,  the  path  is  a  circle  and  the 
pole  O'  coincides  with  the  centre  of  the  hodograph. 

If  c  is  greater  than  unity,  or  r  is  greater  than  d,  then  the  path  is  an  hyperbola,  O'R  is, 
greater  than  QN,  and  the  pole  O'  is  outside  the  hodograph. 

General  Equation  of  a  Conic  Section. — The  general  polar  equation  of  a  conic  section  is 
by  Analytical  Geometry, 


i  -\-  e  cos  0 ' 


where  r  is  the  radius  vector  of  any  point  making  the  angle  6  with  an  axis,   and  A  and  e  are 
constants. 

If  e  =  i,  we  have  a  parabola,  r  is  the  radius  vector  from  the  focus  F, 
as  shown  in  the  figure,  0  is  the  angle  AFP  of  the  radius  vector  with  the 
axis  measured  round  from  the  apex  A.  A  straight-line  path  is  a  special 
case  of  the  parabola  when  the  directrix  passes  through  the  focus. 

If  e  <  i,  we  have  an  ellipse*  r  is  the  radius  vector  from  a  focus  F,  and 
6  is  the  angle  AFP  of  the  radius  vector  with  the  major  axis  CA  measured  round  from  the 
nearer  vertex  A,  as  shown  in  the  figure. 

In  this  case  A  in  equation  (i)  is  the  semi-major  axis  CA, 

CF 

and  e  is  the  eccentricity,  or  e  =  ^  .    The  circle  is  a  special  case 

when  e  =  o. 

If  e  >  i,  we  have  an  hyperbola,  r  is  the  radius  vector  from 
a  focus  F,  and  6  is  the  angle  AFP  of  the  radius  vector  with  the 
major  axis  AC  measured  round  from  the  nearer  vertex  A,  as  shown  in  the  figure. 

In  this  case  A  in  equation  (i)  is  the  semi-major  axis  CA, 

and  e  is  the  eccentricity,  or  e  =  -^  . 

Central  Acceleration  Inversely  as  the  Square  of  the  Dis- 
tance—Equation of  the  Path. — We  have  for  the  path  in  gen- 
eral a  conic  section,  as  proved  on  page  115,  and  for  the  general 
polar  equation  of  a  conic  section,  as  just  explained, 


+  e  cos  9 ' 


(0 


CHAP.  III.  ]  CENTRAL  ACCELERATION  INYERSEL  Y  AS  SQUARE  OF  DISTANCE—  EQUATION  OF  PATH. 

Let  the  notation  be  as  in  the  figure.  Thus  rv  vl 
and  0j  are  the  given  radius  vector,  velocity  and  angle  for 
the  initial  position  Pv  vl  making  the  given  angle  ex  with 
rr  F)r  any  position  P  we  have  r,  v  and  6.  For  the 
apex  A  we  have  0  =  o,  rx  and  vy.  For  0  =  180  we 
have  r'x  and  vy.  Let  /0  be  the  known  central  accelera-  y 
tion  at  a  known  distance  r0. 

For   acceleration    towards  the  centre  we   have,   as 
already  proved,  page  112, 


or,  solving  for  r, 


ri 
If  in  (i)  we  make  6       0V  ^  becomes  vv  r  becomes  rlt  and  we  have 

A(i-e*)  =  ri(i+ecosOJ  .........     (3) 

Substituting  in  (i)  we  have 

^(i+'cosgj 

i  +  ^  cos  e 

If  we  make  0  =  o,  r  becomes  rx,  v  becomes  vy,  and  we  have,  from  (2)  and  (4), 


If  we  make  0  =  180°,  r  becomes  r'x,  v  becomes  v'y,  and  we  have,  from  (2)  and  (4), 


T  _  - 


We  have  also,  by  the  principle  of  equal  areas  (page  III), 

V*  =  vyr*  =  vfi  sin  ei 
From  (7)  and  (5)  we  obtain 


sn  eti     - 

•  •  •         •  • 


From  (7)  and  (6)  we  obtain 


vl  sin  et(i  -  e)  ,  , 

i  +  e  coslT"   '     '     '     '     '  .     '     '     ' 


Substituting  (8)  in  (5),  and  (9)  in  (6),  we  have,  after  reduction, 

22i         ei        €  cos  fy  =  r^2  sin2  e/i  +  r)2  +  (i  +  e  cos 
e  cos     )  =  r2  sin2  ei  -  e?  +  (l  +  e  cos 


n8  KINEMATICS  OF  A  POINT-APPLICATIONS.  [CHAP.  III. 

From  these  two  equations  we  obtain 


r.?'.2  sin2  e. 

-'   '  L 


7oro 


(10) 


Substituting  (10)  and  (i  I  in  (4)  we  have  for  the  general  equation  of  the  path,  in  terms  of 
known  quantities, 

r*v?  sin2  e1 
7>o2 


-f-  cos 


(12) 


where  /0  is  the  known  central   acceleration  at  a  known  distance  r0,  vl  is   the  given  initial 
velocity  at  a  given  distance  r,,  making  the  angle  el  with  rp  and  r  is  the  radius  vector  for  any 
angle  6.     The  quantity  under  the  radical  is  the  value  of  <?. 
The  path  will  be  a  parabola  when  e  =   i,  or  when 


v?  =  -  -2-,    or     Vl  = 


The  path  will  be  an  ellipse  when  e  <  i,  or  when 


.    or 


The  path  will  be  an  hyberbola  when  #•  >   i,  or  when 

*t»  >  3&i,    or     V|  >  y/^'. 


We  see,  then,  that  the  form  of  the  path  depends  solely  upon  the  magnitude  of  tJie  initial 
velocity  and  not  upon  its  direction,  tliat  is  simply  upon  the  initial  speed. 
Prom  (2),  page  117,  making  v  =  o  and  r  infinity,  we  have 


for  the  velocity  of  projection  which  would  carry  the  point  from  r,  to  rest  at  an  infinite  dis- 
tance, or  which  the  point  would  acquire  in  moving  from  rest  at  an  infinite  distance  to  ;,. 

The  path  is  an  ellipse,  parabola  or  hyperbola  according  as  ?',  is  less  than,  equal  to  or 
greater  than  this. 

For  acceleration  away  from  the  centre  we  should  change  the  sign  of /0  in  equation  (2), 
as  pointed  out  page  112.  The  value  of  e,  equation  (n),  in  such  case  is  always  greater  than 
unity.  Hence  for  acceleration  away  from  the  centre  the  path  is  always  an  hyperbola. 


CHAP.  III.] 


PATH  A  PARABOLA. 


119 


Path  a  Parabola.— If  the  path    is   a  parabola,  e=  i,  v?  =  ^J>_  and  equation    (12^ 

ri 

becomes 

2r,  sin2  e. 


i+costf* (13) 

We  have  also,  from  (10),  for  the  angle  Ol  of  the  initial  radius  vector  ^  with  the  axis 
cos  ^  =  2  sin2  e1  —  i.     .      .      .      (I4)  Fig.  (a).  Fig.  (S). 


The  path  is  then  completely  determined. 
The  equation  referred  to  the  axis  and  vertex  as 
origin  is 


=  4     sn 


Path  an  Ellipse  or  Hyperbola.—  If  the  path 
is  an  ellipse  or  hyperbola,  the  eccentricity  in  either 
case  is  given  by  (11),  viz., 


.  /'.        *•.*.'  sin'  et(r,'V~2/0f0').  .......     (16) 

V  ~7W~ 

From  (16),  (3)  and  (10)  we  have  for  the  semi-major  axis  of  the  ellipse  or  the  semi-trans- 
verse axis  of  the  hyperbola 

A  =  ±-g&—l  ...........     (17) 

2Soro   —  r\v\ 

where  the  (-{-)  sign  is  taken  for  the  ellipse  and  the  (  —  )  sign  for  the  hyperbola. 

We  see  that  this  is  independent  of  the  angle  el  or  the  direction  of  the  initial  velocity  v^  ,  and 

is  the  same  for  *\  and  rl  constant  in  magnitude  no  matter  what  the  direction. 

The  semi-conjugate  axis  is  given  by  B  —  A^\  —  e~  for  the  ellipse  or  B  =  A^e2  —  I   for 
the  hyperbola.      From  (16)  and  (17)  we  have,  then, 

n  =     __^Liin  ei  _  /,8N 

"  *±W*f~ri*ff- 

where  the  (-J-)  sign  is  taken  for  the  ellipse  and  the  (  —  )  sign  for  the  hyperbola. 

We  have,  from  (10),  'for  the  angle  0l  of  the  initial  radius  vector  rl  with  the  axis 

cos  0,  =  _  rjrf  si"2  6*  --^  (19) 

^/oV  +  ^i*  sin'  €i(^'*  -  2/o'o2) 

The  elliptic  or  hyperbolic  path  is  thus  determined. 

The  equation  of  the  path  referred  to  the  centre  and  axes  is 

A*B\        .........     (20) 


where  the'(+)  sign  is  taken  for  the  ellipse  and  the  (—  )  sign  for  the  hyperbola. 


120  KINEMATICS  OF  A  POINT—  APPLICATIONS.  [CHAP.  III. 

If  the  path  is  a  circle,  we  have  e  =  o,  e,  =  90°,  /0  =/,  r^  =  r,  v^  =  vr,  and  hence, 
from  (16), 


as  should  be  (p.  79). 

Planetary  Motion  —  Kepler's  Laws.  —  By  long  and  laborious  comparison  of  the  observa- 
tions which  TYCHO  BRAHE  had  made,  through  many  years,  of  the  planets,  especially  of  Mars, 
KEPLER  discovered  the  three  laws  of  planetary  motion  which  are  known  as  KEPLLR'S  LAWS. 
He  gave  these  laws  simply  as  the  expression  of  facts  which  seemed  established  by  the 
observations. 

The  three  laws  are  as  follows: 

I.    Tke  planets  describe  ellipses,  the  sun  occupying  one  of  the  foci. 

II.    The  radius  vector  of  each  planet  describes  equal  areas  in  equal  times.      This  is  known 
as  the  law  of  equal  areas. 

III.  The  squares  of  the  periodic  times  of  the  planets  are  proportional  to  the  cubes  of  their 
mean  distances  from  the  sun. 

The  second  law,  as  we  have  seen  (page  in),  is  a  necessary  consequence  of  tf//  central 
acceleration. 

The  first  law,  as  we  have  just  seen,  follows  if  the  central  acceleration  is  inversely  as  the 
square  of  the  radius  vector. 

The  third  law  is  also  a  direct  consequence  of  such  central  acceleration,  as  we  shall  see 
in  the  next  article. 

Verification  by  Application  to  the  Moon.  —If  col  and  <w2  are  the  mean  angular  velocities 
of  two  planets  and  Tl  ,  T2  their  periodic  times,  then 

27T  27T 

0,     =r  »     =          . 


We  have  also  for  the  mean  central  accelerations,  if  rl  ,  r.2  are  the  mean  radius  vectors, 

1  2 

Hence 

•fi  —   ?j£— • 

But  if  the  accelerations  are  inversely  as  the  squares  of  the  radjus  vectors,  we  have  also 

/       r 2 

Hence  we  obtain 

2  =  5£ 

This  is  Kepler's  third  law,  and,  as  we  see,  it  is  a  direct  consequence  of  central  accelera- 
tion, varying  inversely  as  the  square  of  the  radius  vector. 

Assuming  Kepler's  third  law,  Newton  was  kd  directly  to  this  conclusion.  He  tested  it 
by  application  to  the  moon,  as  follows; 


CHAP.  III.]  PLANETARY  MOTION.  121 

The  moon  moves  in  a  sensibly  circular  orbit,  the-centre  being  at  the  centre  of  the  earth. 

Let  the  mean  radius  of  the  earth  be  r0,  the  acceleration  at  the  surface  beg,  the  distance 
from  centre  of  earth  to  centre  of  moon  be  r,  and  acceleration  of  moon  be/.  Then,  by 
the  hypothesis  of  inverse  squares,  we  have 

f:g:  :r»:r»t      or    /  =  ^g. 

Inserting  r0  =  4000  miles,  r  =  240000  miles,  and  g  —  32.2  ft.-per-sec.  per  sec.,  we 
have  by  hypothesis  for  the  moon's  central  acceleration 

(400o)2 
/=(24oooo)a  X  32.2  =  0.0089  ft.-per-sec.  per  sec. 

But  the  speed  of  the  moon  in  its  orbit  is  v  =  3375  ft.  per  sec. 
Its  radial  acceleration  is  then  in  reality  (page  79) 


240000x5280  =  °-°°89  ft-per'sec-  per  sec 


Newton's  hypothesis  of  inverse  squares  is  thus  verified  in  the  case  of  the  moon. 

As  we  have  seen  (page  116),  the  path  in  general  for  central  acceleration  varying 
inversely  as  the  square  of  the  distance  is  either  an  ellipse,  a  'parabola  or  an  hyperbola 
Instances  of  all  these  paths  are  found  in  the  solar  system.  Thus  the  planets  and  satellites 
move  in  elliptic  orbits,  while  comets  have  paths  elliptic,  parabolic  or  hyperbolic. 

Velocity  of  a  Planet  at  any  Point  of  its  Orbit.  —  We  have  from  equation  (2),  page  112, 


which  gives  the  velocity  for  any  distance  r  when  vl  and  rl  are  given. 

From  equation  (i),  page  1 16,  and  equation  (12),  page  118,  we  have 


From  page  in,  vl—  rv  sin  e,    or     &  -.          — ±— *,    where  /is  the  lever-arm  of  v. 

^2(j     _  et\r 

From  Analytical  Geometry,  for  an  ellipse/2  = -^ —     —  •     Hence  we  have 


or,  from  (2  i), 


Equation  (23)  gives  the  velocity  for  any  distance  r  if  the  semi-major  axis  A  is  known. 
COR.  I. — We  see  that  the  velocity  is  greatest  where  r  is  least,  or  at  perihelion,  and  least 
where  r  is  greatest,  or  at  aphelion. 

COR.  2. — If  a  point  moves  in  a  circle  of  radius  r  with  velocity  v' ,  its  radial  acceleration 

is  V—  (page  79). 


122  KINEMATICS  OF  A  POINT-APPLICATIONS.  LCHAP'  11L 

If  this  acceleration  is  equal  to  the  acceleration  of  the  planet,  we  have 

"'2_/oV- 

7asV' 

or,  from  (21), 


Therefore,  from 

That  is,  the  square  of  the  speed  in  the  ellipse  is  to  the  square  of  the  speed  in  the  circle  as  the 
distance  of  the  planet  from  the  unoccupied  focus  is  to  the  semi-major  axis. 

COR.  3. — If  rl  is  the  perihelion  distance  and  r2the  aphelion  distance,  we  have,  from  (23), 

(or  r=n,  ^^       . 

r  2 -^00      1** 

2 

while  for  r  =  A  we  have 

"  ~^f " 

That  is,  the  speed  at  the  extremity  of  the  minor  axis  is  a  mean  proportional  between  the  speeds 
at  perihelion  and  aphelion. 

Periodic  Time. — The  moment  i\r^  sin  e,  is  equal  to  twice  the  areal  velocity  of  the 
radius  vector  (page  89),  and  this  areal  velocity,  as  we  have  seen  (page  ill),  is  constant. 
Twice  the  area  of  an  ellipse  is  2nA*  \> ' \  —  e2.  Hence  the  periodic  time  is 


sin  e 


(24) 


or,  substituting  the  values  for  A  and  (e},  equations  (17)  and  (16), 

^       _  (25) 

' 


From  (25)  we  see  that  the  periodic  time  is  independent  of  the  angle  ev  or  the  direction 
of  the  initial  velocity  ?',. 

We  can  write  (24)  in  the  form 


*r*  sin3 


T*       "    A(\  -f3)  ' 
or,  from  (21), 

?',2r,5  sin2  e. 


CHAP.  III.]  PL/tNETARY  MOTION.  123 

But,  by  Kepler's  third  law,  we  have  for  two  different  planets  of  periodic  times  T  and  Tl 

T%  /J3  /13  A  3 

-L  Si  fi.  jfi. 


A*> 

Hence 


is  a  constant  quantity  for  all  the  planets. 

Now/o  is  the  acceleration  at  a  distance  r0 ,  and,  since  the  acceleration /at  any  distance 
r  is 

•^"~    r  ;' 

we  see  that /Qr02  is  the  magnitude  of  the  acceleration  at  the  distance  unity,  or  r  —  I. 

Hence  it  follows,  from  Kepler's  third  law,  that  for  all  the  planets  the  acceleration  would 
be  tJu'  same  at  the  same  distance  from  tJie  sun.* 

Value  of  f0  for  Planetary  Motion. — In  all  our  equations  for  central  acceleration  we  see 
that  it  is  necessary  to  know  the  acceleration /0  at  some  known  distance  r0. 

It  will  be  proved  hereafter  (page  206)  that  if  M  is  the  mass  of  the  sun  and  m  the  mass 
of  a  planet,  the  value  of  /0  at  a  distance  r0  equal  to  the  mean  radius  of  the  eartJi  is  given  by 

M  +  m 


where  m0  is  the  mass  of  the  earth  and  g  the  mean   acceleration   of   a  body  at   the    earth's 
surface. 

If  the  two  bodies  are  the  earth  and  a  small  body  of  mass  ;»',  then  /0  =  -         —  g,    or, 

mo 

since  m  is  insignificant  with  respect  to  ;«0  ,  /„  =  g.     If,  in  the  preceding  article,  we  had  used 
the  value  of  f0  given  by  (i),  we  should  have  obtained 

M+m  47t*As  M+m, 

—  —  gr*  =  —~2-      and        —  --  -grf  = 

mQ  l  11  o 

Hence 

r2  _  Jf+ml   A^ 
'  A     ' 


We  see,  then,  that  Kepler's  third  law  is  not  strictly  exact.  The  value  of  /Or0a,  or  the 
acceleration  at  units  distance,  is  not  strictly  constant.  The  more  accurate  expression  is  that 
the  squares  of  the  periodic  times  are  directly  as  the  cubes  of  the  semi-major  axes  and 
inversely  as  the  sum  of  the  masses  of  sun  and  planet. 

The  error  from  this  source  is  insignificant,  the  mass  of  Jupiter,  the  largest  of  the  planets, 
being  less  than  a  thousandth  part  of  that  of  the  sun. 


"Of  all  the  laws,"  says  Sir  John  Herschel,  "to  which  induction  from  pure  observation  has  ever  con- 
ducted man,  this  third  law  of  Kepler  may  justly  be  regarded  as  the  most  remarkable,  and  the  most  pregnant 
with  important  consequences.  When  we  contemplate  the  constituents  of  the  planetary  system  from  the  point  of 
view  which  this  relation  affords  us,  it  is  no  longer  mere  analogy  which  strikes  us.  no  longer  a  general  resem- 
blance among  them  as  individuals  independent  of  each  other,  and  circulating  about  the  sun.  each  according  to 
its  own  peculiar  nature,  and  connected  with  it  by  its  own  peculiar  tie.  The  resemblance  is  now  perceived  to 
be  a  true  family  likeness;  they  are  bound  up  in  one  chain;  interwoven  in  one  web  of  mutual  relation  and 
harmonious  agreement;  subjected  to  one  pervading  influence,  which  extends  from  the  centre  to  the  farthest 
limit  of  that  great  system,  of  which  all  of  them,  the  earth  included,  must  henceforth  be  regarded  as  members." 
—  Outlines  oj  Astronomy, 


124 


KINEMATICS  OF  A  POINT— APPLICATIONS. 


[CHAI-.  111. 


The  motion  of  translation  of  the  planets  is  not  affected  by  their  rotation  on  their  axes, 
and  we  may  treat  them  as  material  points  at  their  centres  of  mass,  so  far  as  translation  is 
concerned. 

The  centre  of  mass  of  the  sun  is  not  strictly  a  fixed  point,  but  both  sun  and  planet 
move  in  orbits  about  a  common  centre  of  mass.  The  sun  is  also  affected  by  the  other 
planets,  and  the  planets  by  each  other. 

The  attraction  of  the  planets  for  each  other  sensibly  modifies  their  orbits. 

Kepler's  laws  are  thus  approximate.  If  we  had  but  two  bodies,  one  fixed  and  the  other 
free  to  move,  then  Kepler's  first  two  laws  would  be  accurate,  and  the  third  would  approach 
accuracy  as  the  mass  of  the  moving  body  becomes  insignificant  with  respect  to  the  mass  of 
the  other. 

Examples. — (i)  Find  the  speed  and  periodic  time  of  a  body  moving  in  a  circle  at  a  distance  from   the 
earth's  centre  of  n  times  the  earth's  radius,  the  central  acceleration  being  inversely  as  the  square  of  the  distance. 

ANS.  v  =  .  n£.        r  =  •. 


(2)  A  body  at  a  distance  of  rt  from  the  centre  of  the  earth  is  projected  in  a  direction  which  makes  an  angle 
of  e,  =  bo"  with  rt  with  a  speed  v\ ,  which  is  to  the  speed  acquired  by  falling  from  an  infinite  distance  as  i  to. 
^3.  Find  the  path,  the  major  axis,  eccentricity  and  periodic  time. 

I  I^Ti 

ANS.  We  have  e,  =  60°,  sin  e,  =     — ,  vt  =  — -=>      l:5— ,  A  =  if. 
\4  4/3     \    '*     • 

The  path  is  an  ellipse.     From  (17),  iA  =  £r,.     From  (16),  e  =     |_i. 

l~  3 

From  (25)  T—  3-^J.     P'_'(  where  ru  is   the    radius  of    the  earth. 

4'*o    ^V  JT 

From  (12)  we  have,  by  making  0  =  o,  the    perihelion   distance  FA 

equal  to  -  r\  (  i  —      I— ).   From  (19)  we  have  for  the  angle  AFPi  —  61 


The  centre  of  the  earth  is  at  the  focus  F. 
(3)  In  the  preceding  example,  let  e,  =90°,  so  that  tfo  body  is  projected  in  a  direc- 
tion at  right  audits  to  r  . 

ANS.  From  (ij),2A  =  —r,,  as  before. 


From  (i 6),  e  =  — . 
From  (25),  T=  3- 


,      y 


'.  as  before. 


From  (12)  we  have,  by  making  0  =  o,  the  perihelion   distance  FA  equal 

to  -r,. 

2 

From  (19)  we  have  for  the  angle  AFPi 

cos  0i  =  -  i,    or    0,  =  180*. 


CHAPTER   IV. 

CENTRAL  ACCELERATION  DIRECTLY  AS  THE  DISTANCE.     HARMONIC  MOTION. 

Harmonic  Motion. — The  motion  of  a  point  moving  in  any  path,  the  acceleration  being 
always  directed  towards  a  fixed  point  and  varying  directly  as  the  distance  from  that  point,  is 
called  HARMONIC  motion. 

If  the  path  is  a  straight  line,  the  motion  is  SIMPLE  HARMONIC ;  if  the  path  is  a  curve, 
the  motion  is  COMPOUND  HARMONIC. 

The  vibrations  of  such  bodies  as  a  tuning-fork  or  a  piano-wire  are  approximate  ex- 
amples of  such  motion,  hence  the  term  "harmonic"  The  vibrations  of  an  elastic  body, 
such  as  a  spring  or  the  air,  are  also  examples  of  such  motion. 

It  is  also  (page  207)  the  motion  of  a  body  under  the  action  of  gravitation,  within  a 
homogeneous  sphere. 

The  motion  of  the  piston  of  a  steam-engine  when  moved  by  a  crank  and  connecting- 
rod  approximates  the  same  motion  if  the  rotation  of  the  crank  is  uniform,  the  approxi- 
mation being  closer  the  longer  the  connecting-rod. 

Harmonic  Motion — Velocity. — Let/0  be  the  known  acceleration  at  a  given  distance  r0, 
and  /  the  acceleration  at  any  distance  r.  Then 

f:/0::r:r0,     or    /  =  -/o- 
ro 

This  gives  the  magnitude  of  the  central  acceleration. 

If /is  towards  the  centre  of  accelefation,  it  is  negative  and  we  have 


From  (I),  page  no,  we  have 

2f.  dp  — L,     where    /= /0, 

2  ro 

Let  Pl  and  P2  be  two  consecutive  points  at  distances  rl  and  rz  from  the  centre  of  accel- 
eration.    Then  if  the  points  are  consecutive,  dp  =  rz  —  rr  and  r  =  — -.      Hence 


For  the  next  two  consecutive  points,  Pz  and  P3 ,  we  have  in  the  same  way 


125 


126  KINEMATICS  OF  A  POINT-APPLICATIONS 

For  the  next  two  consecutive  points,  P3  and  P4 , 


[CUAP.   IV. 


and  so  on. 

Summing  up,  if  r,  is  the  initial  and  r  the  final  distance,  we  have 


Hence 


If  the  distance  is  increasing,  r  is  greater  than  rl ;  if  decreasing,  r  is  less  than  rr 
Equation  (i)  holds  in  both  cases. 

COR.  It  is  evident  that  the  same  equation  holds  for  motion  in  any  path  if  we  take  the 
centre  of  acceleration  at  some  fixed  point  of  the  path  and  measure  all  distances  along  the 
path,  if  the  tangential  acceleration  ft  is  directly  as  the  distance. 

We  have,  in  such  case, 


(2) 


where /0  is  the  tangential  acceleration  at  a  distance  J0,  sl  is  the  initial  and  s  the  final  distance 
from  the  fixed  point  of  the  path,  all  distances  measured  along  the  path. 

Simple  Harmonic  Motion. — If  the  central  acceleration  varies  directly  as  the  distance 
from  the  centre  of  acceleration  and  coincides  with  the  direction  of  the  velocity,  the  motion 
is  rectilinear  and  we  have  simple  harmonic  motion. 
In  this  case  equation  (2)  still  holds. 

Let  a  point  Q  move  with  uniform  speed  r<&  in  a 
circle  of  radius  CQ  =  r  with  uniform  angular  speed  w- 
Then  the  radial  acceleration^  is  towards  the  centre  C  and 
equal  to 

/P  =  roo\ 

The  projection  of/p  upon  a  diameter  CA  is  rat .  cos  QCA. 

A  But  r  cos  QCA  is  the  distance  CP  =  s,  if  P  is  the  projection 
of  Q  upon  the  diameter.  The  acceleration  of  P  along  the  diameter  towards  C  is  then  jci2, 
or  directly  proportional  to  the  distance  s.  The  motion  of  P  is  then  simple  harmonic. 

Hence,  if  a  point  Q  move  in  a  circle  with  uniform  speed,  its  projection  P  upon  a  diameter 
moves  -with  simple  harmonic  motion  along  the  diameter. 

If  v\  is  the  initial  velocity  at  the  initial  position  P^  so  that  CPl  =  slt  equation  (2)  becomes 


(3) 


CHAP.    IV.]  SIMPLE  HARMONIC  MOTIONS.  127 

The  point  P  starts  from  rest  at  A  at  the  distance  CA  =  r.      If  then  we  make  v  =  o  in 
(3)  and  s  =  r,  we  have 

/o 
so 

and  substituting  this  in  (3), 

v*=—  (r2  —  .y2) (4\ 

V 

We  see  from  (4)  that  the  velocity  increases  as  the  distance  s  decreases  till  P  arrives  at 
C,  where  the  velocity  is  a  maximum  and  equal  to 


hence  «  =  ^       .......     (5) 

Then  the  velocity  decreases  and  finally  becomes  zero  when  P  arrives  at  A'  at  the  distance 
—  r  on  the  other  side  of  C. 

From  A  to  A'  is  called  a  VIBRATION,  from  A  to  A'  and  back  to  A  is  called  an  OSCILLA- 
TION. The  radius  r  is  called  the  RANGE  or  AMPLITUDE  of  an  oscillation. 

PERIODIC  TIME.  —  The  time  from  A  back  to  A,  or  the  time  of  an  oscillation,  is  called 


the  PERIODIC  TIME.     Since  the  uniform  speed  is  ra>  —  r-,  the  periodic  time  is 


SQ 
If  /is  the  acceleration  at  any  distance  s,  we  have  for  harmonic  motion 

== 


Hence 


The  periodic  time  depends,  then,  only  upon  the   constant  ratio  -?  =  —3,  and  is  independent 

of  the  range  r  or  amplitude  of  oscillation.      For  this  reason  the  oscillations  are  said  to  be 
ISOCHRONOUS,  or  made  in  equal  times,  no  matter  what  the  range  or  amplitude. 

COR.  —  Since  the  motion  of  a  body  under  the  action  of  gravity  in  a  homogeneous 
sphere  is  harmonic  (page  207),  if  we  put  g  for/0,  and  r0,  the  radius  of  the  earth,  for  j0,  we 
have  from  (3),  for  a  body  falling  under  the  action  of  gravity  in  a  well  or  shaft  assuming 
the  earth  to  be  a  homogeneous  sphere  and  neglecting  the  resistance  of  the  air, 


128  KINEMATICS  Of-'  A  POINT-APPLICATIONS.  [CHAP.  IV. 

This  can  be  written 


If  the  fall    takes    place    near  the  surface,   for  a  short  distance  compared  to  r0,   we  have 
sl  -f-  s  practically  equal  to  2r0,  and  hence 

which  is  the  same  as  for  uniform  acceleration  g  (page  101). 

We  obtained  the  same  result  (page  101)  for  a  body  external  to  the  earth.  The  equa- 
tions of  page  101  hold  good,  then,  in  all  practical  cases,  whether  the  fall  takes  place  above 
the  earth  or  within  the  earth,  for  small  fall  near  the  surface,  neglecting  air  resistance. 

Epoch Phase. — If  Pl  is  the  initial  position  or  the  position  of  P  at  zero  of  time,  the 

time  of  passing  from  A   to  Pl  is  called   the  EPOCH.     The 
9  epoch  may  also    be    defined  with   reference    to  the  auxil- 

iary circle  as  the  angle  ACQV  in  radians.  This  is  the 
I  epoch  in  angular  measure.  The  epoch  in  angular  meas- 
ure is,  then,  the  angle  described  on  the  auxiliary  circle 
in  the  interval  of  time  defined  as  the  epoch. 


The  epoch  locates  the  initial  position  of  P. 

The  fraction  of  the  periodic  time  in  passing  from  A  to  any-  position  "P  is  called  the 
PHASE.  Measured  on  the  circle  is  the  ratio  of  the  angle  ACQ  radians  to  2n  radians. 

The  phase  locates  the  position  of  P  at  any  instant. 

It  therefore  varies  with  the  time  or  with  the  position  of  P.  The  phase  at  the  initial 
position  Pl  multiplied  by  2n  gives,  then,  the  epoch  in  angular  measure,  and  multiplied  by 
the  periodic  time  gives  the  epoch  in  time. 

Examples. — (i)  A  point  whose  motion  is  simple  harmonic  has  velocities  20  and  25  m.  per  sec.  at  distances 
10  and  8  m.  from  the  centre  of  acceleration.     Find  the  period  and  acceleration  at  units  distance. 

ANS.  We  have  400  =  f-(r*  —  100)  and  625  =  ^Vr1  —  64).    Therefore 
st  Jo 


=  /£ 

We  have  also  *-ss  -^-,or/=  j-.     Making  s  =  i,/=  ±  —J  =  ±  6.25  ft.-per-sec.  per  sec. 

(2)    The  period  of  a  simple  harmonic  motion  is  20  sec.  and  the  maximum  velocity  is    to  ft.  per  sec. 

bo 
Find  the  velocity  at  a  distance  of  —  //.  from  the  centre. 

ANS.  We  have  —  =  =  20  sec.    Therefore  -  =  —  .    When  s  =  o,  v  =  10  and  100  =  —  r>,  orr  =  *°°ft. 
//.  st      100  100 

*    J. 


Hence 


Vi  =  ^(1?   -Sr)-    or    "  =  8ft.persec 


(3)  Find  the  mean  speed  of  a  point  in  simple  harmonic  motion  during  the  time  of  moving  from  one  to  the 
other  extremity  of  its  range,  tts  maximum  speed  being  s  ft.  Per  sec. 


CHAP.  IV.] 


COMPOUND  HARMONIC  MOTION. 


I29 


& 


ANS.  The  distance  is  2r.    The  time  .    The  mean  speed  — ^L.    When  s  =  o,  we  have 

t/ 4L  *#*• 

Jo 

25  =  —  r*,    or 


Therefore  the  mean  speed  is  —  ft.  per  sec. 


(4)  If  T  is  the  period  and  r  the  amplitude  of  a  simple  harmonic  motion,  v  the  'velocity  and  s  the  distance 
from  the  centre  at  any  instant,  show  that 


(5)  ^4  /0/«/  has  simple  harmonic  motion  whose  period  is  4  min.  12  sec.     Find  the  time  during  which  its 
phase  changes  from  ^  to  \  of  a  period* 
ANS.   21  sec. 

Compound  Harmonic  Motion. — If  the  central  acceleration  varies  directly  as  the  distance 
from  the  centre  of  acceleration  and  does  not  coincide  with  the  direction  of  the  velocity,  we 
have  motion  in  a  curve  and  the  motion  is  COMPOUND  HARMONIC. 

Any  Compound  Harmonic  Motion  may  be  Resolved  into  Two  Simple  Harmonic  Motions 
at  Right  Angles. — Let  C  be  the  centre  of  acceleration,  and  P  the  position  of  the  moving 
point  at  any  instant.  Let  the  velocity  v  of  P  make  an  angle  a. 
with  the  axis  of  X,  and  let  the  motion  of  P  be  harmonic  so  that 

the  acceleration  of  P  is  —r,  where  /0  is    the  acceleration  at  a 
ro 

known  distance  r0,  and  r  is  the  distance  CP. 

The  velocity  v  may  be  resolved  into  v  cos  a  and  V  sin  a  in 
the  directions  CX  and  CY,  and  the  acceleration  may  be  resolved  c 


•^rcosPCA  or^CA, 
rn  rn 


and 


—  r  cos  PCB  or  -CB,  in  the  same  directions. 


0  ' 0  '9  '0 

The  component  accelerations  are  therefore  directly  as  the  distances  CA  and  CB,  and 
the  component  velocities  are  in  the  directions  of  CA  and  CB.  The  compound  harmonic 
motion  of  P,  whatever  the  direction  of  the  velocity  v,  is  therefore  the  resultant  of  two  simple 
harmonic  motions  in  the  lines  CA  and  CB  at  right  angles. 

If,  then,  any  compound  harmonic  motion  is  resolved  into  two  components  at  right  angles, 
the  component  motions  are  rectilinear  harmonic. 

Conversely,  the  resultant  of  two  rectilinear  harmonic  motions  at  right  angles  is  a  com- 
pound harmonic  motion. 

Composition  of  Simple  Rectilinear  Harmonic  Motions  in  Different  Lines. — Let  the 

point  Q  move  in  a  circle  AQA'  of  radius  r  =  CA  =  CQ 
with  a  constant  angular  velocity  GO.  Then  the  motion 
of  the  projection  P  in  the  line  AA'  is  simple  harmonic 
(page  126). 

Let  the  point  Ql  move  in  the  circle  CBQl  of  radius 
rl  =  CB  =  CQ^,  with  constant  angular  velocity  col. 
Then  the  motion  of  the  projection  Pl  in  the  line  CB 
is  simple  harmonic.  Let  the  angle  BCA  between  the 

planes  of  the  circles  be  a. 
FIG.  i. 


130  KINEMATICS  OF  A  POINT—APPLICATIONS.  [CHAP.  IV. 

Let  the  time  count  from  the  instant  when  Ql  is  at  Bt  so  that  the  epoch  of  1\  is  zero 
(page  128).  At  this  instant  let  the  epoch  of  P  be  e.  Then  e  is  the  difference  of  epoch,  or, 
in  angular  measure,  the  angle  of  Q  above  or  below  A  at  the  beginning  of  the  time.  In  any 
time  /,  Ql  will  have  moved  from  B  through  the  angle  aoj  measured  from  CB,  and  Q  through 
the  angle  cat  ±  e  measured  from  CA. 

By  the  preceding  article  we  can  resolve  the  harmonic  motion  of  Pt  into  a  simple 
rectilinear  harmonic  motion  at  right  angles  to  CA,  and  another  along  CA. 

The  displacement  of  /*,  from  C  for  any  time  /  is  rt  cos  (<»/),  and  tn's  displacement  may 
be  resolved  into  rt  cos  a  cos  (&J)  along  CA,  and  rt  sin  a  cos  (c^/)  perpendicular  to  CA. 
The  displacement  of  P  from  C  in  the  same  time  /  is  r  cos  (oat  ±  e). 

If  a  point  undergoes  these  displacements  simultaneously,  its  resultant  displacement 
along  CA  will  be 

x  =  r  cos  (cot  ±  e)  -f-  rl  cos  a  cos  (<»/), (i) 

and  perpendicular  to  CA 

• 

y  =  rv  sin  a  cos  (oV), (2) 

The  equation  of  the  curve  in  which  the  point  moves,  referred  to  rectangular  co-ordinates 
with  C  for  the  origin,  will  then  be  obtained  by  combining  (i)  and  (2)  so  as  to  eliminate  /. 
Such  combination  (page  129)  gives  always  compound  harmonic  motion  about  C,  the  radius 
vector  from  C  passing  over  equal  areas  in  equal  times  (page  n  i). 

Equations  (i)  and  (2)  enable  us,  then,  to  find  the  curve  resulting  from  the  combination 
of  any  two  simple  rectilinear  harmonic  motions  inclined  at  any  angle  a. 

If  the  component  motions  are  at  right  angles,  a  —  90°.  If  the  amplitudes  are  equal, 
r  =  rr  If  the  periods  are  equal,  GO  =  GHI  ,  the  difference  of  epoch  is  constant,  and,  since 
the  epoch  equals  the  product  of  the  phase  at  zero  of  time  by  2?r  radians  (page  128),  when 
the  periods  are  equal  the  difference  of  phase  is  constant.  When,  then,  the  periods  are 
equal  and  e  =  o,  or  the  epochs  are  equal,  the  phases  are  also  equal  at  any  instant. 

Two  Component  Simple  Harmonic  Motions  in  Different  Lines  with  the  Same  Period. 
— In  this  case  <y  =  oov  and  e  is  constant,  or  the  difference  of  epochs  is  constant  and  difference 
of  phase  at  any  instant  is  constant. 

We  have  then,  from  (i)  and  (2), 

x  =  r  cos  (wt  -{-  e)  -|-  rv  cos  a  cos  (cot),          y  =  r  sin  a  (cos  cat). 
Combining  these  two  equations  by  eliminating  oat,  we  have 

(r,8  sin2  ot)x*  —  2 r,  sin  a(r  cos  e  -|-  r,  cos  ct)xy  -f  (r2  -f  2rr,  cos  e  cos  a  -f  r,2  cos2  at^y* 

=  r^r?  sin2  a  sin2  e.      (3) 

This  is  the  equation  of  an  ellipse  referred  to  its  centre  and  rectangular  axes. 

Hence  if  a  point  has  two  component  simple  harmonic  motions  in  any  directions,  of  any 
amplitudes,  and  any  difference  of  epoch,  if  the  periods  of  the  two  components  are  the  same, 
the  resultant  motion  of  the  point  will  be  harmonic  in  an  ellipse,  the  centre  of  acceleration  at 
the  centre  of  the  ellipse.  The  areal  velocity  of  the  radius  vector  about  the  centre  is 
constant  (page  1 1 1). 

Such  motion  is  called  elliptic  harmonic  motion.  Elliptic  harmonic  motion,  then,  is 
compound  harmonic  motion  when  the  periods  of  the  components  are  the  same. 


CHAP.  IV.] 


COMPOUND  HARMOML   MOTION. 


Equation  (3)  gives  all  cases  of  compound  harmonic  motion  for  equal  periods  of  the 
components. 

It  will  be  instructive  to  derive  from  it  special  cases. 

(a)  Two  COMPONENT  SIMPLE  RECTILINEAR  MOTIONS  IN  DIFFERENT  LINES  WITH  THE 
SAME  PERIOD  AND  PHASE. — In  this  case  we  make  in  (3)  e  —  o,  and  therefore  the  phases 
are  equal,  and  we  have  at  once 

r  4-  r,  cos  a 
x  =  -  — y. 

rv  sin  a 


The  resultant 
R 


FIG.  2. 


This  is  the  equation  of  a  straight  line  passing  through  the  centre  C. 
motion  is  therefore  central  harmonic  in  a  straight  line,  or 
simple  rectilinear  harmonic. 

If  CA  and  CB  are  the  amplitudes  r  and  rv  inclined  at 
the  angle  a,  the  resultant  motion  has  the  amplitude  CR,  in 
direction  and  magnitude  the  diagonal  of  the  parallelogram 
whose  adjacent  sides  are  r  and  rv  inclined  at  the  angle  a. 

Conversely,    a    simple     rectilinear    harmonic    motion 
whose  amplitude  is  CR  may  be  resolved,  by  completing  the  parallelogram,  into  two  others  in 
any  two  directions,  of  the  same  period,  epoch  and  phase. 

If  a  =  90°,  we  have  7  =  —  x.      Therefore  the  projection  of  a   simple   rectilinear  har- 

monic motion  on  any  straight  line  is  also  a  simple  rectilinear  harmonic  motion  of  the  same 
period,  epoch  and  phase. 

If  the  component  motions  are  more  than  two,  they  may  be  compounded  two  and  two, 
and  therefore  any  number  of  component  simple  rectilinear  harmonic  motions  in  any  direc- 
tions, of  the  same  period,  epoch  and  phase,  give  a  single  resultant  rectilinear  harmonic 
motion  of  determinate  direction  and  amplitude,  which  may  be  resolved  into  two  components 
in  any  two  directions,  of  the  same  period,  epoch  and  phase. 

(b]  Two  COMPONENT  SIMPLE  RECTILINEAR  MOTIONS  IN  THE  SAME  LINE  WITH  THE 

SAME  PERIOD  AND  DIFFERENT  EPOCHS  AND  PHASES. 

-^In  this  case  we  make   in  (3)  a  —  o,  and  obtain  at 

once 


2rr  cos  e 


=  o. 


But  since  for  a  =  o,  y  =  o,  (see  Fig.  I,)  we  have 


r2  -f-  2rr\  cos  e 


=  constant. 


In  Fig.  3  the  points  Pand  Pl  move  in  the  line 
A  A'  with  simple  harmonic  motion  and  the  diagonal 
CR  = 


2rrv  cos  e  -|-  rf,  where  e  is  the  constant 
difference  of  epoch  and  phase. 

Since  e  is  constant  and  CR  is  constant,  its  inclination  to  CQ  or  CQl  is  constant.  At 
any  instant  the  resultant  displacement  is  CPl  +  CP  —  CS,  and  the  motion  of  5  is  therefore 
the  resultant  motion  and  is  simple  rectilinear  harmonic,  with  the  amplitude  CR,  the  diagonal 
of  the  parallelogram  on  r  and  rr  The  epoch  and  phase  are  intermediate  between  the 
epochs  and  phases  of  the  components. 

If  the  epochs  and  phases  are  the  same,  e=  o  and  the  amplitude  of  the  resultant  motion 


13  *  KINEMATICS  OF  A  POINT—  APPLICATIONS.  [CHAP.  IV. 

is  r4-rt,  or  the  sum  of  those  of  the  components.  If  the  difference  of  epoch  or  phase  is 
e  =  n  radians,  the  amplitude  is  r  —  r,,  or  the  difference  of  those  of  the  components. 

By  taking  CQt  and  CQ  of  proper  lengths  we  can  make  QCP  and  Q^CQ  what  we  please 
without  changing  CR.  Therefore  any  simple  harmonic  motion  may  be  resolved  into  two  others 
in  the  same  line,  with  any  required  difference  of  phase  and  one  of  them  having  any  desired 
epoch,  the  periods  being  the  same. 

Three  or  more  component  simple  harmonic  motions  in  the  same  line  and  with  the  same 
period  may  be  compounded  two  and  two,  and  the  resultant  will  be  rectilinear  harmonic 
with  the  same  period. 

If  the  periods  are  different,  the  angle  Q^CQ  =  e  will  vary  and  CR  will  vary.  When 
€  =  o,  CR  will  have  its  maximum  value  r  -\-  rr  When  the  difference  of  epoch  e  is  -n  radians, 
CR  has  its  minimum  value  r  —  rr  The  angular  velocity  of  CR  is  also  variable.  The 
direction  of  CR  will  oscillate  back  and  forth  about  CQ,  the  maximum  inclination  being 

sin'1  —  .     The  resultant  motion    is  therefore   not  simple    harmonic,    but   a  more  complex 

motion.  It  is,  as  it  were,  simple  harmonic  with  periodically  increasing  and  decreasing 
amplitude,  and  periodical  acceleration  and  retardation  of  phase  or  epoch. 

(c)  Two  COMPONENT  SIMPLE  RECTILINEAR  HARMONIC  MOTIONS  AT  RIGHT  ANGLES 
WITH  THE  SAME  PERIOD  AND  DIFFERENT  PHASES  OR  EPOCHS.  —  The  general  equation  for 
this  case  is  given  by  (3).  If  the  directions  are  at  right  angles,  we  have  a  =  90°.  Suppose 
in  addition  the  amplitudes  equal,  so  that  r  =  rlt  and  the  difference  of  epoch  e  =  90°.  We 
have  then,  from  (3), 

=  r*. 


Since  the  motion  is  central  harmonic,  according  to  page  129,  the  areal  velocity  of  the 
radius  vector  is  constant;  and  since  the  radius  is  constant,  the  speed  in  the  circle  is  constant. 
We  have  already  seen,  page  126,  that  the  projection  of  the  motion  of  a  point  moving  with 
uniform  speed  in  a  circle,  upon  a  diameter,  gives  rectilinear  harmonic  motion.  The 
projection  upon  two  diameters  at  right  angles  gives,  then,  two  component  rectilinear 
harmonic  motions  of  the  same  period,  with  a  difference  of  epoch  of  90°,  or  of  phase  of  J, 
since,  when  one  component  has  its  greatest  displacement,  the  other  is  at  the  centre  with 
displacement  zero. 

It  follows  also  that  two  component  simple  harmonic  motions  at  right  angles,  with  the 
same  period  and  equal  amplitudes,  differing  in  epoch  by  90°  or  in  phase  by  one  quarter  of  a 
period,  will  give,  as  a  resultant,  uniform  motion  in  a  circle  whose  radius  is  the  common 
amplitude  of  the  components. 

If  the  amplitudes  are  not  equal,  but  a  and  e  still  90°,  and  periods  the  same,  we  have, 
from  (3), 


which  is  the  equation  of  an  ellipse  referred  to  its  centre  and  axes. 

The  resultant  motion  is  therefore  harmonic  in  an  ellipse,  whose  semi-diameters  are  r 
and  r,,  the  centre  at  the  centre  of  the  ellipse. 

The  same  result  is  evidently  obtained  by  projecting  the  circle  in  the  preceding  case 
upon  a  plane,  so  as  to  obtain  the  required  amplitude  r,  ,  r  remaining  unchanged. 

(d)  THREE  OR  MORE  COMPONENT  SIMPLE  RECTILINEAR  HARMONIC  MOTIONS  IN 
DIFFERENT  LINES  WITH  THE  SAME  PERIOD  BUT  DIFFERENT  PHASES  OR  EPOCHS.  —  We 
have  seen  from  equation  (3)  that  the  resultant  of  two  simple  rectilinear  component  harmonic 


CHAP.  IV.]  COMPOUND  HARMONIC  MOTION.  133 

motions  in  any  two  directions,  of  the  same  period  and  different  epoch  or  phase,  is  elliptic 
harmonic  motion. 

We  have  also  seen  from  (a)  that  any  simple  rectilinear  harmonic  motion  may  be  re- 
solved into  two  others  of  the  same  period  and  phase  or  epoch  in  any  two  given  directions. 
Any  number  of  given  simple  rectilinear  harmonic  motions,  then,  of  the  same  period  and 
different  phases  or  epochs  may  each  be  resolved  into  its  own  pair  in  any  two  given  directions. 
These  pairs  constitute  a  number  of  simple  rectilinear  harmonic  motions  in  two  given  lines, 
all  of  the  same  period  and  different  phases  or  epochs. 

According  to  (<£),  all  in  one  line  may  be  compounded  into  one  resultant,  and  all  in  the 
other  line  into  another  resultant,  these  two  resultants  having  the  same  period  and  different 
phases  or  epochs.  The  resultant  of  these  two  is,  according  to  equation  (3),  elliptic  har- 
monic motion. 

Hence  the  resultant  of  any  number  of  component  simple  rectilinear  harmonic  motions 
of  tJie  same  period,  whatever  their  amplitudes,  directions,  phases  or  epochs,  is  elliptic  har- 
monic motion,  the  centre  of  the  ellipse  being  then  centre  of  acceleration,  and  the  radius 
vector  describing  equal  areas  in  equal  times. 

In  special  cases  this  becomes,  as  we  have  seen,  uniform  circular  motion  or  simple 
rectilinear  harmonic  motion. 

Since  the  above  holds  whatever  the  inclination  of  the  two  resultants,  elliptic  harmonic 
motion  may  be  considered  as  the  resultant  of  two  component  simple  harmonic  motions  of 
the  same  period  and  different  epochs  or  phases  at  right  angles. 

Graphic  Representation. — We  may  exhibit  graphically  simple  or  compound  rectilinear 
harmonic  motion  by  a  curve  in  which  the  abscissas  represent  intervals  of  time,  and  the  ordi- 
nates  the  corresponding  distance  of  the  moving  point  from  its  mean  position. 

In  the  case  of  a  single  harmonic  motion  we  have  (page  130)  x  =  r  cos  (oot  ±  e).      If  the 

distance  x  is  to  be  zero  when  t  =  o,  we  must  have  the  epoch  e  =  -  radians,  or  one  fourth  of 
the  periodic  time.  This  gives  x  —  r  sin  cot. 

Since  00  =  -=, ,  where  T  is  the  periodic  time,  /^  r         Xs\ 

we  have  for  t  =  o,  x  =  o ;  for  t  =  \  T,  x  =  r ;  for 
t  -  \T,  x  =  o;  for  t  =  $T,  x  —  -  r\  for  t  =  T, 
x  =  o. 

The  curve  is  the  curve  of  sines,  or  the  curve  which  would  be  described  by  the  point  P 
(page  126)  if,  while  Q  maintained  its  uniform  circular  motion,  the  circle  itself  were  to  move 
with  uniform  speed  in  a  direction  perpendicular  to  CA. 

It  is  the  simplest  possible  form  assumed  by  a  vibrating  string,  when  it  is  assumed  that 
at  each  instant  the  motion  of  every  particle  of  the  string  is  simple  harmonic. 

If  the  rectilinear  harmonic  motion  is  compound,  we  have  (page  130)  in  general 


x  =  r  cos  (  Got  ±  e)  -j-  rt  cos  (oof 


If  the  displacement   of   one  of  the   motions  is  zero   when  t  =  o,  we  have  e  =  —  ;  if 


x  =  r  sin  oof  -f-  rl  sin  (aoj  -f-  mt). 


134  KINEMATICS  OF  A  POINT-APPLICATIONS.  [€HAP.  IV. 

If  the  period  of  one  motion  is  twice  that  of  the  other,  for  instance,  we  have  col  =  20?, 
and 

x  =  r  sin  oot  -\-  rl  sin  (zoot  -\-  nit). 

If  the  difference  of  phase  is  zero,  n  =.  o;  and 
J  if  the  amplitudes  are  equal  also,  we  have 

O^  Ml    \        y^T        V  / 

\  ./  x  =  r  sin  cat  -}-  r  sin  (2  oat). 

This  gives  a  curve  as  shown  in  the  figure. 
Periods  Unequal. — We  have  in  general 

x  —  r  cos  (cot  +  e),     y  =  rl  cos  (<»,/  +  ei) 

for  the  two  component  rectilinear  harmonic  motions  at  right  angles.     The  elimination  of  /  in 
any  case  gives  the  curve  of  resultant  compound  harmonic  motion. 

If  the  periods  of  the  components  are  as  I  to  2,  and  e  is  the  difference  of  the  epochs,  we 
have  for  equal  amplitudes 

x  =  r  cos  (2oot  -\-  e),     y  —  r  cos  cat. 
Eliminating  /, 


which  is  the  general  equation  of  the  curve  for  any  value  of  e. 
Thus  for  e  =  o,  or  equal  epochs, 


which  is  the  equation  of  a  parabola.     For  e  =  —  , 


which  is  also  the  equation  of  a  parabola. 

If  we  make  e  in  succession,  o,  1,2,  etc.,  eighths  of  a  circumference,  we  obtain  a  series 
of  curves  as  shown. 


Period  1:2 


In  the  same  way  we  can  find  the  curve  for  any  ratio  of  periods  and  difference  of  epoch. 
Thus  if  the  periods  are  as  I  to  3  or  2  to  3,  and  we  make  ein  succession  o,  1,2,  etc.,  eighths 
of  a  circumference,  we  obtain  the  following  series  of  curves: 


CHAP.  IV.] 


COMPOUND  HARMONIC  MOTION. 


J35 


Period  1:3 


Period  2:3 


Blackburn's  Pendulum. — The  motion  of  a  pendulum  which  swings  through  a  small  arc 
is,  as  we  shall  see  hereafter  (page  138),  simple  harmonic,  and  the  pro- 
jection of  the  bob  on  a  horizontal  plane   moves  with  simple  rectilinear C D 

harmonic  motion. 

Curves  similar  to  those  just  given  are  therefore  traced  by  Black- 
burn s  pendulum.  This  consists  of  two  pendulums,  CED  and  EB, 
arranged  so  as  to  swing  in  two  planes  at  right  angles. 

Any  difference  of  period  may  be  made  by  adjusting  the  lengths  of 
the  pendulums,  and  they  may  be  started  with  any  difference  of  epoch. 
If  the  bob  B  is  made  to  trace  its  path  on  a  horizontal  plane,  we  have, 
approximately,  the  compound  harmonic  curve. 


CHAPTER  V. 

CONSTRAINED  MOTION  OF  A  POINT. 

Motion  on  an  Inclined  Plane— Uniform  Acceleration. — Let  a  point   have  a  uniform 
acceleration  /  in-  the  direction  AE,  and  let  the  point  be  constrained  to 
move  in  the  straight  line  AB  which  makes  the  angle  a  with  the  horizon. 
The  component  of  the  acceleration   in  the  direction  of  the  motion  is 
then /sin  a. 

The   motion   along  AB   is  then    rectilinear   motion  'under   uniform 
acceleration  /  sin  a,  and  equations  (2)  to  (7),  page  92,   apply  directly  if 
we  substitute /sin  a  in  place  of/. 

If  t/j  is  the  initial  velocity  at  A,  and  v  is  the  velocity  at  B   we  have  from  (7),  page  92, 

tf  —  v*  —  2/1  sin  «, 

where  /is  the  length  of  the  inclined  plane  AB.     But  /  sin  a  =  AE. 

The  speed,  therefore,  gained  in  moving  from  A  to  B  is  equal  to  that  which  would  be 
gained  in  falling  through  AE  with  the  uniform  acceleration  f. 


The  time  in  falling  from  A  to  E  is,  from  ^2),  page  91,  /'  =     — - — '•  and  in  passing  from 


A  to  B,  t  =  ~  ^'.    Hence 
sin  a 


J_ 
AE 


or  the  times  are  proportional  to  the  distances  /  and  AE. 

The  distance  passed  through  along  AB  is,  from  (5),  page  92, 


where  i\  is  the  initial  velocity. 

If  the  point  starts  irom  rest,  we  have  for  the  distance  along  AB 


/=-/sin  a .  P. 


136 


CHAP.  V.] 


CONSTRAINED  MOTION. 


137 


Let  AD  be  the  vertical  diameter  of  a  circle,  and  AB  —  /  any  chord.  Join  DB.     Then 

we  have  AB  —  AD  cos  DAB  =  AD  sin  ABC.       If  AB  =  /,  we  have  A 

/  ~      ~^~~r\  ^ 

I 


also 


sin    ABC.     therefore  AD  =   - 


2AD 


This  is  independent  of  the  position  6"f  the  chord  AB,  and  therefore 
it  is  the  same  for  any  chord  through  A  or  D. 

Hence   for  uniform  acceleration  f,  the  .time  of  descent  down  all 
chords  through  the  highest  and  lowest  points  of  a  circle  are  equal. 

This  property   enables  us  to  find  the  line  of  swiftest  descent  to  a 
given  curve  from  any  point  in  the  same  vertical  plane. 

Thus  if  EG  is  the  curve  and  A  the  point,  draw  AC  par- 
allel to  the  direction  of  /,  and  with  centre  in  A C  describe 
a  circle  passing  through  A  and  tangent  to  the  curve  EG 
at  P. 

Then  AP  is  the  line  of  swiftest  descent  "from  A  to  the 
curve  EG.  For  any  other  point  /  in  EG,  Ap  cuts  the 
circle  in  some  point  q,  and  since  the  time  from  A  to  q  is 
equal  to  that  from  A  to  P,  the  time  from  A  to  /  is.  greater. 

Examples. — (g  =  32.16  ft.-per-sec.  per  sec.     Friction,  etc.,  disregarded.) 

(1)  Find  the  position  of  a  point  on  the  circumference  of  a  vertical  circle,  in  order  that  the  time  of  recti- 
linear descent  from  it  to  the  centre  may  be  the  same  as  the  time  of  descent  to  the  lowest  point.     Acceleration  due 
to  gravity. 

ANS.  30°  from  the  top." 

(2)  The  straight  line  down  which  a  particle  will  slide,  under  the  action  of  gravity,  in  the  shortest  time 
from  a  given  point  to  a  given  circle  in  the  same  vertical  plane,  is  the  line  joining  the  point  to  the  utoper  or 

lower  extremity  of  the  vertical  diameter,  according  as  the  point  is  within  or  without  the  circle. 

(3)  Find  the  line  of  quickest  descent. from  the  focus  to  a  parabola  whose  axis  is  vertical  and  vert  ex  upwards, 
and  show  that  its  length  is  equal  to  that  of  tlie  latus-rectum.     Acceleration  vertical  and  uniform. 

(4)  Find  the  straight  line  of  swiftest  descent  from  the  focus  of  a  parabola  to  the  curve  when  the  axis  is 
horizontal.     Acceleration  vertical  and  uniform. 

(5)  The  times  in  which  heavy  particles  slide  from  rest  down  inclined  planes  of  equal  height  are  propor- 
tional to  their  lengths. 

(6)  Show  that  if  a  circle  be  drawn  touching  a  horizontal  straight  line  in  a  point  P  and  a  given  curve  in 
a  point  Q  below  P,  PQ  is  the  line  of  swiftest  descent  to  the  curve,  under  constant  vertical  acceleration. 

(7)  Find  the  straight  line  of  quickest  descent  from  a  given  point  to  a  given  straight  line,  the  point  and  the 
line  being  in  the  same  vertical  plane.      Acceleration  constant  and  vertical. 

ANS.  From   P,  the  given   point,  draw  a  horizontal  line  meeting  the  given  line  in  C.     Lay  off  along  the 
given  line  CD  equal  to  PC.     PD  is  the  required  line  of  swiftest  descent. 

(8)  A  given  point  P  is  in  the  same  plane  with  a  given  vertical  circle  and  outside  it,  the  highest  point  Q  of 
the  circle  being  below  P.     Find  the  line  of  slowest  descent  from  P  to  the  circle.     Acceleration  constant  and 
vertical. 

ANS.  Join  PQ  and  produce  it  to  meet  the  circumference  in  R.     PR  is  the  line  required. 

Motion  in  a  Curved  Path — Uniform  Acceleration. — Let  ABCD  be  any  curved  path, 

and  Ad  the  direction  of  the  acceleration/".  Any  very  small 
portion  of  the  curve,  AB,  may  be  considered  as  a  straight  line. 
We  have  then,  as  on  page  136,  the  change  of  speed  in  moving 
from  A  to  B,  the  same  as  in  moving  from  A  to-#  with  the  con- 
stant acceleration/.  So,  also,  in  moving  from  B  to  C  the 
change  of  speed  is  the  same  as  in  moving  from  b  to  c  with  the 
constant  acceleration/. 

Hence  the  change  of  speed  in  traversing  any  portion  of  the 


KINEMATICS  OF  A  POINT-APPLICATIONS. 


[CHAP.  V. 


path  AD  is  the  same  as  in  traversing  with  constant  acceleration  /  the  projection  Ad  of  the 
path  on  a  line  in  the  direction  of  the  acceleration. 

If,  then,  v,  is  the  initial  speed  at  A,  and  v  is  the  speed  at  any  point  D,  we  have 


Motion  in  a  Circle— Uniform  Acceleration.— This  is  the  case  of  the  simple  pendulum, 
which  consists  of    a  heavy  particle  attached  to  a  fixed   point    by  a  massless    inextensible 
string. 

Let  C  be  the  point  of  suspension  and  CA  the  radius,  and  let 
the  acceleration  /  be  uniform  and  vertical.  For  any  position  of 
the  point  P  the  angle  ACP  =  0,  and  the  acceleration  may  be 
resolved  into  a  tangential  component  /  sin  8  and  into  a  normal 
component  f  cos  6. 

The  normal  component  has  no  effect  upon  the  motion   in  the 
curve  at  P. 
b  If  the  angle  0  is  very  small,  the  arc  will  not  differ  materially 

from  the  sine,  and  we  have  sin  0  =  —  7-  ,  where  /  is  the  length 


o   the  radius  CA. 

The    tangential    acceleration    at   the  point  P  is  then  ft  = 


/  X  arc  AP 


It  is  therefore 


directly  proportional  to  the  distance  of  P  from  A,  measured  along  the  path. 

The  motion  of  Pis  thus  harmonic  in  the  path,  and  the  periodic  time  is  then  (page  127) 


= *  V 


arc  AP 
/X  arc  A~P' 
T~ 


f 


or  for  the  simple  pendulum  the  time  of  a  vibration  is  /  =  n\  — 

g 

The  periodic  time  is  therefore  for  small  displacements  independent  of  the  amplitude, 
and  therefore  for  small  arcs  the  oscillations  are  isochronous. 

The  time  of  vibration  is  half  the  periodic  time,  or  the  time  between  the  instants  at 
which  the  pendulum  reaches  opposite  ends  of  its  swing.  Thus  the  seconds  pendulum  makes 
a  complete  oscillation  in  2  seconds. 

If  0  is  not  very  small  the  time  is  different,  but  the  variation  is  practically  very  slight. 

COR. — If  the  velocity  of  /*at  any  instant  is  not  wholly  in  the  plane  PC  A,  it  may  be 
resolved  into  two  components,  one  in  the  plane  PC  A  and  the  other  perpendicular  to  it,  and 
both  tangential  to  a  spherical  surface.  Hence,  in  the  case  in  which  0  is  small,  P's  motion 
may  be  resolved  into  two  simple  harmonic  motions  of  the  same  period  ;  and  its  motion  is 
therefore  (page  133)  elliptic  harmonic  motion,  the  period  being  the  common  period  of  the 
components.  The  ellipse  described  will  depend  upon  the  amplitude  and  epoch  of  the 
components  and  therefore  upon  the  magnitude  and  direction  of  the  initial  velocity  of  P. 

If  ft  is  not  very  small,  and  the  component  motions  are  of  different  amplitudes,  the 
periods  will  have  different  values,  and  the  point  P  describes  curves  similar  to  those  given  on 
page  135. 


CHAP,  V.]  CONSTRAINED  MOTION.  139 

If  the  component  motions  are  equal  in  amplitude  and  therefore  in  period  and  differ  in 
phase  by  one  quarter  period,  the  point  P  moves  (page  132)  in  a  circle  about  the  foot  of  the 
perpendicular  on  CA  as  a  centre.  This  is  the  case  of  the  conical  pendulum. 

Examples.  —  (i)  Find  the  time  of  oscillation  of  a  pendulum  loft,  long  at  a  place  at  which  g  =  32  ft.-per- 
sec.  per  sec. 

ANS  1.75  sec. 

(2)  Find  the  length  of  the  seconds  pendulum  at  a  place  at  "which  g  =  jf.p. 

ANS.  3.232  ft. 

(3)  Find  the  length  of  the  pendulum  which  makes  24  beats  in  I  min.  when  g  =  32,2. 

ANS.  20.39  ft- 

(4)  A  seconds  pendulum  was  lengthened  i  per  cent.     How  much  does  it  lose  per  day? 

ANS.  7  min.  8.8  sec. 

(5)  The  length  of  the  seconds  pendulum  being  99.4.14.  cm.,  find  the  value  of  g. 

ANS.  981.17  cm.-per-sec.  per  sec. 

(6)  A  pendulum  j/.&  inches  long  makes  182  beats  in  3  min.     Find  the  value  of  g. 

ANS.  31.78  ft.-per-sec.  per  sec. 

(7)  If  two  pendulums  at  the  same  place  make  25  and  jo  oscillations  respectively  in  I  sec.,  what  are  their 
relative  lengths  ? 

ANS.  1.44  to  i. 

(8)  A  pendulum  which  beats  seconds  at  one  place  is  carried  to  another  where  it  gains  2  sec.  per  day 
Compare  the  value  of  g  at  the  two  places. 

ANS.  As  0.999953  to  i. 

(9)  A  pendulum  which  beats  seconds  at  the  sea-level  is  carried  to  the  top  of  a  mountain,  where  it  loses  40.1 
sec.  per  day.     Assuming  the  value  of  g  to  be  inversely  proportional  to  the  distance  from  the  centre  of  the  earth, 
and  the  sea-level  to  be  4000  miles  from  that  point,  find  the  height  of  the  mountain. 

ANS.  1.86  miles. 

Motion  in  a  Cycloid  —  Uniform  Accelera- 
tion. —  A  cycloid  is  the  curve  traced  by  a  point 
in  the  circumference  of  a  circle  which  rolls 
along  a  straight  line.  pV  —  / 

If  the  circle  EP  rolls  along  the  line  AB, 
the  point  P  being  originally  at  A,  the  path  of 
P  is  the  cycloid  ACB.  ~~t 

If  C  is  the  position  of  P  when  the  diameter  of  the  circle  through  P  is  perpendicular  to 
AB,  the  line  CD  perpendicular  to  AB  is  the  axis  and  C  is  the  vertex  of  the  cycloid. 

Let  the  uniform  acceleration  /  be  always  parallel  to  DC  and  vertical. 

Let  the  moving  point  Q  have  at  Q^  a  speed  zero.      Its  speed  at  £>2  is  then  (page  136) 


Let  t  be  the  time  in  which  the  point  would  with  the  same  acceleration  and  with  initial 
speed  zero  move  from  D  to  C.     Then  CD  =  \P.      Hence 


72  • 


-CD  =  ±  CD(CN,  -  CNj). 
Now  by  a  property  of  the  cycloid 

4CD  .  CN,  =  CQ?     and     $CD  .  CN2  =  CQ*. 


Hence 

f&jtfW'+ZQfi. 


140 


KINEMATICS  OF  A  POINT-APPLICATIONS'. 


[CHAP.  V. 


Now  t*  =  is  a  constant.      Hence  the  motion  of  Q  in  the  cycloid  is  harmonic  (page 

126),  where  -  =^ ,  /0  being  the  tangential  acceleration  of  Q  at  the  distance  J0  measured  along 
the  curve.      If  T  is  the  time  of  a  complete  oscillation,  we  have 


If  /'  is  the  time  occupied  in  moving  from  Ql 


to  C, 


'    = 


or  the  time  of  a  pendulnm  whose  length  is  2  CD,   or  4  times  the  radius  of  the  generating 

circle. 

As  this  involves  only  constant  quantities,  the  time  is  the  same  whatever  be  the  position 

of  the  starting-point  Ql  ,  or  the  oscillations  are  isoch- 
ronous.     Hence  the  cycloid  is  called  a  tnutochrone. 

This  result  is  rendered  of  practical  importance  by 
one  of  the  properties  of  the  cycloid,  viz.,  that  if  a 
flexible  and  inextensible  string  AB  is  fixed  at  the  end 
A  and  wrapped  tightly  round  the  semi-cycloid  AC,  the 
end  B  of  the  string  as  it  unwinds  will  describe  another 
semi-cycloid.  If,  then,  AC  and  AD  are  fixed  semi- 
cycloids,  symmetrical  with  reference  to  the  vertical 
AB,  and  AB  is  a  simple  pendulum,  B  will  describe  a 

cycloid,  and  its  oscillations  will  be  isochronous  whatever  their  extent. 

[Application  of  the  Calculus.—  To  Determine  the   Motion  of  a  Point  Constrained  to  Move  in  a  Cycloid,  the 
Acceleration  being  Constant,  in  the  Direction  of  the  Axis  and 
towards  the  Vertex.]—  By  the  application  of  the  general 
formulas  of  page  92  we  can  deduce  the  results  already 
obtained. 

Let  the  axis  CD  =  ir,  where  r  is  the  radius  of  the 
generating  circle  DP'C.  Let  the  acceleration  /  act 
downward.  Let  CN  =  y,  NP  =  x  and  the  length  of  arc 
CP  =  s.  Let  the  initial  position  he  />,  at  the  height 
CN,  —  h  above  C,  and  the  speed  at  />,  be  v,  =  o. 

We  have 


for  the  speed  at  any  point  given  by  CN  —  y.     Whenj/  =  o,  we  have,  at  the  lowest  point  C,  v  = 
which  is  the  same  as  that  due  to  the  vertical  height  //. 
By  the  property  of  the  cycloid  we  have 


s  =  arc  CP 


Hence 


We  have  then 


ds 


dy  j/HT. 


=  2  chord  Cf. 


CHAP,  v.]  CONSTRAINED  MOTION. 

Integrating,  since  for  /  =  Q,y  —  h,  we  have 


141 


(0 


For  the  time  of  descent  to  the  lowest  point  where  y  =  o,  or  for  the  time  of  one  quarter  of  a  complete 
oscillation, 


The  periodic  time  is  then 


or  the  same  as  a  simple  pendulum  (page  138)  whose  length  is  4  times  the  radius  of  the  generating 
circle  DP'C. 

The  time  is  independent  of  h  and  is  the  same  no  matter  what  the  position  from  which  the  point  begins 
to  descend.  The  oscillations  are  therefore  isochronous  and  hence  the  cycloid  is  called  the  tautochrone. 

The  reason  of  this  remarkable  property  is  easily  seen  by  considering  the  tangential  acceleration. 

In  the  cycloid  the  chord  CP'  is  always  parallel  to  the  tangent  TP.  The  tangential  acceleration 
or  tangential  component  of  y  is  then 


The  tangential  acceleration  is  therefore  directly  proportional  to  the  distance  from  the  vertex  measured 
along  the  path,  and  the  motion  of  P  is  simple  harmonic  (page  125). 
The  periodic  time  is  then  (page  127) 


If  in  (i)  we  makej'  =  — ,  we  have 


-  |/  -j.  , 


or  half  the  time  from  Pi  to  C.    The  time,  therefore,  in 


descending  through  half  the  vertical  space  to  C  is  half  the  time  of  passing  from  P\  to  C. 

[To  Find  a  Curve  such  that  a  Point  Moving  on  it  under  the  Action  of  Gravity  will  Pass  from  any  one  Given  Posi- 
tion to  any  Other  in  Less  Time  than  by  any  Other  Curve  through  the  Same  Two  Points.]—  This  is  the  celebrated 
problem  of  the  "curve  of  swiftest  descent"  first  propounded  by  Bernoulli.  The  following  is  founded  upon 
his  original  solution. 

If  the  time  of  descent  through  the  entire  curve  is  a  minimum,  that  through  any  portion  of  the  curve  is 
a  minimum. 

It  is  also  obvious  that  between  any  two  contiguous  equal  values  of  a  continuously  "varying  quantity,  a 
maximum  or  minimum  must  lie. 

This  principle,  though  simple,  is  of  very  great  power,  and  often  enables  us  to  solve  problems  of  maxima 
and  minima,  such  as  require  not  merely  the  processes  of  the  Differential 
Calculus  but  those  of  the  Calculus  of  Variations.     The  present  case  is 
a  good  example. 

Let,  then,  PQ,  QR  and  PQ',  Q'  R  be  two  pairs  of  indefinitely  small 
sides  of  a  polygon  such  that  the  time  of  descending  through  either  pair, 
starting  from  P,  may  be  equal.  Let  QQ  be  horizontal  and  indefinitely 
small  compared  with  PQ  and  QR.  The  curve  of  swiftest  descent  must 
lie  between  these  paths,  and  must  possess  any  property  which  they  have 
in  common.  Hence  if  we  draw  Qm,  Qn  perpendicular  to  RQ  ,  PQ,  and 
let  v  be  the  speed  down  PQ  or  PQ  (supposed  uniform)  and  i/  that 
down  QR  or  Q1R,  we  have  for  the  time  from  P  to  R  by  either  path 


QR 


PQ  QR 

z> 


or 


Qn       Q'm 


14*  KINEMATICS  OF  A  POINT-  APPLICATIONS.  [CHAP.  V. 

Now  if  6  be  the  inclination  of  FQ  to  the  horizontal,  and  tf  that  of  QR,  we  have  Qn  =  QQ  cos  6, 
Q'n  -  QQ  cos  6'.  Hence 

cos  8  _  cos  6' 

This  is  true  for  any  two  consecutive  elements  of  the  required  curve,  and  therefore  throughout  the  curve 
we  have,  at  any  point,  v  proportional  to  the  cosine  of  the  angle  which  the  tangent  to  the  curve  at  that 
point  make/  with  the  horizontal.  But  7/1  is  proportional  to  the  vertical  distance  //  fallen  through. 

Hence  the  curve  required  is  such  that  the  cosine  of  the  angle  it  makes  with  the  horizontal  line  through 
the  point  of  departure  varies  as  the  square  root  of  the  distance  from  that  line. 

Now  in  the  figure  of  page  140  we  have,  from  the  property  of  a  cycloid, 


cos  CfN  =  cos  TPN  =  cos  CDP  = 

DC 

The  curve  required  is  therefore  the  cycloid.     The  cycloid  has  received  on  account  of  this  property  the 
name  of  Brachistochrone. 


KINEMATICS  OF  A  RIGID  BODY. 


CHAPTER  I. 

ANGULAR    VELOCITY    AND    ACCELERATION    COUPLES.      ANGULAR    AND    LINEAR 
VELOCITY  AND  ACCELERATION  COMBINED. 

Angular  Velocity  Couple. — Two  simultaneous,  equal,  parallel  and  opposite  angular 
velocities,  not  in  the  same  straight  line,  we  call  an  ANGULAR  VE-  & 

LOCITY  COUPLE.  / 

Thus,  if  the  point  P  has  an   angular  velocity  +  GO  about  an  p /• v 

axis  Oa,  so  that  Oa  =  -f~  GO  is  its  line  representative,  and  at  the 
same  time  has  an  angular  velocity  —  oj  about  an  axis  O'b,  so  that 
O'b  =  —  GO  is  its  line  representative,  then,  since  the  line  represen-  / 

tatives  are  equal,  parallel  and  opposite,  and  do  not  coincide,  they 
constitute  a  couple. 

Moment  of  Angular  Velocity  Couple. — We  have  seen  (page 
90)  that  the  moment  OP  X  GO  of  the  angular  velocity  Oa  =  -j-  co 
relative  to  P,  gives  the  linear  velocity  Pa  =  -j-  v  at  right  angles  to 
the  plane  of  Oa  and  OP,  due  to  angular  velocity  about  the  axis  Oa. 
The  direction  of  +  v  is  such  that  when  we  look  along  Oa  in  its  &*  —~*° 
direction,  -f-  v  is  seen  as  clockwise  rotation,  or  towards  the  reader  in  the  preceding  figure. 

In  the  same  way  O'P  X  GO  gives  the  linear  velocity  Pb  —  —  r  of  P  in  a  direction  away 
from  the  reader,  at  right  angles  to  the  plane  of  the  couple. 

The  resultant  linear  velocity  of  P  is  then 

v  =  (O'P  -  OP]  GO=  O'O  X  GO, 

where  O'O  is  the  perpendicular  distance  between  the  line  representatives  Oa,  O'b  of  the  couple. 

This  resultant  velocity  is  at  right  angles  to  the  plane  of  the  couple  and  in  such  a  direc- 
tion that  when  we  look  along  its  line  representative,  rotation  as  indicated  by  the  arrows  of 
the  couple  is  seen  clockwise. 

With  this  convention,  if  the  distance  O'O  =  /,  we  have 

V   =    I  GO 

This  result  holds  no  matter  where  the  point  P  may  be  taken  in  the  plane  of  the  couple. 

If,  then,  P  is  a  point  of  a  rigid  body,  every  point  of  this  body  in  the  plane  of  the  couple 
must  have  the  same  velocity  v  in  the  same  direction.  That  is,  the  body  has  a  velocity  of 
translation. 

143 


144 


KINEMATICS  OF  A  RIGID  BODY.  [CHAP. 


Hence,  if  a  rigid  body  is  acted  upon  by  an  angular  velocity  couple,  the  result  is  a  velocity 
of  translation  for  every  point  of  the  body. 

We  denote  velocity  of  translation  always  by  v. 

Angular  Acceleration  Couple. — Two  simultaneous,  equal  parallel  and  opposite  angular 
accelerations  not  in  the  same  line  we  call  an  ANGULAR  ACCELERATION  COUPLE.  We  have 
then,  in  precisely  the  same  way  and  with  the  same  convention  as  to  direction,  the  acceleration 
for  every  point  of  the  body 

/=/«, 

where  a  is  the  angular  acceleration  and  /  is  the  linear  acceleration  of  translation  of  the  body. 
Hence,  if  a  rigid  body  is  acted  upon  by  an  angular  acceleration  couple,  the  result  is  a  linear 
acceleration  of  translation  for  every  point  of  the  body. 
We  denote  acceleration  of  translation  always  by/. 

Angular  and  Linear  Velocity  Combined. — Let  a  rigid  body  have  an  angular  velocity  GO 
about  an  axis  O'a,  so   that    O'a  =  GO  is  the   line  representative. 
/*v*  Let  O  be  any  point  of  the  body.      If  at  this  point  we  apply  two 

,/^  equal  and  opposite  angular  velocities  Oa  =  GO  and  Ob  =  GO,  both 

parallel  to  O  a  =  GO,  the  previous  motion  of  the  body  is  evidently 
not  affected. 

We  see,  then,  that  the  single  angular  velocity  O'a  =  GO  about 
an  axis  O'a  can  be  reduced  to  the  same  angular  velocity  Oa  =  GO 
about  a  parallel  axis  Oa  through  any  point  O,  and  an  angular 
velocity  couple  O'a  and  Ob. 

Let  /  be  the  distance  between  the  parallel  axes.  Then,  as  we 
have  just  seen,  the  couple  causes  a  velocity  of  translation  ~vn  =  loo  at  right  angles  to  the 
plane  of  the  couple,  so  that  looking  along  the  line  representative  of  vn  in  its  direction,  the 
arrows  of  the  couple  indicate  clockwise  rotation.  In  the  figure  vn  is  at  right  angles  to  the 
plane  of  the  couple  and  away  from  the  reader. 
Hence 

(a)  A  single  angular  velocity  GO  of  a  rigid  body  about  a  given  axis,  can  be  resolved 
into  an  equal  angular  velocity  about  a  parallel  axis  through  any  point  O  of  the  body  at  a 
distance  /,  and  a  normal  velocity  of  translation  VH  =  IGO  of  this  axis  in  a  direction  at  right 
angles  to  the  plane  of  the  two  axes. 

(&)  Conversely,  the  resultant  of  an  angular  velocity  GO  of  a  rigid  body  about  a  given 
axis  and  a  simultaneous  velocity  of  translation  VH  normal  to  that  axis,  is  a  single  equal 

vm 
angular  velocity  about  a  parallel  axis  at  a  distance  /  =  — ,   the  plane  of   the  two  axes  being 

perpendicular  to  vn. 

(c)  If,  then,  a  rigid  body  has  any  number  of  angular  velocities,  each  one  about  a  different 
axis  through  a  different  point,  then  by  (a)  we  can  reduce  each  one  to  an  equal  angular 
velocity  about  an  axis  through  any  point  O  we  please,  and  a  normal  velocity  of  translation 
of  this  axis. 

All  the  angular  velocities  at  this  point  O  can  then  be  reduced  to  a  single  resultant 
angular  velocity  GO  about  a  resultant  axis  by  the  polygon  of  angular  velocities  (page  68),  and 
all  the  normal  velocities  of  translation  can  be  reduced  to  a  single  resultant  velocity  of 
translation  v,  not  necessarily  normal  to  the  resultant  axis,  by  the  polygon  of  linear  velocities 
(page  66). 

The  motion  of  a  rigid  body  in  general  can  then  be  reduced  at  any  instant  to  an  angular 


CHAP.  I.] 


INSTANTANEOUS  AXIS  OF  ROTATION. 


145 


velocity  GO  about  an  axis  through  any  point  O  we .  please,  and  a  velocity  of  translation  v  of 
this  axis.  This  velocity  v  of  translation  is  not  necessarily  normal  to  the  axis. 

The  angular  velocity  GO  has  the  same  magnitude  and  direction  no  matter  what  point  is 
taken,  but  the  velocity  of  translation  v  varies  in  magnitude  and  direction  with  the  position 
of  this  point. 

(d)  This  velocity  of  translation  v  is  not  necessarily  normal  to  the  axis  and  can  in 
general  be  resolved  into  a  component  ~va  along  the  axis  through  O,  and  a  component  vn 
normal  to  this  axis. 

But  by  (b]  we  can  reduce  GO  and  ~vn  to  the  same  angular  velocity  GO  about  a  parallel  axis 

at  a  distance  /  =  — . 

CO 

Instantaneous  Axis  of  Rotation. — This  axis  is  called  the  INSTANTANEOUS  AXIS  OF 
ROTATION  because  it  is  the  axis  without  translation  about  which  at  a  given  instant  angular 
velocity  takes  place. 

Hence,  in  general,  the  motion  of  a  rigid  body,  at  any  instant,  can  be  reduced  to  an 
angular  velocity  GO  about  an  axis  through  any  point  of  the  body,  a  velocity  of  translation  va 
along  this  axis,  and  a  normal  velocity  vn  of  translation  of  this  axis.  Or  to  an  angular 

velocity  GO  about  a  parallel  instantaneous  axis  at  a  distance  /  =  —  and  a  velocity  of  transla- 
tion va  along  this  axis. 

Spin.  Screw-Spin. — Angular  velocity  of  a  rigid  body  about  any  axis  we  call  a  SPIN 
about  that  axis.  Angular  velocity  of  a  rigid  body  about  any  axis  together  with  velocity  of 
translation  along  that  axis  we  call  a  SCREW-SPIN.  The  velocity  of  translation  va  along  the 
axis  we  call  the  VELOCITY  OF  ADVANCE. 

Hence  the  motion  of  a  rigid  body  at  any  instant  can  be  reduced  in  general  to  a  spin  or 
a  screw-spin  about  an  instantaneous  axis,  or  to  a  spin  or  screw-spin  about  a  parallel  axis 
through  any  point  together  with  normal  velocity  of  translation  vn  of  that  axis. 

Spontaneous  Axis  of  Rotation. — The  axis  of  rotation  through  the  centre  of  -mass  of  a 
rigid  body  at  any  instant  is  called  the  SPONTANEOUS  axis  of  rotation.  If  it  has  normal 
velocity  of  translation  vn,  the  instantaneous  axis  of  rotation  is  parallel  to  it  at  the  distance 

/  =  VJ*-,  the  plane  of  these  two  axes  being  perpendicular  to  vn  (page  144). 

The  velocity  of  any  point  is  that  due  to  angular 
velocity  about  the  instantaneous  axis  of  rotation,  or 
angular  velocity  about  the  parallel  translating  spon- 
taneous axis  (page  144). 

If  the  spontaneous  axis  of  rotation  has  no  normal 
velocity  of  translation  vn  ,  the  spontaneous  and  in- 
stantaneous axes  coincide. 

Examples. — (i)  A  vertical  circle  of  radius  r  =  2  ft.  rotates 
about  a  fixed  horizontal  axis  through  the  centre  of  mass  at  right 
angles  to  the  plane  of  the  circle  "with  angular  velocity  of  j 
radians  per  sec.  Find  the  velocity  and  central  acceleration  of 
the  top,  bottom,  forward  and  rear  points,  and  of  any  point  in 
general. 

ANS.  Take  co-ordinate  axes  as  shown  in  the  figure.  Then 
r  •=.  2  ft.,  K>y  =  +  3  radians  per  sec.  Since  the  spontaneous  axis 
O  Y  is  without  translation,  the  instantaneous  axis  coincides  with  it. 

We  have  then  at  the  top  point  a  the  velocity  2/^=4-  rooy  =  +  6  ft.   per  sec. 


At  the  bottom    point 


KINEMATICS  OF  A  RIGID  BODY. 


[CHAP.  1 


b  the  velocity  vx  =  —  raty  =  —  6  ft.  per  sec.     At  the  forward  point  c  the  velocity  t/,  =  —  rooy  =  —  6  ft. 
per  sec.    At  the  rear  point  </lhe  velocity  vt  =  4-  rao,  =  +  6  ft.  per  sec. 

Also,  at  the  top  point  a  the  central  acceleration  /ps  =  —  ro»J  =  —  18  ft.-per-sec.  per  sec.  At  the 
bottom  point  i>  the  central  acceleration  fp,  =  +  rao,  —  +  18  ft.-per-sec.  per  sec.  At  the  forward  point 
f  the  central  acceleration  fftx  =  —  roo'  =  —  18  ft.-per-sec.  per  sec.  At  the  rear  point  d  the  central 
acceleration  fpx  =  +  18  ft.-per-sec.  per  sec. 

Let  +  JT,  •+•  y,  +s  be  the  co-ordinates  of  any  point  P.  Then  we  have  the  component  velocities  of 
that  point  given  by 

vx  =  zwy,      vy  =  o,      vt=   -  xa>y (I) 

The  resultant  velocity  is 


=    OOy 


(2) 


and  its  direction  cosines  are 


— •        cos  ft  =  o,        cos  Y  = 


(3) 


The  central  acceleration  is  /J>  =  reoj  towards  the  centre.    The  radius  r  has  the  direction  cosines  cos  ot=   — 
cos  0  =  o,  cos  ^  =  — .     The  component  central  accelerations  are  then 


/puc  =   -  /p-    =    - 


The  resultant  central  acceleration  is 


and  its  direction  cosines  are 


fPy=0, 


(4) 


(5) 


/p 


-  -,        cos  /S 


(6) 


(2)  /?  vertical  circle  of  radius  r  =  •?//.  rolls  on  a  horizontal  straight  line.  The  centre  moves  parallel  to 
that  line  with  a  velocity  of  6  ft.  per  sec.  Find  the  angular  velocity  ;  the  velocity  and  central  acceleration  of 
the  toft,  bottom,  forward  and  rear  points,  and  of  any  point  in  general. 

ANS.   Take  co-ordinate  axes  as  shown  in  the  figure.     Then  r  =  2  ft.,  vx  =    +  6  ft.  per  sec.     The  spon- 
taneous axis  is  O  Y. 

Since  the  circle  rolls,  the  instantaneous  axis  b  Y' 
passes  through  the  bottom  point  b  parallel  to  the  spon- 
taneous axis. 

We  have  then  for  velocity  of  translation 


vx  =  r<oy,   or    ooy  =  —  =  +  3  radians  per  sec. 

The  velocity  of  any  point  is  that  due  to  translation 
and  angular  velocity  ta  about  O  Y,  or  to  angular  velocity 
ta  only  about  bV. 

We  have,  then,  at  the  top  point  a  the  velocity 
vx  =  T/X  +  riDy  =  2  r<0y  =  -f  12  ft.  per  sec. 

At  the  bottom  point  b  the  velocity   is  vx  =  vx  —  rooy 
=  o.     At  the   forward   point  c  we   have  the  component 
velocities  vx  =  vx  =  rooy  =  +  6  ft.  per   sec.,  and   vt  = 
—  rtav  =  —  6  ft.  per  sec.     The  resultant  velocity  is  then 
f  •  =  6  4/2^  ft.  per  ser,  making  an  angle  of  45*  below  OX.  as  shown  in  the  figure. 

At  the  rear  point  d  we  have  the  component  velocities  vx  =  vx  =  rtoy  =  +  6  ft.  per  sec.,  and  v,  = 
+  rtOy  =  +  6  ft.  per  sec.  The  resultant  velocity  is  then  v  =  6  f/2  ft.  per  sec.,  making  an  angle  of  45°  above 
OX.  as  shown  in  the  figure. 


CHAP.  I.] 


SINGULAR  AND  LINEAR    VELOCITY  COMBINED— EXAMPLES. 


147 


The  central  acceleration  of  any  point  is  that  due  to  angular  velocity  about  OY  considered  as  without 
translation  (page  145). 

We  have  then,  just  as  in  the  preceding  example,  at  the  top  point  a  the  central  acceleration  fp,  —  —  r<so\ 
—  —  18  ft.-per-sec.  per  sec.  At  the  bottom  point  b,  fpz  =  +  rooy  =  +  18  ft.-per-sec.  per  sec.  At  the 
forward  point  c ,  fpx  =  —  fooy  =  —  18  ft.-per-sec.  per  sec.  At  the  rear  point  d^fpx  =  +  too*  =  +  18  ft.- 
per-sec.  per  sec. 

Let  +  x,  +  z  be  the  co-ordinates  of  any  point  P  for  the  origin  O.  Then  we  have  for  the  component 
velocities  of  that  point 

Vx  =  ~-Vx    +   ZOJy  =  roOv  +    ZK)y  ,  Vy   =  O,  Vz  =   —  XOOy (l) 


Equations  (i)  give  the  component  velocities  for  any  point  in  general. 
The  resultant  velocity  of  any   point  is  then 


rf 


where  r1  is  the  radius  vector  of  the  point  relative  to  b. 
The  direction  cosines  of  v  are 


v* 

cos  a  =  —, 
v 


COS  ft  =  O, 


(2) 


(3) 


These  equations  reduce  to  (i),  (2),  (3)  of  the  preceding  example  if  vx  =  o. 

The  radius  r  for  the  point  P  has  the  direction  cosines  cos  a  =  £ ,  cos  y  —  I,    The  central  acceleration 

is/p  =  rooy  towards  O.      The  component  central  accelerations  for  any  point  are  then,  just  as  in  the  preceding 
example, 

fp*  =  —  *<*>} ,        fpy  =  o,        f(#  =  -  za>l (4) 

The  resultant  is 

and  its  direction  cosines  are 

j>  *  __ 

(6) 


These  equations  are  the  same  as  (4),  (5),  (6)  of  the  preceding  example. 

(3)  Let  a  vertical  circle  of  radius  r  =  2  ft.  roll  on  a  horizontal  plane.  The  centre  moves  with  a  "velocity  of 
6  ft.  per  sec.  At  the  same  time  let  the  plane  of  the  circle  rotate  about  the  vertical  diameter  with  an  angular 
velocity  of  2  radians  per  sec.  downwards.  Find  the  angular  velocity  about  the  horizontal  axis  ;  the  velocity 
and  central  acceleration  of  the  top,  bottom,  forward  and  rear  points,  and  of  any  points  in  general. 

ANS.  Take  the  co-ordinate  axes  as  shown  in  the 
figure.  We  have  then  r  =  2  ft.,  o>:  =  —  2  radians 
per  sec.,^*  =  +  6  ft.  per  sec.  Since  the  circle  rolls, 

we  have  rooy  =  ~vx  or  ooy  =  —  =  +3  radians  per  sec., 

and  the  instantaneous  axis  passes  through  the  bottom 
point  b.  The  velocity  of  any  point  is  that  due  to 
translation  and  angular  velocity  ooy  about  OF  and  <»» 
about  OZ,  or  to  translation  and  angular  velocity  GO 
about  the  spontaneous  axis  OA,  or  to  angular  velocity 
oo  only  about  the  instantaneous  axis  bA'  parallel  to  OA. 
We  have  then  at  the  top  point  a  the  velocity 
vx  =  ~vx  +  rooy  =  2rooy  =  +12  ft.  per  sec.  At  the 
bottom  point  b  the  velocity  is  vx  =  vx  -  rcay  =  o.  At 
the  forward  point  c  we  have  the  component  velocities 
vx  =  v'x  =  rooy  =  +  6  ft.  per  sec.,  vy  =  rooz  =  —  4  ft. 
per  sec.,  vz  =  —  rwy  =  —  6  ft  per  sec.  The  resultant 

f  Vy  +  Va  =  2  |/22 


velocity  is   then    v  =  \v\ 

sec.,  and  its  direction  cosines  are  given  by 

vx  •? 

cos  a    =  -  =  +  -4=. 


per 


KINEMATICS  OF  A  RIGID  BODY. 


[CHAP.    1. 


At  the  rear  point  d  we  have  the  component  velocities  vx  =  vx  =  r<oy  =  -f  6  ft.  per  sec.,  v,  =  —  >  <at 

=  -f  4  ft.  per  sec.,  v,  =  +  rooy  =  +  6  ft.  per  sec  '1  he 
resultant  velocity  is  as  before  v  =  24/22  ft.  per  aec.,  and 
its  direction  cosines  are 

'•••     .  '    . 

cos"=*  =  '  1* 


The  central  acceleration  of  any  point  is  that  due  to 
angular  velocity  about  OA  considered  as  'without  transla- 
tion (page  145),  or  to  angular  velocity  ooy  about  OY  and 
ea,  about  OZ  without  translation. 

We  have  then  at  the  top  point  a  the  deflecting  accel- 
eration (page  80)  /„  =  —  rcoy  =  -  18  ft.-per-sec.  per  sec. 
and  the  deviating  acceleration  (page  80)  /ay  =  ra)yoot  = 
—  12  ft.-per-sec.  per  sec.  The  resultant  central  accelera- 
tion is  then/p  =  tffrl  +  Jay  =  6  Vrj  ft.-per-sec.  per  sec., 


cos  a  =  o, 


cos  /»='—= -= 

Jf  3*"3 


rl   +  Jay  =  6 
and  its  direction  cosines  are 

2  /„  3 

=  —  -  —  '  cos  Y  =  J~  =  --  y= 

/1  JP  Vi3 


At  the  bottom  point  b  the  deflecting  acceleration  (page  80)  is/«  =  -f  rooy  —  +  18  ft.-per.-sec.  per  sec., 
and  \\\t  deviating  acceleration  is  /„»  =  —  rcay<ax  =  +  12  ft.-per-sec.  per  sec.  The  resultant  central  accelera- 
tion is  then/p  =  4//V1  +/„!  =6  4/13  ft.-per-sec.  per  sec.,  and  its  direction  cosines  are 


cos  a  =  o, 


cos  ft  —  '*jr- 


+  —  -=.  , 
3  4/13 


cos  y  =       =.  +  —=.  • 
JP  4/13 


At  the  forward  point  c  the  central  acceleration  is/p«  =  —  rooy  =  —  18  ft.-per-sec.  per  sec.     At  the  rear 
point  </the  central  acceleration  is/p,  =  +  roo*  =  +  18  ft.-per-sec.  per  sec. 

Let  +  x,  +z,  be  the  co-ordinates  of  any  point  P.     Then  we  have  the  component  velocities  of  that  point  : 


v,  =  vx  +  z<ay  =  r<ay  + 

The  resultant  velocity  of  any  point  is  then 

v=    ^fv\  +  v  J  +  v\  —  \/[(z 
where  r1  is  the  radius  vector  of  the  point  relative 
The  direction  cosines  of  v  are 


(i) 


(2) 


(3) 


Those  equations  reduce  to  equations     (i),  (2),  (3)  of  the  preceding  example  if  oot  =  o. 

The  radius  r  for  the  point  P  has  the  direction  cosines  cos  a  =  -,  cos  y  =  —  •   The  deflecting  accelera- 

tion is/r  =  rooy  towards  O  for  rotation  about  OY,  and  xoo\  towards  OZ  for  rotation  about  OZ.     The  com- 
ponent deflecting  accelerations  are  then 

fr*  -   -  XOOy   -  XGO\  ,  fry  =  0.  fr,    =    -  ZOO,. 

The  deviating  acceleration  is/,,  =  zoayoo,.     The  components  of  the  central  acceleration  are  then 
/P*  =  -xa>y  -  jcool  , 


The  resultant  is 

ff  =  V 

The  direction  cosines  of/p  are 

cos  a  =S'X. 
JP 

These  equations  reduce  to  (4).  (5).  (6)  of 


=  Z00yt0t 


fft=  -  zoo, 


=, 

JP  Jo 

nrerrding  example  if  aot  =  o. 


(4) 


(6) 


CHAP.  I.] 


ANGULAR  AND   LINEAR    VELOCITY  COMBINED— EXAMPLES. 


149 


(4)  Let  a  vertical  circle  of  radius  r  =  2  ft.  roll  on  a  horizontal  plane  with  an  angular  velocity  of  j  radians 
per  sec.,  while  its  centre  describes  a  horizontal  circle  of radius  j  ft.  with  angular  velocity  about  a  fixed  vertical 
axis.  Find  the  angular  velocity  about  the  fixed  axis  ;  the  velocity  and  central  acceleration  of  the  top,  bottom, 
forward  and  rear  points,  and  of  any  point  in  general. 

ANS.  Take  the  co-ordinate  axes 
as  shown  in  the  figure.  Let  O ' Z'  be  the 
fixed  axis.  Then  r  =  2  ft.,  x.  =  3  ft., 
<»„  =  +  3  radians  per  sec.  Since  the 
circle  rolls,  the  bottom  point  b  is  on  the 
instantaneous  axis.  The  instantaneous 
axis  must  also  pass  through  the  point  O1 . 
Hence  Ob  is  the  instantaneous  axis,  and 
OA  parallel  to  it  is  the  spontaneous 
axis.  We  have  then 

rtxty  -  -  ycoz, 
or 

rooy 
ooz= =~  —  —  2  radians  per  sec. 

ooz  is  therefore  negative,  as  shown  in  the 

figure.     The  velocity  of  translation  of  -^_ 

the  centre  is  given  by 

vx  =  raoy  =  -\-  6  ft.  per  sec. 

The  velocity  at  any  point  is  that  due  to  translation  and  angular  velocity  co  about  the  spontaneous  axis 
OA,  or  to  angular  velocity  K>  only  about  the  instantaneous  axis  O'b. 

We  have  then  at  the  top  point  a  the  velocity  vx  =  z/*  +  rooy  =  +  2ry  =  +  12  ft.  per  sec.  At  the 
bottoii)  point  b  the  velocity  is  vx  =  vx  —  ra)y  =  o.  At  the  forward  point  c  we  have  the  component  velocities 
v x  =  vx  =  rooy  =  +  6  ft.  per  sec.  vy  =  rooz  =  —  4  ft.  per  sec.,  v,  =  —  rooy  =  —  6  ft.  per  sec.  The  result- 
ant velocity  is  then  v  =  \'v\  +  vy  +  vl  =  2  4/22  ft.  per  sec.,  and  its  direction  cosines  are  given  by 


Vx  3 

cos  a  —  —  =  + 
v 


4/22 


|/22 


cos  Y  =  —  =  —  •  —  T=. 

V  4/22 


At  the  rear  point  d  we  have  the  component  velocities  vx  =  vx  =  roay  =  +  6  ft.  per  sec.,  vy  =  —  rco2  = 
+  4  ft.  per  sec.,  vz  =  +  rooy  =  +  6  ft  per  sec.  The  resultant  velocity  is,  as  before,  v  =  2  ^22  ft.  per  sec., 
and  its  direction  cosines  are 


cos  a  =  —  =    H ^= 


3_ 

|/22 


4/22  v  4/22 

The  central  acceleration  of  any  point  is  that  due  to  angular  velocity  about  OA  considered  as  without 

translation  (page  145).  or  to  anglar  velocity 
coy  about  O  Y  and  GOZ  about  OZ,  without 
translation.  In  either  case  O  and  hence 
every  point  has  acceleration  XGO\  towards 
O'Z',  owing  to  rotation  about  OZ'. 

We  have  then  at  the  top  point  a,  the 
deflecting  acceleration  (page  80)  frz  = 
—  mo*  =  —  1 8  ft.-per-sec.  per  sec.  due  to 
rotation  about  OY,  the  deviating  acceler- 
ation (page  So) fay  =  raoyao,  =  —  12  ft.- 
per-sec.  per  sec.  due  to  rotation  of  the 
plane  about  OZ,  and  the  deflecting  accel- 
eration f^  =  -  xv>l  =  -  12  ft.-per-sec. 
per  sec.  due  to  rotation  about  O'Z'.  The 
comconent  central  accelerations  are  then 
fpz  =  -  ra))  =  -  1 8  ft.-per-sec.  per  sec. 
and/p,  =fry  +  fay  =  my™*  -  J™**  =  -  24 
ft.-per-sec.  per  sec.  The  resultant  is  then 


/p  =  4//p2  +  fpy  =   30  ft.-per-sec.  per  sec.,  and  its  direction  cosines  are 


cos  a  =.  o, 


cosy 


=••  =  -  0.6. 

yp 


150  KINEMATICS  OF  A  RIGID  BODY.  [CHAP.  I. 

At  the  bottom  point  b  the  deflecting  acceleration  due  to  rotation  about  OY  is  fn  =  +  roay.  The 
deviating  acceleration  due  to  rotation  of  the  plane  about  OZ  is  fay  =  —  r<oya>t.  The  deflecting 
acceleration  due  to  rotation  about  O'Z'  is/ry  =  -  xoo\.  The  component  central  accelerations  are  then 
ff,  =  rtOy  =  +  1 8  ft.-per-sec.  per  sec.,/^  =  fay  +/ry  =  —  rooyco,  -yv>\  =  o.  The  resultant  central  accelera- 
tion is  then/p,. 

At  the  forward  point  c  the  deflecting  acceleration  due  to  rotation  about  O  Y  isfrx  =  —  rooiy,  and  due  to 
rotation  about  O'Z,  fry  =  —yv>l.  The  deflecting  acceleration  due  to  rotation  of  the  plane  about  OZ  is 
frs  =  —  rooj.  The  component  central  accelerations  are  then  fpx  =  —  ra>y  —  real  =  —  26  ft.-per-sec.  per  sec., 
fn  =  -  To»;  =  —  12  ft.-per-sec.  per  sec.  The  resultant  acceleration  is  then  fa  =  ^fp$  +  fp%  =  2  4/205 
ft.-per-sec.  per  sec.,  and  its  direction  cosines  are 

fax  13  fay  6 

cos  a  =  ->-  = =,          cos  ft  =  --—  =  —      ,  —  cos  y  =  o. 

/p        4/205  yp        4/205 ' 

At  the  rear  point  d  we  have,  in  the  same  way,  fpx  =  +  rao}  +  roo\,fw  —  —yu>\,  or/p*  =  +  26  ft.-per- 
sec.  per  sec.,  fn  =  —  12  ft.-per-sec.  per  sec.,/p  =  24/205  ft.-per-sec.  per  sec.,  and  the  direction  cosines  are 

cos  a  =  H L==,  cos  ft  = =,  cos  y  =  o. 

4/205  4/205 

Let  +  x,  +  y,  +  z  be  the  co-ordinates  of  any  point  P.     Then  we  have  the  component  velocities 
The  resultant  velocity  is 

where  r'  is  the  radius  vector  of  the  point  from  b. 
The  direction  cosines  of  v  are 

cos  a  =  —  ,          cos  /$=—£,          cos  y  =  -- (3) 

Those  equations  are  the  same  as  in  the  preceding  example.  For  the  component  deviating  accelerations 
due  to  rotation  of  the  plane  about  OZ  we  have 

fax    =  O,  fay    =   i-G^H  ,  faz  =  O. 

For  the  component  deflecting  accelerations  due  to  rotation  about  OY,  OZ,  and  O'Z'  we  have 

fr*    =    —  XODy*  ~   -fO^8,  fry    =    -  *<*)?,  frz    =    -   TO?/. 

Hence 


y (4) 

The  resultant  is 

fa  =  4/ypTl-/P>+yV. (5) 

and  its  direction  cosines  are 

yp  yp  yp 

These  equations  reduce  to  (4).  (5),  (6)  of  the  preceding  example  if  x  =  o. 

Angular  and  Linear  Acceleration  Combined.— Let  a  rigid  body  have  an  angular 
acceleration  at  about  an  axis  O'a  so  that  O'a  =  a  is  the  line 
representative.  Let  O  be  any  point  of  the  body.  If  at  this 
point  we  apply  two  equal  and  opposite  angular  accelerations 
Oa  =  a  and  Ob  —  or,  both  parallel  and  equal  to  O'a  =  a,  the 
motion  of  the  body  is  evidently  not  affected. 

|  We  see,  then,  that  the  single  angular  acceleration  O'a  =  a 

about  an  axis  O'a  can  be  reduced  to  the  same  angular  accoJera- 
tion  Oa  —  a  about  a  parallel  axis  Oa  through  any  point  O,  and 
an  angular  acceleration  couple  O'a  and  Ob. 

I g*      »g  Let  /  be  the  distance   between   the   parallel  axes.     Then 


CHAP.  I.]  ANGULAR  AND  LINEAR  ACCELERATION  COMBINED.  151 

(page  144)  the  couple  causes  acceleration  of  translation  fn  =  la  at  right  angles  to  the 
plane  of  the  couple,  so  that  looking  along  the  line  representative  of  fn  in  its  direction,  the 
arrows  of  the  couple  indicate  clockwise  rotation.  In  the  figure/,  is  at  right  angles  to  the 
plane  of  the  couple  and  away  from  the  reader.  Hence 

(a)  A  single  angular  acceleration  a  of  a  rigid  body  about  a  given  axis  can  be  resolved 
into  an  equal  angular  acceleration  about  a  parallel  axis  through  any  point  of  the  body  at  a 
distance  /,  and  a  normal  acceleration  of  translation  fn  =  la  of  this  axis  in  a  direction  at 
right  angles  to  the  plane  of  the  two  axes. 

(£)  Conversely,  the  resultant  of  an  angular  acceleration  a  of  a  rigid  body  about  a 
given  axis  and  a  simultaneous  acceleration  of  translation/,  normal  to  that  axis,  is  a  single 

equal  angular  acceleration  about  a  parallel  axis  at  a  distance  /=  — ,  the  plane  of  the  two 

axes  being  perpendicular  to/",,. 

(c)  If,  then,  a  rigid  body  has  any  number  of  angular  accelerations,  each  one  about  a 
different  axis  through  a  different  point,  then  by  (a)  we  can   reduce  each 'one  to  an  equal 
angular  acceleration  about  an  axis  through  any  point  we  please,  and  a  normal  acceleration 
of  translation  of  this  axis. 

All  the  angular  accelerations  at  this  point  can  then  be  reduced  to  a  single  resultant 
angular  acceleration  a  about  a  resultant  axis  by  the  polygon  of  angular  accelerations  (page 
83),  and  all  the  normal  accelerations  of  translation  can  be  reduced  to  a  single  resultant 
acceleration  of  translation  /  not  necessarily  normal  to  t;he  resultant  axis,  by  the  polygon  of 
linear  accelerations  (page  76). 

The  change  of  motion  of  a  rigid  body  in  general  can  then  be  reduced,  at  any  instant, 
to  an  angular  acceleration  n  about  an  axis  through  any  point  we  please,  and  an  acceleration 
of  translation  /  of  this  axis.  This  acceleration /of  translation  is  not  necessarily  normal  to 
the  axis. 

The  angular  acceleration  a  has  the  same  magnitude  and  direction  no  matter  what  point 
is  taken,  but  the  acceleration  of  translation  f  varies  in  magnitude  and  direction  with  the 
position  of  this  point. 

(d)  This  acceleration  of  translation  f  is  not  necessarily  normal  to  the  axis  and  can  in 
general    be   resolved  into  a  component/   along  the  axis  of  or  and  a  component  fn  normal 
to  this  axis. 

But  by  (fr)  we  can  reduce  a  and/,  to  the  same  angular  acceleration  a  about  a  parallel 

fn- 

axis  at  a  distance  /  =z  — 
a 

INSTANTANEOUS  Axis  OF  ACCELERATION. — This  axis  is  the  INSTANTANEOUS  AXIS  OF 
ACCELERATION  because  it  is  the  axis  -without  acceleration  about  which  at  a  given  instant 
angular  acceleration  takes  place. 

Hence,  in  general,  the  change  of  motion  of  a  rigid  body  at  any  instant  can  be  reduced 
to  an  angular  acceleration  a  about  an  axis  through  any  point  of  the  body,  an  acceleration 
of  translation/,  along  this  axis,  and  a  normal  acceleration/,  of  translation  of  this  axis.  Or 

fn 

to  an  angular  acceleration  a  about  a  parallel  instantaneous  axis  at  a  distance  /  =  ~  and  an 
acceleration  of  translation  /,  along  this  axis. 


IS*  KINEMATICS  OF  A  RIGID  BODY.  [CHAP.  I. 

Twist — Screw-Twist. — Angular  acceleration  of  a  rigid  body  about  any  axis  we  call  a 
TWIST.  Angular  acceleration  about  any  axis  together  with  acceleration  of  translation  along 
that  axis  we  call  a  SCREW-TWIST. 

Hence  the  change  of  motion  of  a  rigid  body  at  any  instant  can  be  reduced  in  general 
to  a  twist  or  screw-twist  about  an  instantaneous  axis  ;  or  to  a  twist  or  screw-twist  about  a 
Parallel  axis  through  any  point  together  with  a  normal  acceleration  of  translation  fH  of  that 
axis. 

SPONTANEOUS  Axis  OF  ACCELERATION. — The  axis  of  acceleration  through  the  centre  of 
mass  of  a  rigid  body  at  any  instant  is  called  the  spontaneous  axis  of  acceleration.  If  it  has 
normal  acceleration  of  translation  /„  ,  the  instantaneous  axis  of  translation  is  parallel  to  it  at 

the  distance  /  =  — ,  the  plane  of  these  two  axes  being  perpendicular  to^,   (page  151). 

The  acceleration  of  any  point  is  that  due  to  angular  acceleration  about  the  instan- 
taneous axis  of  acceleration,  or  angular  acceleration  about  the  parallel  translating  sponta- 
neous axis  (page  151). 

If  the  spontaneous  axis  of  acceleration  has  no  normal  acceleration  of  translation  fH  ,  the 
spontaneous  and  instantaneous  axes  coincide. 


CHAPTER    II. 

ROTATION  AND  TRANSLATION— ANALYTIC  RELATIONS. 

Components  of  Motion. — We  have  seen  (page  144)  that  the  motion  of  a  rigid  body  at 
any  instant  can  be  reduced  to  a  resultant  angular  velocity  GO  about  an  axis  through  any 
point  of  the  body  and  a  resultant  velocity  of  translation  v  of  this  axis  or  of  the  body. 

The  motion  of  the  body  at  any  instant  is  then  known,  if  we  know  at  that  instant  the 
velocity  of  translation  of  the  axis  through  some  given  point  of  the  body  and  the  angular 
velocity  of  the  body  about  this  axis. 

We  always  take  the  given  point  at  the  centre  of  mass  of  the  body,  and  denote  the  com- 
ponents of  the  velocity  of  translation  v  of  the  axis  through  it,  along  the  co-ordinate  axes, 
by  ~vx,  T>y,  Dz,  and  the  components  of  the  angular  velocity  w  by  GOX  ,  ooy  ,  coz. 

The  motion  of  the  body  at  any  instant  is  then  known  when  these  six  quantities  are 
known : 


These  six  quantities  are  therefore  called  the  COMPONENTS  OF  MOTION  of  the  body. 

Components  of  Change  of  Motion. — We  have  seen  (page  151)  that  the  change  of 
motion  of  a  rigid  body  at  any  instant  can  be  reduced  to  a  resultant  angular  acceleration  a 
about  an  axis  through  any  point  of  the  body  and  a  resultant  acceleration  of  translation  /  of 
this  axis. 

The  change  of  motion  of  the  body  at  any  instant  is  then  known,  if  we  know  at  that 
instant  the  acceleration  of  translation  of  the  axis  through  some  given  point  of  the  body  and 
the  angular  acceleration  of  the  body  about  this  axis. 

We  always  take  the  given  point  at  the  centre  of  mass  of  the  body,  and  denote  the 
components  of  the  acceleration  of  translation  f  of  the  axis  through  it,  along  the  co-ordinate 
axes,  by  fx,  fy  ,  fz ,  and  the  components  of  the  angular  acceleration  a  by  ax  ,  <xy ,  at. 

The  change  of  motion  of  the  body  at  any 
instant  is  then  known  when  these  six  quanti- 
ties are  known : 


These  six  quantities  are  therefore  called  the 
COMPONENTS  OF  CHANGE  OF  MOTION  of  the 
body. 

Motion  of  a  Point  of  a  Rigid  Body— 
General  Analytic  Equations.  —  Let  O  be  the 
centre  of  mass  of  a  rigid  body.  Take  any  co- 
ordinate axes  X,  Y,  Z  through  the  centre  of 
mass  O  as  origin,  and  let  the  co-ordinates  of 
any  point  P  of  the  body  relative  to  O  be  x,  v,  z. 

Let  the  point  P  have  the  angular  velocity 
oo  about  any  axis  Oa  .through  the  centre  of  mass 


i53 


154 


KINEMATICS  OF  A  RIGID  BODY.  [CHAP.  II. 


O,  and  let  the  components  of  w  along  the  co-ordinate  axes  be  wx ,  ooy ,  <»,.  Let  the  axes 
X,  Y,  Z  be  fixed  in  the  body  and  move  with  it. 

ROTATION— CENTRE  OF  MASS  FIXED.— Let  the  centre  of  mass  0  be  fixed.  Then 
s<o,  is  the  velocity  of  P  parallel  to  OX  in  a  positive  direction  due  to  angular  velocity  ooy 
about  OY,  and  ycot  is  the  velocity  parallel  to  OX  in  a  negative  direction  due  to  angular 
velocity  a>M  about  OZ.  In  the  same  way  we  have,  parallel  to  OY,  xoo,  positive  and  zwx 
negative,  and  parallel  to  OZ,  yoo,  positive  and  xooy  negative. 

If,  then,  vx,  Vy,  v,  are  the  component  velocities  of  the  velocity  v  of  the  point  Pdue  to 
rotation  about  the  axis  Oa  through  the  fixed  point  O,  we  have  for  rotation  only  about  an 
axis  Oa  through  the  fixed  centre  of  mass 

VM  =  zooy  —  y&>. ,         Vy  =  xoo,  —  ZGOX  ,         vg  =  yoox  —  xwy (i) 

ROTATION — ANY  POINT  FIXED. — Let  any  point  O'  of  the  body  be  fixed,  and  take 
the  co-ordinate  axes  OX1,  O' Y ,  O' Z1 ,  parallel  to  OX,  OY,  OZ.  Let  the  point  P  have  the 
angular  velocity  GO  about  an  axis  O'a'  through  the  fixed  point  O' ,  and  let  the  components  of 
«  along  the  co-ordinate  axes  X' ,  Y' ,  Z'  be  aix ,  coy  ,  ov  Let  the  co-ordinates  of  the  centre 
of  mass  O  relative  to  O'  be  x,  y ,  ~s.  Then  the  co-ordinates  of  any  point  P  will  be  x  -\-  x, 
~y  _|_  yt  -3  _j_  z,  where  x,  y,  z  are  the  co-ordinates  of  P  relative  to  the  centre  of  mass  O. 

We  have  then,  from  (i),  for  the  component  velocities  of  the  velocity  v  of  the  point  P 
due  to  rotation  only  about  an  axis  O'a'  through  any  fixed  point  O'  of  the  body 


••„  =  (s  -h  *)«,  -  ( J+  y)wf .   v,  =  (x  +  x)wM  -  (J  +  2)00, ,  v,  -  (y+y}oox  -(*  +  *)«,.  (2) 

If,  in  these  equations,  we  make  x  =  o,  y  =  O,  z  =  O,  we  have  for  the  component 
velocities  vx,  v1t  v,  of  the  centre  of  mass  O  due  to  rotation  about  the  axis  O'a'  through  the 
fixed  point  O' 

vx  =  ~ZGoy  —  y<at ,          vy  =  lew,  —  ~ZGOX  ,          vt  =  yoox  —  *a>y (3) 

ROTATION  ABOUT  TRANSLATING  Axis  THROUGH  CENTRE  OF  MASS — Let  the  axis 
Oa  through  the  centre  of  mass  O  have  the  velocity  of  translation  v  in  any  direction,  and  let 
^*  >  Vy,  vt  be  the  components  of  ~v.  The  point  P  will  have,  then,  component  velocities  of 
translation  in  addition  to  the  component  velocities  given  by  equations  (i )  for  rotation  only 
about  an  axis  through  the  fixed  centre  of  mass.  Hence,  for  rotation  and  translation 
combined,  we  have  the. component  velocities  of  P  : 

v,  =  vx  +ZGoy  —  yoof ,         vy—Vy-\-  xwt  —  z<ax  ,         v,  =  vt  +  yoox  —  xcoy.    .     .  (4) 

As  we  have  seen,  page  145,  the  motion  of  a  body  in  general  can  be  reduced  to  angular 
velocity  about  an  axis  through  the  centre  of  mass  and  a  velocity  of  translation  of  the  body. 
Equations  (4),  then,  are  general  and  include  all  cases. 

Thus  if  v,  =  o,  Vy  —  o,  vm  —  o,  we  have  rotation  only  about  an  axis  through  the  fixed 
centre  of  mass,  as  given  by  equations  (i);  if  in  addition  we  put  J-  -}-  x,  y  -f- y,  ~z  -\-  z  for 
x,  y,  z,  we  have  equations  (2)  for  rotation  only  about  an  axis  through  any  fixed  point.  If 
we  put  x,y,  J  for  x,y,  z,  we  have  equations  (3)  for  the  component  velocities  of  the  centre  of 
mass  due  to  rotation  about  an  axis  through  any  fixed  point 

RESULTANT  VELOCITY. — In  all  cases  the  resultant  velocity  of  Pis  given  by 

^VST+V;  "+"«*• ••  •'  •   (s) 


CHAP.  II.]  MOMENT  OF  VELOCITY.  155 

Its  line  representative  passes  through  P,  and  its  direction  cosines  are 

cos  a  —  VJL  cos  ft  =  -,          cos  y  =  - (6) 

V  V  V 

RESULTANT  ANGULAR  VELOCITY. — The  resultant  angular  velocity  is 


GO  =     ^GOX    -f-   G3|    -f-    GOZ (7) 

Its  line  representative  passes  through  the  centre  of  mass  O,  and  its  direction  cosines  are 

M*  wi  <*>*  /o\ 

cos  a  =  — ,          cos  8  =  -^,          cos  y  =  — (8) 

GO  '  GO'  GO 

MOMENT  OF  VELOCITY. — Let  the  moment  of  the  velocity  of  P  about  the  axis  O' a' 
through  any  point  O'  be  M'v.  Then  for  the  component  moments  M'vx  ,  M'vy ,  M'vz  about  the 
axes  X ,  Y' ,  Z'  of  the  component  velocities  vx ,  vy ,  vz  of  the  point  P  due  to  translation  and 
rotation  we  have 

about      O'X' M^x  =  vz(y+y)  —  vy(  z~  -\-  z),  ~] 


where  vx,  vy,  vz  are  given  by  equations  (4).      Substituting  these  values,  we  have  for  rotation 
and  translation 


-  x(y 
*  (*  +  s)<»,-  y(s- 

^(^  -f  ^)w,-  4^ 

For  rotation  only  ~vx  =  o,  ^  =  o,  ^  =  o.    For  axis  through  centre  of  mass  ~x  =  o,  J7  —  o,  F  —  o. 
VELOCITY  ALONG  THE  Axis  OF  ROTATION.  —  For  the  velocity  of  translation  va  along 
the  axis  of  rotation  we  have 


a  =  vx  cos  a  --  vy  cos  y 
or,  from  (8), 

-=y,«,  +    r,»,+  W 

GO 

This  is  called  the  axial  velocity  or  velocity   of  advance.      It  is   the  velocity  with  which 
the  body  moves  along  the  axis  of  rotation. 

VELOCITY  NORMAL  TO  THE  Axis  OF  ROTATION.  —  The  components,  along  the  axes,  of 

the  axial  velocity  are 

T>a  cos  a,          ~va  cos  /3,          va  cos  y. 

If  we  subtract  these  from  the  components  vx  ,  vy,  vt  of  the  velocity  v  of  translation,  we 
have  the  components  of  the  velocity  of  translation  vn  normal  to  the  axis  of  rotation: 

VGflx       I 

vnx  =  vx  —  v    cos  a  —  vv  — 


vny  —  vy  —  va  cos  ft  =  vy  —  ~-y 


_ 
vnz  =  vz  —  va  cos  y  =  vz  — 


'5* 


KINEMATICS  OF  A  RIGID  BODY. 


[CHAP.  II. 


Equations  (7),  (8)  and  (i  i)  give  the  screw-spin  (page  145)  about  the  axis  of  rotation  through 
the  centre  of  mass.  Equations  (7)  and  (8)  give  the  angular  velocity  (»about  the  axis  of 
rotation  and  the  direction  of  this  axis,  and  equations  (u)  give  the  velocity  va  along  this 
axis.  Equations  (12)  give  the  components  of  the  velocity  of  translation  normal  to  this  axis. 
INSTANTANEOUS  Axis  OF  ROTATION. — Let  xt  y,  z  be  the  co-ordinates  of  any  point  of 
the  instantaneous  axis  of  rotation.  Since  the  normal  velocity  of  any  point  of  this  axis  is 
zero,  we  have,  from  equations  (4)  (page  1 54), 

These  are  the  equations  of  the  projections  of  the  instantaneous  axis  of  rotation  on  the  three 
co-ordinate  planes. 

Let  JT,,  yt.  2,  be  the  intercepts  of  these  projections  on  the  co-ordinate  axes.     Then  in 
equations  (13),  making  y  =  o  and  z  =  o  in  the  last  two,  we  have 


making  x  =  o  and  z  =  o  in  the  first  and  last, 


Vnx 

y*  =  w. 


making  y  =  o  and  x  —  o  in  the  first  two, 


(H) 


GOX 


Let  the  perpendicular  from  the  centre  of  mass  O  upon  the  axis  of  rotation  be  />,  and  let 

/*»  Pyt  Pt  De  its  projections  on  the  co-ordinate  axes. 
Let  the  intersection  of  p  with  the  axis  be  O',  so  that 
00'  =p. 

^  Let    us    consider  the   projection  of  the  axis  on  the 

plane  YZ.  In  the  figure  p,  —  Ob  and  py  —  Oc.  We 
have,  then, 


where  sl  and  j/,  are  the  intercepts  Oa  and  Od  given   by  equations  (14).      We   have  also  the 
distance 


Hence 


Substituting  the  value  of  py,  we  obtain 


A 


Substituting  the  values  of  £,  and  _y,   from  (14),  we  have 

=  P-M,  _       _Vn 

*    *  HI*        I         /V)Z  '  '1  '  ft"    - 


CHAP.  II.] 


THE  INVARIANT  FOR   COMPONENTS  OF  MOTION. 


157 


In  the  same  way  we  can  find  px  and  flx,  px  and  py  on  the  other  two  co-ordinate  planes. 
We  thus  have 


/,=  - 


Py=~ 


05) 


Equations  (15)  give  the  position  of  the  instantaneous  axis  of  rotation. 
We  have  also,  from  (13)  and  (8), 

vx=va  cos  a  —  a)(px  cos  ft  —  py  cos  y) ,  GOX  =  GO  cos  a, 
vy  =  va  cos  ft  —  Go(px  cos  Y  —  pz  cos  a),  ooy=  GO  cos  /?, 
vf  =  va  cos  Y  —  Go(pv  cos  a  —  px  cos  ft),  G?  =  GO  cos  y. 


(16) 


When,  therefore,  the  components  of  motion,  vx ,  vy ,  v, ,  £».,.,  c^ ,  GOZ  ,  are  given  for  the 
centre  of  mass  O,  we  have  GO  from  (7),  the  direction  of  the  instantaneous  axis  of  rota- 
tion from  (8),  and  the  position  of  this  axis  from  (15)-  We  have  also  the  velocity  va  along 
this  axis  from  (11),  and  the  normal  velocity  vn  from  (12). 

On  the  other  hand,  if  the  position  and  direction  of  the  instantaneous  axis  of  rotation  are 
given,  together  with  the  velocity  va  along  it  and  the  angular  velocity  GO  about  it,  the  compo- 
nents of  motion  are  given  by  equations  (16). 

THE  INVARIANT  FOR  COMPONENTS  OF  MOTION. — From  (11)  we  have 

VaGO  =  VXGOX -\-VyG0y  -}-^VzGOz 


But  whatever  point  we  take  as  origin,  GO  does  not  change,  and  the  velocity  va  along 
the  instantaneous  axis  of  rotation  does  not  change.  This  quantity  (17)  is  therefore  called 
the  invariant  of  the  components  of  motion. 

If  the  invariant  is  zero  in  any  case,  we  must  evidently  have  either  va  or  GO  zero.  If  GO  is 
not  zero  but  the  invariant  is  zero,  then  the 

velocity  va  along  the  axis  must  be  zero.      In  this  •<---x* 

case  the  condition 

VXGOX  ~h  ^y^v  +  V,<0,  =  O 


is  the  condition  for  a  spin,  or  angular  velocity 
only,  about  the  instantaneous  axis  of  rotation. 

If  v "a  is  not  zero  but  the  invariant  is  zero, 
then  GO  must  be  zero,  and  we  have  translation 
only. 

Change  of  Motion  of  a  Point  of  a  Rigid 
Body — General  Analytic  Equations. — Let  O  be 
the  centre  of  mass  of  a  rigid  body.  Take  any  co- 
ordinate axes  X,  Y,  Z  through  the  centre  of  mass 
O,  and  let  the  co-ordinates  of  any  point  P  be  x, 
y,  z. 

Let  the  point  P  have  angular  acceleration  a 
about  any  axis  Oa,  and  let  the  components  of  a 
along  the  co-ordinate  axes  be  a 


Let  the  axes  X,  Y,  Z  be  fixed  in  the  body  and    fAx 
>/>  inith  it  X 


-a' 


-Y' 


move  with  it. 


I58 


KINEMATICS  OF  A  RIGID  BODY. 


[CHAP.  II. 


TANGENT  ACCELERATION.  —  Let  the  axis  Oa  be  fixed.  Then  zciy  is  the  tangent 
acceleration  of  P  parallel  to  OX  in  a  positive  direction  due  to  angular  acceleration  ay  about 
OY,  and  yan  is  the  tangent  acceleration  parallel  to  OX  in  a  negative  direction  due  to 
angular  acceleration  at  about  OZ.  In  the  same  way  we  have,  parallel  to  OY,  xat  positive 
and  saM  negative,  and  parallel  to  OZ,  ynx  positive  and  xay  negative. 

If,  then,  ftxiftyt  ftt  are  the  component  accelerations  of  the  tangent  acceleration  ft  of  the 
point  P  due  to  angular  acceleration  a  about  the  fixed  axis  Oa,  we  have 


=  sa,  —  ya. 


ft,  =  xat  —  zax  ,         ft.  =  yet,  -  xay. 


For  any  other  axis  through  any  point  O'  we  have  x  -f-  xt  "y-\-  y,  z  -\-z\n  place  of  x,y,  z, 
and  hence 


/te  =  (*  4-  *K  -  CP 


/*== 


V 


These  equations  are  general.     For  axis  through  the  centre  of  mass  we  have  x  =  o, 

y  =  O,   2  =  O. 

Wz  CENTRAL  ACCELERATION. — The  central  acceleration  is 

due   to    rotation    only,    and    is    not    affected    by   translation. 

^"*<"2   Let,  then,  vx,  vy,  vt  be  the  component  velocities  of  the  point  P 
due  to  rotation  only  about  an  axis  O'a'  through  any  point  O'. 

Then  vtcoy  is  the  central  acceleration  of  P  parallel  to  O'X' 
in  a  positive  direction,  and  vyGox  is  the  central  acceleration 
parallel  to  O'X'  in  a  negative  direction.  In  the  same  way 

— Y'     we  have,  parallel  to  O'  Y' ,  vx<vt  positive  and  vf<ax  negative,  and 
parallel  to  O' Z' ' ,  vyc*ix  positive  and  vxwy  negative. 

If,  then,  /pv./p>,  /p,  are  the  component  central  accelera- 
tions of  the  point  P  due  to  rotation  only  about  an  axis  O'a' 
through  any  point  O',  we  have 

where  vxt  vy,  vt  are  given  by  equations  (i)  page  154. 

Substituting  these  values  of  vx,  t'y,  vnt  we  have  for  axis  through  centre  of  mass 

ffy    =  ZOOyGO,  4  XOOyCOx  —  yw*    —  yco*, 

/P,  =  xw,vx+y(lov>y  —  zoo*  —  sco*. 

For  axis  through  any  point  O'  we   have  only  to  put  ~x  4-  x,  y  +  y,  z  \-  z  in  place  of 
x,  y,  z,  and  we  have  then,  in  general, 


-  (x  +  x)<*>ya>x  —  (y+y}co*  —  (y  +  y)<*l, 
fp,  =  (*  4  •*)<».<»,  4-  ( y  +?)<*>,<*>,  -  (  *  4-  -)«!  -  ( z  +  z)coj. 

These  equations  are  general  for  axis  through  any  point.      For  axis  through  centre  of 
mass  we  have  x  —  o.  y  =  o,  2  =  o. 


CHAP.  II.]  DEFLECTING  AND  DEVIATING  ACCELERATIONS.  159 

DKFLECTING  AND  DEVIATING  ACCELERATIONS. — We  see,  from  page  80,  that  in 
equations  (2)  all  terms  containing  &?J,  o^,  col  give  the  component  deflecting  accelerations, 
the  other  terms  give  the  component  deviating  accelerations.  If,  then,/r  is  the  deflecting  or 
radial,  and  fa  the  deviating  or  axial,  acceleration,  we  have  for  the  components  of  the 
deflecting  acceleration 


(4) 


(3) 

frt   ~    (  ^4~   2)  <*%   (  ' 

and  for  the  components  of  the  deviating  acceleration 

fa*  =  (y  -i-f)™*™,  +  ( *. 

.     fay    =    (  Z  +  Z)B>y<0,  +    (  X 

faz=(x+*}<*>,GOx  + 

All  these  equations  are  general.     For  axis  through  centre  of  mass  we  have  ~x=  o,  y  =  o, 

RESULTANT  ACCELERATION. — Let  the  centre  of  mass  O  have  the  acceleration  of 
translation  f  and  the  components  fx,  fy,  fz.  Then  the  components  of  the  resultant 
acceleration /are 

/,  =Z  +/P,  +/,  =  f*  +f«*  -h /«  +  A, 


f,  =7,  +/<>«  +A  =7,  +/«  +/«  +A; 

or  substituting  the  values  oifax,  fay,  fa*,  fr**  fry,  /«  ,  /<*  ,  A'  /<«  from  equations  (4), 
(3)  and  (i),  we  have 


f*  =  7*  +  ( 

fy  =  fy+  (*~+  *)«/».  +'(*+*)»^^ 

^ 


These   equations    are    general.      For   axis   through   centre    of    mass    we    have    x  =  o, 
=  o,  5  =  o.    In  any  case  the  resultant  acceleration  of  P  is  given  by 


/=     Vf**    +fyZ  +/?•    '    •.         •        •        •       .-        •        •        •      -.        (6) 

Its  line  representative  passes  through  P,  and  its  direction  cosines  are 

.......     (7) 


t  , 

RESULTANT  ANGULAR  ACCELERATION.  —  The  resultant  angular  acceleration  is  given  by 

'a?  +  «     +^?  .....      .....      (8) 


160  KINEMATICS  OF  A  RIGID  BODY.  [CHAP.  II. 

Its  line  representative  passes  through  O,  and  its  direction  cosines  are 

cos  a  =  ^,        cos  0  =  -r,        cos  y  =  -£  ........     (9) 

MOMENT  OF  ACCELERATION.  —  Let  the  moment  of  the  acceleration  of  P  about  the 
axis  O'a'  through  any  point  O'  be  M'r  Then  for  the  component  moments  M'fx,  M'fy,  Mft 
about  the  axes  X  ',  V,  Z'  Ave  have 

about  O'X'  ......  M',*= 

"       0'Y>  ......  ^= 

"      O'Z'  ......  M*= 

where  fx,  /„  /,  are  given  by  equations  (5).     Substituting  these  values,  we  have  for  rotation 
and  translation 

M'fx  =  f.(y  +y)-ffc  +  **)_+(?+  y)C*  +  *)(».<»*  -  «,)-(«+*)(  f  +  *)(«,<»,+  «0  1 
4-  (* 


For  rotation  only  fx  =  o,  ^  =  o,  yj  =  o.  For  axis  through  centre  of  mass  J  —  o, 
J  =  o,  5"=  o. 

ACCELERATION  ALONG  THE  Axis.  —  For  the  acceleration  of  translation  /,  along  the 
axis  of  angular  acceleration  we  have 

fa  =  fx  cos  a  +  fy  cos 
or,  from  (9), 


This  is  called  the  axial  acceleration  or  the  acceleration  of  advance.  It  is  the  acceleration 
with  which  the  body  moves  along  the  axis  of  angular  acceleration. 

ACCELERATION  NORMAL  TO  THE  Axis  OF  ANGULAR  ACCELERATION.  —  The  com- 
ponents along  the  co-ordinate  axes  of  the  axial  acceleration  are 

fa  cos  a  ,         ft  cos  fi,         fa  cos  y. 

If  we  subtract  these  from  the  components  fx  ,  fy  ,  f,  ,  we  have  the  components  of  the 
acceleration  of  translationy,,  normal  to  the  axis  of  angular  acceleration, 


=  A  -  fa  cos  a  =  fjc—fa°^, 


/„=/„  -/.cos  /*=/,-/„ 


.-  =/.-/-  cos  y  =  /,-/,  — '• 


(12) 


CHAP.  II.] 


INSTANTANEOUS  AXIS  OF  ACCELERATION. 


161 


Equations  (8),  (9)  and  (11)  give  the  screw-twist  about  the  axis  of  angular  acceleration 
through  the  centre  of  mass.  Equations  (8)  and  (9)  give  the  angular  acceleration  and  the 
direction  of  the  axis  of  angular  acceleration,  and  equation  (i  i)  gives  the  acceleration  fa  along 
this  axis.  Equations  (12)  give  the  components  of  the  acceleration  of  translation  normal  to 
this  axis. 

INSTANTANEOUS  Axis  OF  ACCELERATION. — Let*,  y,  z  be  the  co-ordinates  of  any  point 
of  the  instantaneous  axis.  Since  the  normal  acceleration  for  any  point  of  this  axis  is  zero, 
we  have 

7n*  +  My-?"*    =  0»  fny   +   XOi*    ~   *<**    =   O,  fnz  +  yUje    -    XOiy    =   O.  (13) 

These  are  the  equations  of  the  projections  of  the  instantaneous  axis  of  acceleration  on 
the  three  co-ordinate  planes. 

Let  x^  yv,  zl  be  the  intercepts  of  these  projections  on  the  co-ordinate  axes.  Then,  in 
equations  (13),  making^  =  o  and  z  =  o  in  the  last  two,  we  have 


_<_^. 

a' 


making  x  =  o  and  z  =  o  in  the  first  and  last, 


^=^.= 


/» 

ctr 


making  y  =  o  and  x  =  o  in  the  first  two, 


fny 


04) 


Let  the  perpendicular  from  the  centre  of  mass  O  upon  the  axis  of  angular  acceleration 
be/,  and  let/^,  py,  pz  be  its  projections  on  the  co-ordinate  axes.  Let  the  intersection  of/ 
with  the  axis  be  O',  so  that  OO'  =  p.  Then,  just  as  on  page  157,  we  have 


P*  =  ~ 


Py  ~  ~ 


fny<* 


(15) 


ot  +  ct-y    "    a*  +  a* 
We  have  also,  from  (13)  and  (9), 

/*•  =  fa  cos  a  —  a(pz  cos  /?  —  py  cos  y), 
fy=fa  cos  /3  —  <x(px  cos  Y  —  P*  cos  «), 
/2  =  /a  cos  Y  —  vt(py  cos  a  —  px  cos  /?), 


ax  =  a  cos  a, 
ciy  •=.  a  cos  fi, 
az  =  a  cos  7, 


(16) 


When,  therefore,  the  components  of  change  of  motion  fx,fy,fs,  ax->  <*y ,  <*z  are  given 
for  the  centre  of  mass  O,  we  have  a  from  (8),  the  direction  of  the  instantaneous  axis  of 
acceleration  from  (9),  and  the  position  of  this  axis  from  (15).  We  have  also  the  acceleration 
fa  along  the  axis  from  (n),  and  the  normal  acceleration/,  from  (12). 

On  the  other  hand,  if  the  position  and  direction  of  the  instantaneous  axis  of  acceleration 
are  given,  together  with  the  acceleration  fa  along  it  and  the  angular  acceleration  a  about  it, 
the  components  of  change  of  motion  are  given  by  (16). 

THE  INVARIANT  FOR  COMPONENTS  OF  CHANGE  OF  MOTION. — From  (n)  we  have 


faa  =  fxax  +  fyay 


(17) 


162 


klN£M/1TICS  Of  A  RIGID  BODY. 


fCHAP.    11. 


But  whatever  point  we  take  as  origin,  a  does  not  change  and  the  acceleration  /,  along 
the  instantaneous  axis  of  acceleration  does  not  change.  This  quantity  (17)  is  therefore 
called  the  invariant  of  the  components  of  change  of  motion. 

If  in  any  case  the  invariant  is  zero,  we  must  evidently  have  either/,  or  a  zero.  If  « 
is  not  zero  but  the  invariant  is  zero,  then  the  acceleration  /,  along  the  axis  must  be  zero. 
In  this  case  the  condition 


*(**  -\-fytXy 


=  o 


is  the  condition   for  a  twist,  or  angular  acceleration  only  about  the  instantaneous  axis  of 
acceleration. 

If  fa  is  not  zero  but  the  invariant  is  zero,  then  a  must  be  zero  and  we  have  no  angular 
acceleration. 

Examples. — ( i)  A  baseball  rotates  about  an  axis  through  its  centre  of  mass,  with  angular  velocity  GO,  and  its 
centre  of  mass  has  a  -velocity  v  making  an  angle  with  oo.     Find  the  resultant  motion. 

ANS.  Take  v  coinciding  with  the  axjp  of  Y,  and  the  plane 
of  v  and  GO  as  the  plane  of  XY. 

The  components  of  motion  are 

vx  =  o,         vy  =  v,         v,  =  o,         va  —  v  cos  ft, 

00X  =    00  COS   a,  GOy  =   00  COS  ft',  CO,  =  O, 

where  a  and  ft  are  the  angles  of  GO  with  the  axes  of  X  and  Y. 

For  the  position  of  the  instantaneous  axis  of  rotation  we 
have,  from  equations  (12),  page  155,  and  equations  (15),  page 
Y     157. 

Vnx  =  ~  V  COS  ft,  V,,y  =:~V  —  V  COS4  ft,  Vnt  =  O, 


Show  how  this  can  be. 


where  va  is  the  component  of  the  velocity  along  G^  or  va  = 
z7  cos  a. 

The  motion  of  the  ball  is  then  a  screw-spin  consisting  at  any  instant  of  rotation  GO  about  an  axis  O'a' 
parallel  to  the  axis  of  rotation  at  the  centre  of  mass  O  at  the  distance  pz  from  the  centre  of  mass,  and  a 
velocity  of  translation  Wa  along  this  axis.  Or  rotation  to  about  the  axis  Oa,  througli  the  centre  of  mass, 
velocity  va  along  this  axis  and  z7«  of  translation  of  this  axis,  where  ~-vn  —  —  "v  cos  ft. 

Disregarding  the  action  of  gravity,  the  centre  of  mass  O  moves  in  the  resultant  of  z/"a  and  VH  ,  or  along 
OY  with  uniform  velocity  v. 

Since,  owing  to  gravity,  the  ball  falls  vertically,  the  centre  of  mass  O  moves  in  a  vertical  curve,  the 
plane  of  which  intersects  the  plane  XY  in  a  straight  line  OY. 

(2)  Kail-players  assert  that  this  intersection  is  not  a  straight  line,  but  a  curve. 

ANS.  No  account  has  been  taken  of  the  resistance  of 
the  air. 

ist.  CURVATURE  NORMAL  TO  THE  PLANE  OF  GO  andv. 
—  Owing  to  rotation  ao  about  the  axis  Oa  and  the  velocity 
vn  of  translation  of  this  axis,  the  velocity  of  all  points  of  the 
advancing  quadrant  Obc  will  be  greater  than  for  correspond- 
ing points  of  the  receding  quadrant  Ocd. 

The  normal  pressure  «  on  the  advancing  quadrant  Obc 
will  then  be  greater  than  the  normal  pressure  n'  on  the  re- 
ceding quadrant  Ocd. 

The  resultant  pressure  N  at  the  centre  O  makes  an 
angle  with  T-n  and  its  component  NH  normal  to  VH  and  to 
the  plane  of  oo.  and  v  is  always  away  from  the  advancing 
quadrant  Obc  and  causes  curvature  in  this  direction. 

It  is  generally  supposed  that  the  curvature   is  due  to  the  ball  rolling  upon  a  cushion  of  compressed  air 


CHAP.  II.  J 


EXAMPLES. 


163 


in  front  (5f  it,  but,  as  we  see,  the  direction   of  Nn  is  always  opposite  to  the  direction  in  which  the  ball  would 
thus  tend  to  roll. 

3d.  CURVATURE  IN  THE  PLANE  OF  a>  AND  v. — The  air  causes  a  retardation  of  v~a  and  17n,  that  is  of 
translation  along  the  axis  and  translation  of  the  axis. 

If  these  retardations  were  proportional  to  the  velocities  there 
would  be  no  curvature  in  the  plane  of  GO  and  v'. 

But  these  retardations  are  more  nearly  proportional  to  the 
squares  of  the  velocities.  Hence  the  greater  component  is  retarded 
proportionally  more  than  the  lesser,  and  we  have  curvature  in  the 
direction  of  the  lesser.  If,  then,  ~v  makes  an  angle  with  GO  less  than 
45°,  we  have  curvature  Oa  in  the  direction  of  z7M  ;  if  greater  than  45°, 
we  have  curvature  Ob  in  the  direction  of  z>a- 

Thus  by  giving  the  ball  a  spin  the  pitcher  is  able  to  make  it  curve 
right  or  left  in  the  plane  of  GO  and  v,  and  at  the  same  time  a  curve  at 
right  angles  to  this  plane  has  also  been  proved. 

We  havejthen,  an  acceleration/  right  or  left  in  the  plane,  and  an 
acceleration /„  parallel  to  Nn  in  the  first  figure  at  right  angles  to  this  plane. 

The  curvature  will  always,  then,  be  opposite  in  direction  to  the  direction  in  which  the  ball  would  tend  to 
roll  on  the  forward  cushion  of  compressed  air.  So  far  as  any  such  action  exists  it  decreases  the  curvature. 

If,  then,  TJ  is  at  right  angles  to  oo,  or  makes  an  angles  of  45°  with  it,  there  is  no  curvature  in  the  plane  of 
v  and  on  at  that  instant,  but  in  all  cases  and  at  every  instant  there  is  curvature  at  right  angles  to  this  plane 
opposite  to  the  direction  in  which  the  ball  would  tend  to  roll  on  the  forward  cushion  of  compressed  air,  and 
diminished  more  or  less  by  this  tendency. 

(3)  A  vertical  circle  of  radius  r  =  2 ft.  rotates  about  a  fixed  horizontal  axis  through  the  centre  of 
mass  at  right  angles  to  the  plane  of  the  circle  with  angular  velocity  of  j  radians  per  sec.  Find  the  velocity 
and  central  acceleration  for  any  point  in  general. 

ANS.  (Compare  with  example  (i),  page  146.)  Take  co-ordinate  axes  as  shown  in  the  figure.  Then 
r  =  2  ft.,  a>  =  ooy  =  +  3  radians  per  sec.,  oox  =  o,  o>2  =  o,  vx  =  o,  ~vy  =  o 
vz  =  o.  The  components  of  motion,  then,  are  l^nown.  Let  x  and  z  be 
the  co-ordinates  of  any  point  P.  For  all  points  y  =  o.  Then,  from 
equations  (i),  page  154,  we  have 

Vx    =   ZOOy  ,  Vy  =  Q,  Vz    =    —    XOOy (i) 


The  resultant  velocity  is 


v  =  4/Vx  + 
Yand  its  direction  cosines  are 


j  +  v\  = 


Vx  Z  a  Vy 

cos  a  —  —  =  +  — ,         cos  p  =  —  =  o, 
v  r  v 


V,  X 

COS  Y  =  -    = • 

v  r 


(2) 


(3) 


/  These  are  the  same  equations  as  on  page  146. 

From  equations  (15),  page  157,  we  see  that  the  axis  OY  is  the 
instantaneous  axis,  or  the  spontaneous  and  instantaneous  axes  of 
rotation  coincide.  From  equations  (17),  page  157,  we  see  that  the  invariant  is  zero,  and  we  have  a  spin 
only  about  the  instantaneous  axis.  , 

From  equations  (2),  page  158,  we  have   for   the   component   central    accelerations,    since  J  =  o,  y  =  o, 
z  =  o,  and/  —  o, 


fftx    —    —   XK>1   ,  fpy    =   O,  fpz    =   ZGOy. 


(4) 


The  resultant  central  acceleration  is 


(5) 


and  its  direction  cosines  are 


cos  a  =    ,    = , 


cos  ft  -  fy-  =  o, 
IP 


(6) 


OF  A  RIGID  BODY. 


[CHAP.  It. 


These  are  the  same  equations  as  in  example  (i),  page  146.  From  equations  (4),  page  159,  we  see  that 
the  component  deviating  accelerations  are  /„.,  =  o,  fay  =  o,  /„,  =  o,  or  the  plane  of  rotation  does  not  change 
in  direction. 

(4)  A  vertical  c  irclt  of  radius  r  =  2  ft.  rolls  on  a  horizontal  straight  line.  The  centre  moves  parallel  to 
that  line  with  a  velocity  of  6  ft.  per  sec.  Find  the  angular  velocity  and  the  ^elocity  and  central  acceleration 
at  any  point. 

ANS.  (Compare  example  (2),  page  146.)  Take  co-ordinate  axes  as  shown  in  the  figure.  Then  r  =  2  ft. 

and  the  components  of  motion  are 


=  +  6  ft.  per  sec.,        vy  =  o, 
=  o,  ooy  =  <a, 


v,  =  o. 
a),  =  o. 


Since  the  circle  rolls,  rooy  =  ~vx  ,  and  hence 

ooy  =  —  =  +  3  radians  per  sec. 

From  equations  (4).  page  1  54,  the  component  velocities  of  any 
point  P  whose  co-ordinates  are  .r,  z,  are 


(i) 


The  resultant  velocity  is 


where  r1  is  the  radius  vector  of  the  point  relative  to  the   bottom 
point  b. 


The  direction  cosines  of  v  are 


'(/—  ..  if.. 

cos  a  =  — -,          cos  ft  =  -*  =  o, 

V  V 


V* 

cos  y  =  — . 


(3) 


These  are  the  same  equations  as  in  example  (2),  page  146. 

From  equations  (15),  page  151,  we  have  for  the  co-ordinates  of  the  instantaneous  axis  of  rotation 


P*  = 


The  axis  b  Y'  through  the  bottom  point  b  parallel  to  O  Y  is  then  the  instantaneous  axis  of  rotation. 

From  equation  (17),  page  157,  we  see  that  the  invariant  is  zero  and  we  have  a  spin  only  about  the 
instantaneous  axis. 

From  equations  (2),  page  158,  we  have  for  the  component  central  accelerations,  since  x  =  o,  ~y  =  o, 
F  =  o  and  y  =  o. 


The  resultant  is 


and  its  direction  cosines  are 


/'  =    tftf  +  ft  = 


(4) 

(5) 


(6) 


These  are  the  same  equations  as  in  example  (2),  page  146. 

From  equations  (4),  page  159,  we  see  that  the  component  deviating  deflections  are /a*  =  o,fay  =  o, 
fa  =  o.  or  the  plane  of  rotation  does  not  change  in  direction. 

(5)  Let  a  vertical  circle  of  radius  r  =  2ft.  roll  on  a  horizontal  plane.  The  centre  moves  with  a  velocity  of 
6ft.  per  sec.  At  the  same  time  let  the  plane  of  the  circle  rebate  about  a  vertical  diam,-tfr  -with  an  angular  velocity 
nf  2  radians  per  sec.  down-wards.  Find  the  angular  velocity  about  the  horizontal  axis,  and  the  velocity  and 
central  acceleration  of  any  point. 


CHAP.  II.] 


EXAMPLES. 


ANS.  (Compare  example  (3),  page  147.)    Take  co-ordinate  axes  as  shown  in  the  figure.     Then  r  =  2  ft. 
and    the  components  of   motion  are  TJX  —  +  6  ft.  per  sec.,  vy  =  o, 
~vz  =  o,  tax  —  o,  ooy,(az  =  —  2  radians  per  sec. 

Since  the  circle  rolls, 


rooy  =  z^,    or     ooy  =  —  =  +  3  radians  per  sec. 

'From  equations  (4),  page  154,  the  component  velocities  of  any 
point  P  whose  co-ordinates  are  x,  z,  are 


vx  =  vx  +  zooy  =  rooy  + 
The  resultant  velocity  is 


vx  =  — 


(0 


l  =  \/[(z 


where  r1  is  the  radius  vector  of  the  point  relative  to  the  bottom 
point  b. 

The  direction  cosines  of  v  are 


cos  a  =  —  cos  6  =  — , 

v  '         v 


cosr  =  £ 


(2) 


These  are  the  same  equations  as  in  example  (3),  page  147. 

From  equations  (15),  page  157,  we  have  for  the  co-ordinates  of  the  instantaneous  axis  of  rotation 


13 

The  instantaneous  axis  passes  through  this  point  and  is  parallel  to  the  spontaneous  axis  Oa,  It  there- 
fore passes  through  the  bottom  point  b. 

From  equations  (17),  page  157,  we  see  that  the  invariant  is  zero  and  we  have  a  spin  only  about  the 
instantaneous  axis. 

From  equations  (2),  page  158,  we  have  for  the  component  central  accelerations,  since  x  =  o,_y  =  o, 
z  =  o  and^'  =  o, 


The  resultant  is 


/p,  =  -  xoo;  -  xaol ,          fpy=  2 

fp   =    Vfpl    +fpy    +/pl    = 


z     =      -     ZQOy. 


(4) 


(5) 


and  its  direction  cosines  are 


cos  a=, 

Jp 


cos  ft  =        ,        cos  y=f- 
fp  fp 


(6) 


These  are  the  same  equations  as  in  example  (3),  page  147. 

From  equations  (3),  pa^e  159,  we  have  the  component  deflecting  accelerations 

frx    -    -   Xto}    -   XUl  ,  fry   =   O,  fn    —    -   ZOOy  , 

and  from  equations  (4),  page  159,  we  have  the  component  deviating  accelerations 

fax   =   O,  fay  —   ZOJyK)z,  faz  =  O. 

The  plane  of  rotation  therefore  changes  in  direction. 

(6)  Let  a  vertical  circle  of  radius  r  —  2  ft  roll  on  a  horizontal  plane  with  an  angular  velocity  of  j 
radtans  per  sec.,  while  its  centre  describes  a  horizontal  circle  of  radius  3  ft.  with  angular  velocity  about  a  fixed 
axis.  Find  the  angular  velocity  about  the  fixed  axis  ;  the  velocity  and  central  acceleration  of  any  point. 


i66 


KINEMATICS  OF  A  RIGID  BODY. 


[CHAP.  II. 


ANS.      (Compare  example  (4),  page  149.)      Take  co-ordinate  axes  as  shown  in  the  figure.      Let  (7Z'  be 

the  fixed  axis.  Then  r  =  2  ft.,  J  =  3  ft.,  and  the 
components  of  motion  are  i/i  =  +  6  ft.  per  sec.. 
"Vy  =  o,  vt  =  o,  oox  —  o,  ooy  =  -(-  3  radians  per  sec.,  <»,. 
Since  tlie  circle  rolls  and  the  centre  rotates 
about  (yZ',  we  have 

rojy  =  v~x  =   —Jaox  or  a?,  = — r  =  —  2  radians. 

ajj  is  therefore  negative  as  shown  in  the  figure. 
t  From  equations  (2),  page   154,  the    component 

velocities  of  any  point  P  whose  co-ordinates  are  x, 
z,  are,  since  x  =  o,  z  =  o  and  y  =  o, 


The  resultant  velocity  is 


=    —    XU)y. 


vy  +  vl 


where  r'  is  the  radius  vector  of  the  point  from  the  bottom  point  b.    The  direction  cosines  of  v  are 

cos  a  =  ^,         cos  ft  =  —  ,         cos  y  —  ~ 


(2) 


(3) 


These  equations  are  the  same  as  in  example  (4),  page  149. 

From  equations  (15),  page  157,  we  have  for  the  co-ordinates  of  the  instantaneous  tixis  of  rotation 

raol  1  8  , 

pl  =  '     -       =  ~    ft- 


The  instantaneous  axis  passes  through  this  point  and  is  parallel  to  the  spontaneous  axis  Oa.  It  there- 
f-.-.re  passes  through  O  and  the  bottom  point  b. 

From  equations  (17),  page  157,  we  see  that'the  invariant  is  zero,  and  we  h;ive  a  spin  only  about  the 
instantaneous  axis. 

From  equations  (2),  page  158,  we  have  for  the  component  centntl  accelerations,  since  x  =  o,  5"=o  find 


—  saa* 


The  resultant  is 


and  its  direction  cosines  are 


, 

Jp 


cos/5=4r. 

Jt> 


These  are  the  same  equations  as  in  example  (4),  page  149. 

From  equations  (3),  page  159,  we  have  the  component  deflecting  accelerations 


(4) 


(5) 


(6) 


and  from  equations  (4),  page  159,  we  have  the  component  deviating  accelerations 

fa,  =  O,  fay  •=.  ZGOyWt  ,  fat  =  O. 

The  plane  of  rotation  therefore  changes  in  direction. 


CHAP.  II.] 


EULER'S   GEOMETRIC  EQUATIONS. 


167 


Euler's  Geometric  Equations.— Let  OX,  OY,  OZ  be  rectangular  co-ordinate  axes 
fixed  in  the  body  and  therefore  rotating 
with  it,  and  let  the  body  rotate  about 
some  axisjfcm/  in  the  body  and  therefore 
making  invariable  angles  with  these  axes, 
so  that  the  component  angular  velocities 
are  o?r ,  cov ,  GJZ. 

Let  OXV,  OYlt  OZl  be  rectangular 
co-ordinate  axes  whose  directions  in  space 
are  invariable.  For  instance,  the  axis 
OZl  may  be  always  directed  towards  the 
north  pole  or  parallel  to  the  earth's  axis, 
then  Xl  Y1  is  the  plane  of  the  celestial 
equator. 

Let  the  point  O  be  taken  as  the 
centre  of  a  sphere  of  radius  r.  Let  Xl , 
Yl ,  Ziy  X,  Y,  Z  be  the  points  in  which 
this  sphere  is  pierced  by  the  fixed  and  moving  axes. 

Let  the  axes  OX,  OY,  6>Z  have  the  initial  positions  OXlt  OYlt  OZr      Turn  the  body, 

1st,  about  OZ^  through  the  angle  X^Z^P  =  $,  so  that  OXl  moves  to  OP,  and  OYl  to  ON. 

2d,  about  ON  through  the  angle   PNE  =  6,  so  that  OP  moves  to  OE,  and  OZ^  to  OZ. 

3d,  about  OZ  through  the  angle  EOX  =  0,  so  that  OE  moves  to  OX,  and  ON  to  OY. 

It  is  required  to  find  the  geometric  relations  between  0,  0,  i/>  and  <&x,  ooy,  ooz.       These 
geometric  relations  are  called  EULER'S  GEOMETRIC  EQUATIONS. 

The  line  O  N  is  called  the  line  of  nodes  y  0js  the  obliquity  and  $  the  precession. 

The  angular  velocity   of  Z  perpendicular  to  the   plane   of  ZOZ^  or  about  OZ^  at  any 

d>b 
instant  is  -=-.     This  is  called  the  angular  velocity  of  precession.      The  angular  velocity  of  Z 

7  f\ 

along  ZZ^  or  about  ON  is  at  the  same  instant  -j-.     This  is  called  the  angular  velocity  of 

nutation.      The  angular  velocity  of  X  relative  to  E,  or  Y  relative  to  N,  at  the  same  instant 
d<f> 

Draw  ZD  perpendicular  to   OZr       Then  ZD  =  r  sin  0,  and  the  linear  velocity  at  any 

d& 

instant  of  Z  perpendicular  to  the  plane  of  ZOZ^  is  r  sin  0 .  j-,  and  along  ZZ^  at    the  same 

instant  it  is  r -r  •      The  linear  velocity  at  the  same  instant  of  Z  along  YZ  is  roax,  and  along 
ZX  it  is  roo  . 

We  have,  then,  directly  from  the  figure, 

dB 

r  —r-  —  roov  cos  0  4-  rco,  sin  0, 
dt 


r  sin  ti  .  —j-   =  roOy  sin  d>  —  roo^  cos  0, 


or,  since  r  cancels  out, 


de  .     , 

——  =  coy  cos  0  -f-  K)X  sin  <p, 

sin  &  .  -j-  =  &>y  sin  0  —  oox  cos  0. 


(0 


1 68  KINEMATICS  OF  A  RIGID  BODY. 

Combining  these  two  equations,  we  have 

oo,  =  -77  .  sin  0  —  -7-  sin  0  cos  0, 
a/  at 


[CHAP.  II. 


(2) 


-  , 


.  cos  0  +  -     sin  0  sin  0. 


We   have  also  the  linear  velocity  of  E  perpendicular  to  the  plane  of  Z^OE,  equal  to 

dtp  dd> 

r  cos  0  '-ft,  and  of  X  relative  to  E  along  EX,  r-,f  .     We  have,  then,  for  the  velocity  of  X 

along  XY 

dtp    QS  e      ^d<t> 
r(°*  ~  T  dt  °  ~*  dt ' 

or 


(3) 


Equations  (2)  and  (3)  are  Euler's  Geometric  Equations.  They  give  the  relations 
between  wx,  coy,  cot  and  the  angular  velocity  -,-  of  OZ  about  OZ'  ',  -,  of  OZ  about  ON, 

d(f> 

the  line  of  nodes,  —  ,-.-  of  O  Y  relative  to  ON. 
at 

Auxiliary  Angles.  —  From  the  spherical  angles  of  the  figure,  page  167,  considering  A7  as 
a  vertex  in  each,  we  have  for  the  direction  cosines  of  the  moving  axes,  with  reference  to  the 
fixed,  , 

cos  XOXl  —  —  sin  if}  sin  0  -{-  cos  ?/'  cos  0  cos  ft, 

cos  YOXl  =  —  sin  ip  cos  0  —  cos  »/?  sin  0  cos  0, 

cos  ZOX^  =  sin  6  cos  ?/;, 

cos  XOYl  =  cos  if;  sin  0  -|-  sin  0  cos  0  cos  ft, 
cos  F<9K,  —  cos  ?/•  cos  0  —  sin  ^  sin  0  cos  0, 
cos  Z0K,  =  sin  H  sin  ^-, 

cos  XOZ^  =  —  sin  0  cos  0, 
cos    YOZX  =  sin  0  sin  0, 
cos   Z<9Z,  =  cos  0. 

For  the  angles  which  the  axes  Z,  ,  Z  and  ON  make  with  the  axes  X,   Y  and  Z  we  have 

cos  ZflX  =  —  cos  0  sin  0, 
cos  Z,(9K=  sin  0  sin  0, 
cos  Z,(9Z  =  cos  0, 

cos  ZOX  =  o, 

cos  Z6>F  ~  o,  j.  /-\ 

cos  ZOZ=  i, 


cos  A^A'  =  sin  0, 
cos  AY?  Y  —  cos  0, 
cos  NOZ  =  o. 


DYNAMICS.    GENERAL   PRINCIPLES. 


CHAPTER   I. 

FORCE.     NEWTON'S  LAWS  OF  MOTION. 

Dynamics. — We  have  given  in  the  preceding  pages  the  principles  of  KINEMATICS,  or 
the  measurable  relations  of  space  and  time  only,  that  is  of  pure  motion.  But  we  have  to 
deal  in  nature  with  material  bodies  and  force. 

That  science  which  treats  of  the  measurable  relations  of  force  and  of  those  measurable 
relations  of  matter,  space  and  time  involved  in  the  study  of  the  change  of  motion  of  bodies 
due  to  force,  is  called  DYNAMICS  (Svva^is,  force]. 

Material  Particle. — We  have  already  seerf(page  20)  that,  whatever  the  constitution  of 
matter  may  be,  we  can  consider  a' body  as  composed  of  an  indefinitely  large  number  of  in- 
definitely small  PARTICLES,  so  small  that  each  may  be  treated  as  a  point. 

We  denote  the  mass  of  such  a  particle  by  in,  and  the  sum  of  the  masses  of  all  the  par- 
ticles of  a  body  or  the  entire  mass  of  a  body  by  in,  so  that 

m  =  2m. 

Impressed  Force. — It  is  a  fact  of  universal  experience  that  no  particle  of  matter  is  able 
of  itself  to  change  its  own  motion.  If,  then,  a  particle  is  at  rest  it  must  remain  at  rest  unless 
acted  upon  from  without.  If  it  is  moving  at  any  instant  in  a  given  direction  with  a  given 
speed,  it  cannot  change  either  its  speed  or  direction,  that  is,  its  velocity  is  uniform,  unless 
acted  upon  from  without.  This  action  from  without  to  which  the  change  of  velocity  in  any 
case  can  always  be  attributed  we  call  IMPRESSED  FORCE. 

Newton's  First  Law  Of  Motion. — A  body  is  a  collection  of  particles.  If  there  are  no 
impressed  forces,  each  particle  must  then  be  at  rest  or  move  with  uniform  speed  in  a  straight 
line,  and  hence  the  body  itself  has  motion  of  translation  in  a  straight  line. 

This  fact  was  expressed  by  Newton  as  follows : 

Every  body  continues  in  its  state  of  rest  or  of  uniform  motion  in  a  straight  line,  except  in 
sj  far  as  it  may  be  compelled  to  change  that  state  by  impressed  forces. 

This  is  known  as  "  Newton's  first  law  of  motion."  It  implicitly  defines  force  as  that 
which  causes  change  of  motion  of  matter. 

Inertia. — We  may  also  express  this  law  by  saying  that  all  matter  is  inert,  that  is,  has 
no  power  of  itself  to  change  its  state  of  rest  or  motion.  This  property  of  matter  we  call 
INERTIA,  and  Newton's  first  law  we  may  call  the  "  law  of  inertia."  We  recognize,  then, 
not  only  extension  and  impenetrability,  but  also  inertia  as  essential  properties  of  matter. 
That  is,  matter  occupies  space,  two  bodies  cannot  occupy  the  same  space  at  the  same  time, 
and  all  matter  is  inert. 

169 


170  DYNAMICS.    GENERAL  PRINCIPLES.  [CHAP.  I. 

Force  proportional  to  Acceleration. — Acceleration  /  of  a  point  we  have  already  illus- 
trated and  defined  (page  76)  as  time-rate  of  change  of  velocity  when  the  interval  of  time  is 
indefinitely  small.  We  can  only  measure  force  by  its  effects.  These  effects  are  apparently 
different  in  different  cases,  but  in  all  cases  when  analyzed  they  are  found  to  consist  either  in 
change  of  velocity  of  particles  or  changes  of  form  or  volume  of  a  body.  As  change  of  form 
or  volume  of  a  body  is  due  to  change  of  relative  position  and  therefore  change  of  velocity 
of  the  particles,  we  see  that  in  all  cases  the  effect  of  force  is  to  cause  acceleration.  The 
impressed  force  on  a  particle  must  then  be  proportional  to  the  acceleration  /  of  tjie  particle, 
and  have  the  same  direction. 

Force  proportional  to  Mass. — Consider  a  body  composed  of  a  number  N  of  particles, 
and  let  the  acceleration  of  each  particle  be /and  in  the  same  direction,  so  that  the  body  has 
motion  of  translation.  Then  the  force  on  each  particle  is  proportional  to /and  the  entire 
force  on  the  body  is  proportional  to  N f  and  in  the  direction  of  f.  But  the  number  N  of 
particles  is  proportional  to  the  entire  mass  m  of  the  body  (page  20).  The  force  on  the 
body  is  then  proportional  to  the  mass  as  well  as  the  acceleration. 

We  have,  then,  for  a  particle  of  mass  m  or 'a  translating  body  of  mass  m  the  impressed 
force  F  in  the  direction  of /and  given  by 

F=cmf. (i) 

where  c  is  a  constant. 

Unit  of  Force. — Equation  (i)  expresses  the  fact  that  force  is  proportional  both  to  mass 
and  to  acceleration. 

We  see  from  (i)  that  we  shall  always  have 

>=m/, (2) 

if  we  take  c  =  equal  to  unity  and 

[/J]  =  [,„]  X  [/]. 

That  is,  equation  (2)  holds,  provided  we  take  as  our  unit  of  force  that  uniform  force  which 
will  gii-e  one  unit  of  mass  one  unit  of  acceleration  in  the  direction  of  the  force. 

This  is  called  "Gauss's  absolute  unit"  or  the  "absolute  unit  of  force,"  because 
it  furnishes  a  standard  force  in  any  system,  independent  of  the  force  of  gravity  at  different 
localities. 

In  the  foot-pound-second  or  "  F.  P.  S."  system,  then,  the  absolute  unit  of  force  is 
that  uniform  force  which  will  give  a  mass  of  one  Ib.  a  change  of  velocity  in  the  direction  of 
the  force  of  one  ft.-per-sec.  in  a  second.  This  has  been  called  by  Prof.  James  Thompson 
the  POUNDAL.  It  is,  then,  the  English  absolute  unit  of  force. 

The  French  absolute  unit  of  force  is  that  uniform  force  which  will  give  one  kilogram  a 
change  of  velocity  in  the  direction  of  the  force  of  one  metre-per-second  per  second. 

In  the  centimeter-gram-second  or  "  C.  G.  S."  system  the  absolute  unit  of  force  is  the 
uniform  force  which  will  give  one  gram  a  change  of  velocity  in  the  direction  of  the  force  of 
one  centimeter-per-sec.  per  sec.  This  is  called  the  DYNE. 

Weight  of  a  Body. — The  weight  of  any  body  is  the  force  with  which  the  eartli  attracts 
it.  This  force  must  vary,  then,  with  the  acceleration  g  due  to  gravity,  and  this,  as  we  have 
seen  (page  100),  varies  with  the  locality.  The  weight  of  a  body,  then,  varies  with  the  locality, 
while  its  mass  of  course  remains  invariable. 

If,  then,  m  is  the  mass  of  a  body  and  IV  its  weight,  we  have  from  (2),  for  the  weight  in 
absolute  units, 

W '=  tt absolute  units. 


CHAP.  I.]  GRAVITATION  UNIT  OF  FORCE.  171 

But  if  m  is  taken  as  one  unit  of  mass,  then  W\s  numerically  equal  to  g,  or 

ONE    LB.    WEIGHS  g  POUNDALS, 
ONE    GRAM    WEIGHS  g  DYNES, 
according  to  the  system  we  use. 

Since  g  (page  100)  is  about  32  ft.-per-sec.  per  sec.,  the  average  weight  of  one  Ib.  is 
about  32  poundals,  or 

ONE   POUNDAL   IS    THE    WEIGHT   OF  ABOUT   HALF   AN   OUNCE. 
Strictly  speaking,  it  is  the  weight  of  — th  part  of  a  Ib.,  where  g  must  be  taken  for  the 

C>   ^ 

locality. 

Example.— An  athlete  throwing  a  hammer  of  100  Ibs.  at  New  Haven  and  at  Edinburgh  throws  a  heavier 
weight  at  the  latter  place,  by  about  4  poundals,  or  the  weight  of  2  ounces  more.  The  mass  is  the  same 
and  would  "  weigh  "  the  same  on  an  equal  armed  balance  in  both  places.  But  a  spring-balance  would  show 
a  greater  strtech  at  Edinburgh. 

Gravitation  Unit  of  Force. — We  have  seen  (page  170)  that  the  absolute  unit  of  force  is 
that  force  which  will  give  one  unit  of  mass  one  unit  of  acceleration.  -In  the  F.  P.  S.  system 
this  is  the  poundal;  in  the  C.  G.  S.  system  the  dyne.  It  is  the  unit  used  in  physical  measure- 
ments and  generally  when  small  quantities  are  of  importance  and  great  accuracy  is  desired. 

In  ordinary  mechanical  problems  this  unit  is  inconveniently  small.  Also  very  great 
accuracy  is  not  a  requisite.  It  is  therefore  usual  in  mechanics  to  express  a  force  at  any 
locality  by  comparing  it  with  the  weight  of  the  unit  of  mass  at  that  locality.  The  weight  of 
the  unit  of  mass  at  tlie  locality  is  then  taken  as  the  unit  of  force.  Such  a  unit  is  evidently 
not  constant,  but  varies  with  the  locality.  The  variation  is  so  small,  however,  that  it  can  be 
disregarded  in  ordinary  mechanical  problems. 

Thus  if  the  mass  of  a  translating  body  is  m  Ibs.  and  the  acceleration  of  the  centre  of  mass 
is/",  we  have  for  the  force,  if /"is  taken  in  ft.-per-sec.  per  sec., 

F  •=•  m/"  poundals. 

But  since  at  any  locality  where  the  acceleration  of  gravity  is  g  the  weight  of  one  Ib.  is  g 
poundals,  if  we  take  this  as  the  unit  of  force  we  have 

m/ 

F  =  —  pounds, 

<D 

where  g  has  a  variable  value  according  to  the  locality.  The  result  is  that  for  the  same  mass 
m  and  acceleration  f  we  have  a  slightly  varying  force  according  to  locality,  which  is  of 
course  absurd,  strictly  speaking.  The  variation,  however,  being  small,  is  disregarded. 

Some  writers  avoid  this  objection  by  taking  a  constant  value  for  g,  say  32^.  That  is, 
the  weight  of  the  unit  of  mass  at  some  prescribed  locality  is  taken  as  the  unit  of  force.  This 
is  indeed  a  constant  force,  but  still  a  force  which  depends  upon  locality,  whereas  the  absolute 
unit  is  independent  of  locality,  as  it  should  be. 

Whichever  method  we  adopt,  either  the  weight  of  the  unit  of  mass  at  any  locality  or 
at  some  prescribed  locality,  this  weight  is  called  the  GRAVITATION  UNIT  OF  FORCE  and  is 
expressed  in  "pounds." 

When,  then,  we  speak  of  a  "  force  of  ten  pounds"  or  a  force  of  ten  kilograms,"  we  mean 
the  force  of  gravity  at  a  given  place  upon  a  mass  of  ten  Ibs..  or  ten  kilograms.  The  expres- 
sion is  strictly  incorrect,  because  "pound"  and  "kilogram"  denote  mass  and  not  force. 


172  DYNAMICS.    GENERAL  PRINCIPLES.  [CHAP.  I. 

The  expression  is  thus  a  brief  and  allowable  locution  for  the  phrase  "  force  of  attraction  of 
the  earth  for  a  mass  of  ten  lbs."at  a  specified  locality. 

A  "  force  of  ten  pounds"  means,  then,  a  force  of  10^  poundals,  where  g  is  the  accelera- 
tion of  gravity  in  feet-per-sec.  per  sec.  at  the  place  -considered.  A  "  force  of  ten  grams" 
means  a  force  of  log  dynes,  where  g  is  the  acceleration  of  gravity  in  centimeters-per-sec. 
per  sec.  at  the  place  considered. 

In  all  cases,  then, 

F=mf 
gives  force  in  poundals  or  dynes  according  to  the  system  used,  and 

F  =  ^ 

g 

gives  force  in  gravitation  measure,  that  is  in  "pounds"  or  "grams."  Or,  expressed  in 
words  instead  of  symbols, 

MASS  (in  Ibs.)  X  ACCELERATION  (in  ft.-per-sec.  per  sec.}  =  FORCE  IN  DIRECTION  OF 
ACCELERATION  (in  poundals). 

If  we  divide  the  force  thus  found  by  g  in  ft.-per-sec.  per  sec.,  we  obtain  the  force  in 
gravitation  units  ((pounds)  for  the  locality  for  which  g  is  taken. 

MASS  (in  grams)  X  ACCELERATION  (in  centimeter s-per- sec.  per  sec.)  =  FORCE  IN  DIREC- 
TION OF  ACCELERATION  (in  dynes). 

If  we  divide  the  force  thus  found  by  g  in  centimeters-per-sec.  per  sec.,  we  obtain  the 
force  in  gravitation  units  (grams)  for  the  locality  for  which  g  is  taken. 

Thus  if  a  mass  of  25  Ibs.  has  an  acceleration  in  any  direction  of  6.4  ft.-per-sec.  per  sec.,  the  force  in  that 
direction  is  25x6.4  =  1 60  poundah,  or  160  times  the  force  necessary  to  give  a  mass  of  one  Ib.  an  accel- 
eration of  one  foot  per  sec.  in  a  second.  This  is  a  definite  and  constant  force  no  matter  what  the  locality. 

If^  for  any  given  locality  is  32  ft.-per-sec.  per  sec.,  we  can  speak  of  this  as  a  force  of  "-  -  pounds  or  a 
"/(free  of  5  pounds,"  meaning  the  force  of  gravity  upon  a  mass  of  5  Ibs.  at  the  locality  for  which  g  is  32. 

Momentum  and  Impulse. — We  have  seen  (page  76)  that  the  acceleration  f  in  any 
direction  is  equal  to  the  limiting  time-rate  of  change  of  velocity  in  that  direction.  Thus  if 
7-,  is  the  initial  and  v  the  final  velocity  in  any  given  direction  in  the  indefinitely  small  time 
dt,  we  have  for  the  acceleration  in  that  direction 

/=" 


dt 
If,  then,  m  is  the  mass  of  the  translating  body,  the  force  is 

F=      (-^~\     or     Fdt  =  m'v-vj (3) 

The  product  of  the  mass  ;;/  of  a  particle  or  the  ma?s  m  of  a  translating  body  by  its 
velocity  v  in  any  direction  is  called  its  MOMENTUM  'in  that  direction. 

The  product  of  a  force  Fin  any  direction  by  the  indefinitely  small  time  dt  is  called  the 
IMPULSE  in  that  direction. 

Newton's  Second  Law  of  Motion. — Newton  expressed  the  relation  of  equation  (3) 
as  follows: 

Change  of  motion  is  "proportional  to  the  motive  force,  and  takes  place  in  the  direction  of 
t'le  straight  line  in  which  the  force  acts. 


CHAP.  I.j  MEASUREMENT  OF  MASS.  173 

By  "motion"  Newton  here  refers,  not  to  velocity  but  to  mass-velocity,  or  what  we 
have  just  designated  as  "momentum,"  and  his  "change  of  motion"  is  change  of  momen- 
tum. This  law,  then,  is  the  statement  of  equation  (3),  and,  using  modern  terms,  we  can 
restate  it  as  follows : 

Change  of  momentum  in  any  direction  is  proportional  to  the  impressed  force  in  that  direc- 
tion, and  equal  to  the  impulse  of  the  impressed  force  in  that  direction. 

The  first  law  defines  force.      The.  second  law  tells  us  how  to  measure  force. 

Measurement  of  Mass. — Newton's  second  law  also  tells  us  how  to  measure  mass.  Thus 
if /"is  the  acceleration  in  any  direction  of  a  translating  body  of  mass  m,  the  impressed  force 
F  in  the  direction  of  the  acceleration  is 

F=mf. 

If,  then,  any  two  bodies  are  known  to  be  acted  upon  by  the  same  force  and  have  fhe 
same  acceleration,  their  masses  must  be  equal. 

Equal  masses  are  those  to  which  the  same  force  gives  the  same  acceleration. 

Now  we  know  by  experiment  that  the  attraction  of  the  earth  gives  to  all  bodies  falling 
in  vacuum  at  any  given  locality  the  same  acceleration. 

Let,  then,  a  mass  A  be  suspended  by  a  spring  at  any  given  place 
and  cause  an  elongation  A,  of  that  spring. 

At  the  same  place  let  another  mass  B  be  suspended  from  the 

same  spring,  and  suppose  the  observed  elongation  A  is  the  same  """/p1  ,  I~B~ 
as  before.  Then  the  force  of  gravity  on  each  body  is  the  same,  for  this  is  proportional  to 
the  elongation  of  the  spring.  Also,  this  force  acting  upon  the  bodies  produces,  as  we  know 
by  experiment,  the  same  acceleration.  Hence  in  both  cases  the  same  force  gives  the  same 
acceleration  and  the  masses  A  and  B  are  equal. 

Also,  when  two   bodies  A  and  B  exactly  balance  on  an  equal-armed  balance,  we  also 
— I     know  that  the  force  of  gravity  on  each  is  the  same  (page  134). 

We  have,  then,  two  bodies  to  which  the  same  force  gfves  the  same  ac- 
£3  celeration,  and  hence  the  masses  are  equal. 
A  By  means  of  the  balance,  then,  we  have  a  convenient  means,  universally 

adopted,  by  which  we  can  readily  duplicate  the  standard  mass  and  fractions  of  it.  Then, 
by  finding  how  many  standard  masses  balance  any  given  body,  we  can  determine  its  mass 
relatively  to  the  standard. 

Mass  Independent  of  Gravity. — The  intensity  of  the  force  of  gravity,  or  the  attraction 
of  the  earth  for  a  body,  varies  with  the  locality  and  the  height  above  sea  level.  But  evi- 
dently the  mass  of  a  body  or  the  quantity  of  matter  it  contains  remains  unchanged  by- 
locality,  and  would  be  the  same  even  if  beyond  the  attraction  of  the  earth. 

The  equal-armed  balance,  however,  will  correctly  determine  the  mass  in  all  localities, 
because  if  two  bodies  exactly  balance  in  one  locality  they  will  balance  in  any  other,  since 
the  force  of  gravity,  whatever  it  may  be,  is  always  the  same  for  each  wherever  they  balance. 

In  the  case  of  the  spring,  however,  in  the  first  experiment  of  the  preceding  article,  the 
graduation  is  only  correct  for  a  given  locality.  A  certain  elongation  which  corresponds  to 
the  weight  of  a  pound  in  one  locality  will  not  hold  for  another  locality.  The  spring 
measures  the  weight  or  attraction  of  the  earth  for  a  body  in  the  given  locality,  the  balance 
measures  the  mass  correctly  whatever  the  locality. 

When  we  speak  of  the  mass  of  one  pound,  we  refer,  then,  to  a  definite  quantity  of  matter. 

When  we  speak  of  the  weight  of  a  pound,  we  refer  to  a  variable/*?;^,  viz.,  the  attraction 
of  the  earth  at  a  given  locality  for  a  mass  of  one  pound. 


174  DYNAMICS.    GENERAL  PRINCIPLES.  [CHAP.  I. 

Notation  for  Mass. — It  is  customary  when  the  mass  of  a  body  is  2,  3  or  4  times  that 
of  the  standard,  to  write  it  2  Ibs.,  3  Ibs.,  4  Ibs.  Here  the  abbreviation  "  Ibs."  stands  for 
the  latin  word  LIBRA  (balance)  and  thus  indicates  that  the  determination  has  been  made  by 
the  balance. 

To  avoid  confusion,  it  will  be  as  well  to  adhere  to  this  notation.  Thus  "  4  Ibs."  means 
a  mass,  while  "  4  pounds  "  means  the  weight  of  4  Ibs.  at  some  specified  locality.  The 
expression  "  4  Ibs."  should  really  read  "4  libras,"  but  as  the  word  "  libra"  has  never  come 
into  use,  no  one  would  understand,  and  "4  Ibs."  must  be  read,  therefore,  "  four  pounds." 
But  as  the  abbreviation  "  Ibs."  is  in  common  use,  we  can  at  least  make  the  distinction  to  the 
eye  if  not  to  the  ear. 

In  the  C.  G.  S.  system  the  distinction  is  complete,  for  we  speak  of  "4  grams"  when 
we  mean  mass,  and  "  4  dynes  "  when  we  mean  force,  grams  being  measured  by  the  balance, 
and  dynes  by  the  spring  or  "  dynamometer." 

The  term  •'  weighing"  which  is  applied  in  common  language  to  the  operation  of  balanc- 
ing should  not  be  allowed  to  mislead.  "  Weighing  "  a  body  by  a  balance  always  determines 
its  mass  and  not  its  weight.  This  latter  is  the  force  of  gravity  upon  it. 

Newton's  Third  Law  of  Motion. — When  one  body  presses  against  or  pulls  another,  it 
is  itself  pressed  or  pulled  by  this  other  with  an  equal  and  opposite  force. 

When  one  body  has  its  momentum  changed  by  the  action  of  another,  the  other  body 
has  the  same  change  of  momentum  in  the  opposite  direction. 

When  one  body  attracts  another,  this  other  attracts  it  with  equal  and  opposite  force. 

Considering  the  entire  phenomenon  of  this  mutual  action  and  reaction  between  two 
bodies,  we  have  Newton's  third  law:  , 

To  every  action  there  is  always  an  equal  and  opposite  reaction  ;  or  the  mutual  actions  of 
any  two  bodies  are  always  equal  and  opposite. 

Remarks  on  Newton's  Laws. — These  three  laws  of  motion  were  enunciated  by  Newton 
in  1687.  Simple  as  they  appear,  the  science  of  Dynamics  made  no  essential  progress  until 
they  were  recognized. 

These  laws  are  statements  of  facts  of  nature,  not  a  /rz^deductions,  and  they  are  veri- 
fied by  the  accord  of  the  results  deduced  from  them  with  observed  phenomena.  The  proof 
thus  furnished  in  Astronomy,  Dynamics  and  Applied  Mechanics  is  of  such  a  nature  that 
these  laws  are  regarded  as  rigorously  true,  and  deductions  made  from  them  are  accepted 
even  when  such  deductions  cannot  be  directly  verified  by  experiment. 

Stress. — From  Newton's  third  law  we  see  that  the  exertion  of  force  upon  a  body  is 
only  one  side  of  the  entire  phenomenon,  which  includes  the  simultaneous  exertion  of  equal 
and  opposite  forces  between  two  bodies. 

When  we  fix  our  attention  upon  one  only  of  these  bodies  and,  disregarding  the  other, 
consider  only  its  action  upon  the  first,  we  call  this  action  the  impressed  force  upon  the  first. 
But  when  we  have  both  bodies  in  mind  and' wish  to  be  understood  as  viewing  this  force 
as  one  of  the  two  mutual,  equal  and  opposite  actions  between  the  bodies,  or  between  two 
particles  of  a  body,  we  call  it  a  STRESS. 

Stress,  then,  is  always  an  internal  force,  while  impressed  force  or  force  in  general  is 
always  external  to  the  body  or  system  under  consideration. 

Motion  of  Centre  of  Mass. — Suppose  a  body  to  rotate  about  an  axis  through  the  centre 
of  mass  O  with  angular  velocity  GO  and  angular  acceleration  a  at  any  instant. 

For  any  particle  distant  r  from  O  we  have  the  central  acceleration  (page  78) 

/„  -=  ro?2 


CHAP.  I.]  MOTION  OF  CENTRE  OF  MASS.  175 

and  the  tangential  acceleration  (page  85) 

ft  =  rot.. 

If  the  mass  of  the  particle  is  m,  we  have  then,  from  equation  (2), 
page  170,  the  central  force 

O 
Fr  —  mfr  —  mrof 


and  the  tangential  force 

Ft  •=.  mft  =  mrot. 

But  if  O  is  the  centre  of  mass,  for  every  particle  of  mass  m  at  a  mra 

distance  +  r  on  one  side  of  O  there  is  a  particle  of  equal  mass  at  the  same  distance  —  r  on 
the  other  side.  The  tangential  forces  are  then  parallel,  equal  and  opposite  in  pairs,  and  the 
central  forces  are  equal  and  opposite  in  pairs. 

Hence  for  a  rotating  body  the  algebraic  sum  at  any  instant  of  all  the  particle  forces  in 
any  direction  is  zero  if  tJie  axis  of  rotation  passes  through  the  centre  of  mass. 

If,  then,  a  body  has  rotation  only  about  an  axis  through  the  centre  of  mass,  we  have  the 
force  Fin  any  direction  at  any  instant 

F=  2mf=o. 

Suppose,  now,  a  body  to  have  motion  of  translation  only.  Then  every  particle  of  mass  m 
has  the  same  acceleration  in  the  same  direction  at  the  same  instant,  and  if /"is  the  accelera- 
tion of  the  centre  of  mass,  we  have 

F=  2mf—mf. 

Now  we  have  seen,  page  145,  that  the  motion  of  a  body  in  general  consists  at  any 
instant  of  angular  velocity  GO  about  an  axis  throngh  the  centre  of  mass,  and  velocity  of 
translation  v  of  the  centre  of  mass.  Also  the  change  of  motion  at  any  instant  consists  of 
angular  acceleration  a  about  an  axis  through  the  centre  of  mass,  and  acceleration  of  transla- 
tion/of the  centre  of  mass.  We  have  also  just  proved  that  for  the  rotation  alone  F  =  o, 
and  for  the  translation  alone  F  =  m/". 

In  all  cases,  then,  the  motion  of  the  centre  of  mass  of  a  body  is  the  same  as  if  it  were  a 
particle  of  mass  equal  to  that  of  the  body,  all  the  forces  acting  upon  the  body  being  transferred 
without  change  in  direction  or  intensity  to  this  particle. 

It  is  this  property  which  makes  the  centre  of  mass  of  such  importance  in  mechanics. 
So  far  as  the  motion  of  the  centre  of  mass  of  a  body  is  concerned,  we  can  consider  it  as  a 
particle  of  mass  equal  to  the  mass  of  the  body  and  acted  upon  by  all  the  forces  which  act 
upon  the  body,  each  force  u'nchanged  in  magnitude  and  direction. 


CHAPTER  II. 

RESOLUTION  AND  COMPOSITION  OF  FORCES. 

Line  Representative  of  Force. — We  see,  then,  that  the  force  on  a  particle  or  on  a  body 
acts  in  the  direction  of  the  acceleration  and  is  proportional  to  the  acceleration  of  the  par- 
ticle or  of  the  centre  of  mass  of  the  body. 

Force,  then,  has  magnitude  and  direction  and  is  therefore,  like  acceleration  itself,  a 
vector  quantity  and  can  be  represented,  like  acceleration  (page  76),  by  a  straight  line. 

— .  Thus  the  length  of  the  line  AB  represents  the  magnitude  of  the 

^B    force  F  =  m/.     Its  point  of  application  is  A,    and    its  direction  of 
action  is  indicated  by  the  arrow  and  is  always  the  same  as  that  of  the  acceleration  /. 

Resolution  and  Composition  of  Forces, — We  have,  then,  the  triangle  and  polygon  of 
forces  just  the  same  as  for  accelerations  (page  76),  and  we  can  resolve  a  force  into  two  com- 
ponents in  any  two  directions,  or  we  can  find  the  resultant  of  any  number  of  forces,  just  as 
for  accelerations. 

We  have  also  relative  force,  just  as  we  have  relative  acceleration  (page  76),  with 
similar  notation. 

Examples. — (i)  A  mass  of  20  Ibs.  rests  upon  a  horizontal  plat  form  which  has  an  acceleration  f  =  30  ft.-per- 
sec.  per  sec.  Find  its  pressure  on  the  platform  when  the  acceleration  is  downwards  and  upwards.  (Take 
g  =  3*.) 

ANS.  Acceleration  of  the  mass  relative  to  the  earth  is  denoted  by  ra.E  =g)  downwards  ;  of  the  plat- 
form relative  to  the  earth  by  PE  =  30  down.     Acceleration  of  the  earth  relative  to  the    ^ 
platform   is   then   EP  =  30  up.      Hence   acceleration    of   mass    relative    to   platform    is 
m/J  =  g  —  f  =  2  ft.-per-sec.  per  sec.     The  pressure  is  then  m(g  —f)  =  40  poundals,  or 

—  =  1.2;  pounds  gravitation  measure.  g 

If    the  acceleration  /  is  upwards,  we  have  in  the  same  way  the  pressure  m  (g  +/) 
=  1240  poundals,  or  ~-  =  38.75  pounds  gravitation  measure. 

(2)  A  mass  hung  from  a  spring-balance  in  an  elevator  at  rest  registers  exactly  10  Ibs.      When  the  elevator 
starts,  it  is  observed  to  register  for  a  moment  10.23  Ibs.     Find  the  acceleration  of  the  elevator  at  that  instant. 

ANS.  Acceleration/=-£  =  0.8  ft.-per-sec.  per  sec.  upwards. 

(3)  A  mass  of  20  Ibs.  rests  on  a  horizontal  plane  which  is  made  to  ascend,  first,  with  a  constant  velocity  of 
i  ft.  per  sec. ;  second,  with  a  velocity  increasing  at  the  rate  of  i  ft.-per-sec.  per  sec.     Find  in  each  case  the  pres- 
sure on  the  plane,     (g  =  jf.) 

ANS.  In  the  first  case  the  pressure  is  the  weight  of  20  Ibs.  or  640  poundals.  In  the  second  case  the 
acceleration  relative  to  the  plane  is  g  + 1  =  33  ft.-per-sec.  per  sec.  Hence  pressure  is  20  x  33  =  660 

.  .          660 
poundals,  or  —  =  2Oft  pounds. 

d)  A  string  balance  is  graduated  correctly  for  a  place  where  g  =  32.2.  It  is  transported  to  a  place  where 
g  -  J2.  and  when  a  mass  is  hung  on  it  there  it  registers  r.6  Ibs.  Find  the  correct  value  of  the  mass. 

176 


CHAP.  II.]  EXAMPLES.  177 

ANS.  If  m  is  the  actual  mass,  m  x  32  is  the  actual  weight  in  poundals  of  that  mass  at  the  place  where 
it  is  weighed. 

If  the  balance  is  graduated  correctly  for  a  place  where  g  =  32.2,  the  mass  it  indicates  x  32.2  ought  to 
equal  the  actual  weight.  Hence 

m  x  32  =  1.6  x  32.2,    or    m  =  1.61  Ibs. 

(5)  Let  a  uniform  force  of  2  pounds  act  on  a  body  0/40  Ibs.  mass  for  half  a  minute.     Find  the  velocity 
acquired  and  the  space  passed  through,     (g  =  32^) 

ANS.  Since  the  force  is  uniform,/  is  constant  in  direction  and  magnitude.     The  force  is  the  weight  of 

2  Ibs.  or  2g  poundals.     By  the  equation  of  force,  4o/  =  -2g.     Hence  f  =  —  =1.6  ft.-per-sec.  per  sec.     Since 
/  is  uniform,  the  equations  (page  92)  apply  and  we  have 

v  =ft  =  48//.  per  sec.,        s  =  {//•  =  72o//. 

(6)  A  body  acted  upon  by  a  uniform  force  describes  in  ten  seconds,  starting  from  rest,  a  distance  of  25  ft. 
Compare  the  force  with  the  weight  of  the  body,  and  find  the  -velocity  acquired,    (g  =  32.) 

ANS.  s  =  -ft*  (page  92).     Hence/  =  -5  =  -—  =  0.5  ft.-per-sec.  per  sec.;  v  —ft  =  5//.  per  sec.  ;  F  = 

F       /      0.5       i 

m/,  or  ^—  =  =f-  —  —  =  7-  . 
m^      g       32       64 

(7)  A  force  equal  to  the  weight  of  one  Ib.  acts  upon  a  mass  of  18  Ibs.  free  to  slide  on  a  smooth  horizontal 
plane.      The  force  is  parallel  to  the  plane.      When  the  distance  described  is  50  ft.  find  the  time  and  the  velocity 
acquired,     (g  =  32.) 

p"      1  6 
ANS.  The  force  F  =  i  x.  g  poundals.     Since  F  =  m,  we  have  i  x^-=  18  x/,     or  /=-^  =  —  ft.-per- 

sec.  per  sec.     From  equations  page  92,  v  =//,  s  =  %ff*,  hence  /  =  7^  sec.,  v  —  I3i//.  per  sec. 

(8)  Forces  of  20  and  30  units  acting  on  two  bodies  produce  accelerations  of  40  and  50  units  respectively. 
Show  that  the  masses  are  as  10  to  12. 

(9)  Two  forces  produce  in  two  bodies  accelerations  of  25  and  30  units  respectively.     Show  that  if  the  masses 
are  equal,  the  forces  are  as  5  to  6  ;  and  if  the  forces  are  equal,  the  masses  are  as  6  to  5. 

(10)  A  balloon  is  ascending  vertically  with  a  velocity  which  is  increasing  at  the  rate  of  3  ft.-per-sec.  per  sec. 
Find  the  apparent  weight  of  i  Ib.  weighed  in  the  balloon  by  means  of  a  spring-balance,     (g  =  J2.2.) 

ANS.  1.093  pounds. 

(i  i)  A  mass  m  lies  on  a  smooth  horizontal  plane.  -A  uniform  horizontal  force  F  is  continuously  applied. 
How  long  will  it  take  to  move  the  mass  s  ft.  from  rest  ?  Take  m  =  2240  Ibs.  ,  F  =  28  pounds,  s  =  3  ft.  (g=jz.) 

ANS.  F  -  m/  poundals  ;  hence/  =  —  ,  or/  =  ^f^  -    ~g  ft.-per-sec.  per  sec. 
^g  =  32  ft.-per-sec.  per  sec.,  we  have/  =    -ft.-per-sec.  per  sec. 


Since/  is  uniform,  we  have  (page  92) 


=  5  sec. 


(  1  2)  Let  the  mass  m  «=  2240  Ibs.  be  moved  by  a  rope  which  passes  over  the  edge  of  the  plane  on  a  pulley  and 
sustains  a  mass  P  =  28  Ibs.  at  its  other  end.     Disregarding  all  friction  and  mass  of 
pulley  and  rope  and  sup-hosing  the  rope  perfectly  flexible  and  inextensible,  find  how 
long  it  will  take  to  move  the  mass  m  a  distance  s  =  S  ft.  from  rest,     (g  =  32.) 

ANS.  The  student  should  carefully  compare  this  example  with   the  preceding 
and  following. 

Here  the  tension  on  the  rope  is  F  =  m/  poundals,  where/  is  the  acceleration  I_L 

of  m.     Since  P  has  the  same  acceleration  downward,  the  resultant  acceleration  of 
p**g—f-    Hence  the  tension  on  the  rope  is  also  P(g  —  /)  poundals.     Therefore 

W  =  P(S  -/>,    or   /  =  ^  =  ^  =  g  ft,per-sec.  per  sec. 


i78 


DYNAMICS.    GENERAL   PRINCIPLES. 


[CHAP.  II. 


Or  we  may  obtain  the  same  result  as  follows  :  The  moving  force  is  the  weight  of  P  or  thejittraction  of 
gravity  for  P,  or  Pg  poundals,  or  the  weight  of  28  Ibs.,  as  in  Ex.  1 1.    The  mass  moved  is  P  -f  m.     Hence 

(P  +  m)/  =  Pg,     or   /  =  -p^jg 
We  have  for  uniform  acceleration  (page  192) 


,  =  !/(•.    or     5  =  2- 
The  tension  on  the  rope  is  m/  or  P(g  —f)  or 


t    or    /  =  5.051  sec. 


Pm 


^J^=  -  224°  x  -3-  poundals,  or  the  weight  of  .   =  ==£  =  27!?  lbs. 

P  +  m  81  /*+  » 

(13)  Two  masses  P  =  2 '4°  lbs.  and  Q  =  2212  lbs.  are  hung  by  means  of  a  perfectly  flexible  inextensible 
robe  <<vtr  a  pulley.  Disregarding  all  friction  and  the  mass  of  pulley  and  rope,  how  long  will  it  take  for  each 
mass  to  move  through  s  =  5  ft.  from  rest  f  (f  =  32.) 

ANS.  The  student  should  carefully  compare  this  with  the  two  preceding  examples. 
The  tension  on  the  descending  side  is  P(g—f),  on  the  ascending  side  Q(g+f), 
where/  is  the  acceleration.     Hence 

P(g  -/)  =  Q(g  +  /)>    or    /  =  -^p-^r  =  -JTfjj  ft.-per-sec.  per  sec. 

Or  we  may  obtain  the  same  result  as  follows  :  The  weight  of  P  is  Pg  poundals. 
The  weight  of  Q  is  Qg  poundals.  The  moving  force  is  Pg  —  Qg  or  (P  —  Q)g 
poundals,  or  the  weight  of  28  lbs.,  as  in  Ex.  u  and  12.  The  mass  moved  is  P  +  Q. 
Hence 


Since  s  =  -/f,  we  have 


5  =  r  x 


or/  =  7'04  sec- 


The  tension  on  the  rope  is  Q(g  +  /)  or  P(g  —f)  or  -jr~Q  poundals,  or  the  weight  of  -p~rn  =  2225-9 lbs- 

NOTE. — The  moving  force  in  Ex.  11,  12,  13  is  the  weight  of  28  lbs.  In  Ex.  11  the  mass  moved  is 
m  —  2240  lbs.,  hence  28<f  —  m/".  In  Ex.  12  the  mass  moved  is  P  +  m  =  2268  lbs.,  hence  28,^  =  (P  +  m)/. 
In  Ex.  13  the  mass  moved  is  (P  +  Q)  =  4452  lbs.,  hence  28^-  =  (P  +  0/.  In  all  cases,  moving  force  —  mass 
movt-d  x  acceleration. 

The  pressure  on  the  axle  is  the  sum  of  the  two  tensions,  or  (P  +  Q)g  —  (P  —  Q)/.  If  the  pulley  is  not 
allowed  to  rotate,  the  pressure  upon  the  ax  e  would  be  the  weight  of  P  and  Q,  or  (P  +  Q)g.  The  pressure 
on  the  axle  during  motion  is  therefore  less  than  when  at  rest. 


CHAPTER   III. 

CONCURRING  FORCES.  TWO  NON-CONCURRING  FORCES.  MOMENT  OF  A  FORCE. 
RESOLUTION  AND  COMPOSITION  OF  MOMENTS.  RESULTANT  OF  TWO 
PARALLEL  FORCES. 


Concurring  and  Non-Concurring  Forces. — Forces  which  act  at  the  same  point  are  called 
CONCURRING  forces.  If  in  the  same  plane,  they  are  COPLANAR  concurring  forces.  Forces 
which  act  at  different  points  are  called  NON-CONCURRING  forces.  If  in  the  same  plane,  they 
are  coplanar. 

Resultant  for  any  number  of  Concurring  Forces. — Let  any  number  of  forces^,  F2,F3, 
etc.,  act  at  a  common  point  P,  Fig.  (a).  Lay  off  these 
forces  in  order  so  as  to  obtain  the  force  polygon 
OF^^FS,  Fig.  (£).  Then  the  line  OF3  necessary  to 
close  the  polygon,  taken  as  acting  the  opposite  way 
round,  or  from  O  to  F3,  gives  the  direction  and  magni- 
tude of  the  resultant  F. 

This  resultant  must  of  course  act  at  the  same  point 
P  as  the  forces  themselves. 

We  see,  then,  that  any  number  of  forces  acting  upon 
the  same  point,  whether  in  the  same  plane  or  not,  can 
be  reduced  to  a  single  resultant  force  acting  at  this  point. 

We  see  also  from  Fig.  (£)  that  the  component  On  or  nF3  of  the  resultant  F  in  any 
direction  is  equal  to  the  algebraic  sum  of  the  components  of  the  forces  in  that  direction. 

We  see  also  that  if  the  forces  are  all  parallel  the  force  polygon,  Fig.  (6),  becomes  a 
straight  line,  and  the  resultant  F  is  parallel  to  the  components  and  equal  in  magnitude  to 
their  algebraic  sum,  or 

F=  2F. 

Analytical  Determination  of  Resultant  for  Concurring  Forces. — The  same  equations 
Z  and  conventions  hold  for  finding  the  resultant  force  as 

Nf  for  acceleration   (page  77)  or  velocity  (page  67)       We 

have  therefore  only  to   replace  v  in  equations  (i),  (2), 
(3),  (4),  page  67,  by  F  with  the  proper  subscripts. 
~*"F»  We  have,  then,  for   the    components  of   the  re- 

sultant 


Fx  =  Fl  cos  al  -{-  F2  cos  a2  -j-  F3  cos  a3 
_Y  -f  .  .  .  =  2F  cos  a, 

Fy  —  Fl  cos  /^  4-  F2  cos  fi.2  4-  F3  cos  /33 

-f-  •  •  •  =  2F  cos  /?, 

Fz  =  Fl  cos  Xi  +  F2  cos  y2  +  F3  cos  y3 

4-  .  .  .  =  2F  cos  y. 


179 


l8o  DYNAMICS.    GENERAL  PRINCIPLES. 

For  the  magnitude  of  the  resultant 


For  the  direction  cosines  of  the  resultant 


F  F  F 

cos  a  =      ,     cos  ft  =      ,     cos  y  =  -- 


(2) 


(3) 


We  have  also 


Fr  —  Fx  cos  a  -f-  Fy  cos 


F,  cos 


(4) 


Examples.  —  Students  should  solve  examples  (/)  #«</  (2), 


68,  for  forces  instead  of  velocities. 


.B 


Moment  of  a  Force. — Let  AB  be  the  line  representative  of  a  force  F  acting  at  the 
point  A. 

Take  any  point  O  and  draw  Oa  perpendicular  to  AB,  inter- 
secting AB  produced  if  necessary,  at  a.  Let  the  length  of  this 
perpendicular  OA  be  p.  Then  the  product  Fp  is  called  the 
MOMENT  of  the  force  F  relative  to  O.  The  point  O  is  called  the 
CENTRE  OF  MOMENTS,  and  the  perpendicular  /  is  called  the 
LEVER-ARM  for  F  relative  to  O. 

Line  Representative  of  Moment. — This  moment  has  both  magnitude  and  direction  and 
can  therefore  be  represented  by  a  straight  line  and  an  arrow,  just  like  moment  of  acceler- 
ation, page  89. 

Thus  the  line  representative  of  the  moment  in  the  preceding  figure  is  a  straight  line 
OM,  passing  through  the  centre  of  moments  O,  at  right  angles  to  the  plane  of  BAO,  whose 
magnitude  is  Fp,  with  an  arrow  so  directed  that  when  we  look  in  the  direction  of  the  arrow 
the  rotation  indicated  by  the  direction  of  F  relative  to  O  is  seen  clockwise. 

When  we  speak  of  the  "direction"  of  a  moment  we  mean  the  direction  of  its  line 
representative. 

Thus  if  the  force  F  lies  in  a  vertical  east  and  west  plane  and  points  east,  then  if  O  is 
below  F,  the  moment  is  Fp  and  its  direction  north.  If  O  is  above  F,  the  moment  is  Fp  and 
its  direction  south. 

Resolution  and  Composition  of  Moments. — We  see,  then,  that  a  force  moment  is  a 
vector  quantity  and  all  the  principles  of  pages  88,  89  hold. 

We  have,  then,  component  and  resultant  moments  and 
the  triangle  and  polygon  of  moments,  just  as  for  displace- 
ment (page  54),  velocity  (page  66),  acceleration  (page  76) 
and  force  (page  176).  Also,  the  moment  of  a  resultant  force 
is  equal  to  the  algebraic  sum  of  the  moments  of  its  com- 
ponents. 

Moment  about  an  Axis. — Hence,  just  as  on   page  88, 

if  OZ  is  a  given  axis  and  AB  a  force  F  acting  at  the  point  A, 
the  moment  of  F  relative  to  the  axis  OZ  is  the  same  as  the 
moment  A' B'  X  /  of  the  component  A' B  in  a  plane  perpen- 
dicular to  the  axis  OZ  relative  to  the  point  of  intersection  O 
of  the  axis  and  plane. 


CHAP.  III.] 


RESULTANT  OF  TWO  PARALLEL  FORCES. 


181 


Significance  of  Force  Moment. — Force  is  proportional  to  acceleration  and  acts  in  the 
same  direction  (page  176).  We  have  seen,  page  89,  that  the  moment  of  an  acceleration  is 
twice  the  areal  acceleration  of  the  radius  vector.  Hence  the  moment  of  a  force  is  proportional 
to  the  areal  acceleration  of  the  radius  vector. 

m        f=mf  Thus  if  m  is  the  mass  of  a  particle,  /  its  acceleration,  and  F  the 

*"    force,  we  have 

F=  mf. 

The  moment  fp  of  the  acceleration  is  twice  the  areal  accelera- 
6  tion  of  the  radius  vector  Om. 

The  moment  Fp  =  mfp  of  the  force  is  therefore  proportional  to  the  areal  acceleration 
of  the  radius  vector. 

Unit  of  Force  Moment. — We  must  evidently  take  for  the  unit  of  force  moment  one 
unit  of  force  with  a  lever-arm  of  one  unit  of  length.  If  we  take  feet  and  poundals,  our  unit 
is  then  one  "  poundal-foot,"  or  one  poundal  with  a  lever-arm  of  one  foot.  If  we  take  feet 
and  pounds,  our  unit  is  one  " pound- foot '."  So,  also,  we  may  have  the  "  dyne-foot"  or  the 
' '  kilogram-foot. 

Thus  a  force  of  10  pounds  with  a  lever-arm  of  3  feet  gives  a  moment  of  30  pound-feet 
or  y^g  poundal- feet. 

Resultant  of  two  Non-Concurring  Forces. — Let  two  forces  Fv  F2  act  at  the  points  Al , 


A2,Fig.  (a}.      Then  the  magnitude  and  FIG.  (a). 

direction  of  the  resultant  F  are  found  by 

the  triangle  of    forces,    Fig.  (&),  just  as 

for  concurring  •  forces,    page    179.      But 

the  position  of  the  resultant  in  the  plane 

of  Fl  and  F2  has  still  to  be  determined. 

Take  a  point  O  anywhere  in  this 
plane  as  ,a  centre  of  moments,  and  draw 
the  lever-arms/j,  pz  and/ 

Then,  since  the  moment  of  the  result- 
ant relative  to  any  point  is  equal  to  the 
algebraic  sum  of  the  moments  of  the  components,  we  have  in  general 


FIG.  (b). 


F2pz 


(0 


[Regard  must  be  paid  to  sign, 
have  for  the  case  of  the  figure 


Thus  if  we  take  counter-clockwise  rotation  as  positive,  we 


Equation  (i)  holds  good  no  matter  where  the  centre  of  moments  O  is  taken  in  the  plane 
of  the  forces.  Let  us  then  take  it  at  P,  the  point  of  intersection  of  Fl  and  F2  prolonged. 
For  this  point  the  lever-arms  pr  p2  are  zero.  Since  we  must  always  have  the  moment  of 
the  resultant  equal  to  the  algebraic  sum  of  the  moments  of  the  components,  Fp  =  o  and 
the  lever-arm  /  must  be  zero.  The  resultant  F  must  therefore  pass  through  the  point  P, 
and  the  system  reduces  to  two  concurring  forces  acting  at  P.  Hence 

(i)  A  force  acting  at  any  point  can  be  considered  as  acting  at  any  point  in  its  line 
of  direction, 


l82 


DYNAMICS.     GENERAL   PRINCIPLES. 


[CHAP.  III. 


FIG.  (a). 


FIG. 


F-F-F 


(2)  TJie  resultant  of  two  non-concurring  forces  lies  in  the  plane  of  the  forces,  passes 
through  the  point  of  intersection  of  the  forces  produced,  and  can  also  be  considered  as  acting  at 
any  point  in  its  line  of  direction. 

Resultant  of  Two  Parallel  Forces.— Let  us  take  next  two  parallel  forces,  Flt  F2,  acting 

at  the  points  Al  and  A2 , 
either  in  the  same  direc- 
tion, Fig.  (a),  or  in  op- 
posite directions,  Fig.  (&). 

The  resultant  F  is  par-' 
allel  to  the  forces  and  given 
in  magnitude  by 

where  Fl  and  F2  are  to  be 
taken  v/ith  their  proper  signs,  (+)  if  acting  up,  (— )  if  acting  down.  Thus,  in  Fig.  (#), 
F  =  F,  -h  F2 ,  and  in  Fig.  (b)  F  =  Fl  -  F2. 

In  order  to  find  the  position  of  the  resultant  F,  join  the  points  of  application  Al  and  A2 
and  let  O  be  the  point  where  the  resultant  Fr  intersects  the  line  A^A2.  Through  O  and  Av 
draw  Oc  and  A^a  perpendicular  to  F2,  and  let  the  angle  cOA2  or  aAvA2  be  denoted  by  a. 

If  now  we  take  A2  or  any  point  on  the  line  representative  of  F2  as  a  point  of  moments, 
we  have  the  moment  of  F2  for  this  point  zero.  The  moment  of  Fl  is  F^  X  A^a  = 
Fl  X  A^jCos  «.  The  moment  of  F  is  F  X  Oc  =  F  X  OA2  cos  a.  Since  the  moment  of 
the  resultant  is  equal  to  the  algebraic  sum  of  the  moments  of  the  components,  we  have 


Hence 


FX  OAZ  cos  a  =  Fl  X  AVA2  cos  a. 

p 
F  X  OA2  =  Fl  X  A^A2,    or     OA2  =  TT  .  . 


(2) 


If  we  take  A{  or  any  point  on  the  line  representative  of  Fl  as  a  point  of  moments,  we 
have  the  moment  of  F^  for  this  point  zero.  The  moment  of  F  is  F  X  OAl  cos  a.  The 
moment  of  F2  is  Ft  X  A1A2  cos  a.  We  have  then 

F  X  0^!  cos  a=  F2X  AVAZ  cos  a. 
Hence 


FxOAl=FtxAlAtt    or 


(3) 


If  we  take  any  point  on  the  line  representative  of  F  as  a  point  of  moments,  the  moment 
of  F  for  this  point  is  zero.  Hence  the  algebraic  sum  pf  the  moments  of  the  components  for 
this  point  must  also  be  zero.  We  have  then 


or         = 


(4) 


We  see  from  (2)  and  (3)  that  the  distances  OAl  and  OA2  depend  only  upon  the  magni- 
tudes of  Fl  and  F2  and  the  distance  A^A2  between  their  points  of  application,  and  not  at  all 
upon  the  angle  a  or  upon  the  common  direction  of  Fl  and  F2.  For  the  same  forces  Ft  and 
F^  and  the  same  points  of  application  Al  and  A2  the  resultant  F  will  then  always  pass 
through  the  same  point  O,  no  matter  what  the  direction  of  the  parallel  forces  may  be.  This 
point  O  is  therefore  called  the  CENTRE  of  the  two  parallel  forces. 


CHAP.  III.]  EXAMPLES.  183 

We  have  then  the  following  principle : 

The  resultant  of  two  parallel  forces  Flt  F2,  acting  at  the  extremities  of  a  straight  line 
A^A2,  is  in  their  plane  and  equal  in  magnitude  to  their  algebraic,  sum.  It  acts  parallel  to  the 
forces  in  the  direction  of  the  larger  force,  and  always  acts  at  a  point  O  on  the  straight  line  A^A2 
or  on  this  line  produced,  which  divides  this  line  into  segments  inversely  as  the  forces.  Or  the 
products  of  the  forces  into  the  adjacent  segments  are  equal. 

This  principle  is  known  as  the  "law  of  the  lever."  When  the  forces  act  in  the  same 
direction,  as  in  Fig.  (a),  the  resultant  lies  within  the  components.  When  the  forces  act  in  op- 
posite directions,  as  in  Fig.  (b),  the  resultant  lies  outside  the  components  and  on  the  side  of  the 
larger. 

Examples. — (i)  Two  parallel  forces  F\,  Ft,  of  17  and  33  pounds  respectively ,  act  in  the  same  direction, 
and  their  points  of  application  A\,  A*  are  8  ft.  apart.  Find  the  resultant  and  the  distances  OAi,  OA,  of  its 
point  of  application. 

Axs.  ^=50  pounds  parallel  to  the  forces,  and  acting  in  the  same  direction.  OAi  =  5.28  ft.,  OAt 
—  2.72  ft. 

(2)  Find  the  resultant  and  the  point  O  when  the  forces  in  the  preceding  example  act  in  opposite  directions. 
ANS.  F  =  1 6  pounds  in  the  direction  of  the  larger  force.     OA\  =  16.5  ft.,  OAi  —  8.5  ft. 

(3)  Two  parallel  forces  F\ ,  F*  of  12.5  and  23  pounds  act  in  the  same  direction  upon  two  points.     The 
resultant  acts  at  a  distance  of  4  ft.  from  F\.     Find  the  distance  between  the  forces. 

ANS.  6  ft. 

(4)  Resolve  a  force  F  =  52  pounds  into  two  parallel  forces  acting  in  the  same  direction,  F\  and  Ft,  (a) 
when  the  distances  from  F  are  2  and  3  ft.;  (b)  when  F\  =  20  pounds  at  a  distance  of  2  ft. 

ANS.  00  Fi  =  31.2  pounds,  Ft  =  20.8  pounds.     (&)  Ft  —  32  pounds  at  a  distance  from  F  oi  1.25  ft. 

(5)  Resolve  a  force  F  =  20  pounds  into  two  parallel  forces  Ft,  F»,  one  of  which,  F\ ,  acts  opposite  to  Ft :  (a) 
when  the  forces  are  distant  from  F  8  and  3  ft.  ;  (b)  when  F\  is  jo  pounds  and  distant  from  F  6  ft. 

ANS.  (a)  Fi  =  12  pounds,  Ft  =  32  pounds,     (b)  Ft  =  50  pounds  at  a  distance  of  3.6  ft. 


CHAPTER   IV. 

FORCE-COUPLES.    EFFECT  OF  FORCE-COUPLE   ON  A  RIGID  BODY. 

Force-Couple. — Two  equal  and   parallel  forces  acting  in  opposite  directions  and  not  in 

the  same  line  constitute  a  FORCE-COUPLE. 

Thus  the  two   equal    parallel    forces  -\- F,  —  F, 
acting  at  Al  and  A2 ,  constitute  a  force-couple. 

The  perpendicular  distance  /  between  the  forces 
is  called  the  ARM  of  the  couple. 

The  plane  of  the  two  forces  is  the  plane  of  the 
couple. 

Moment  of  a  Force-Couple. — Take  any  point  O 
in  the  plane  of  the  couple  on  the  left  of  —  F,  distant  x  from  —  F.  For  any  and  every 
such  point  we  have  the  moment  (taking  counter-clockwise  rotation  positive) 


+  F 

V 

0 

• 

p 

to 

_.  * 

A, 

Take  any  point  O  in  the  plane  of  the  couple  on  the  right  of  +  F,  distant  x  from  -f-  F. 
For  any  and  every  such  point  we  have  the  moment 


Take  any  point  O  in  the  plane  of  the  couple  between  the  forces  distant  x  from  -f-  F 
and  p  —  x  from  —  F.  For  any  and  every  such  point  we  have  the  moment 

Fx  +  F(p  -x}  =  Fp. 

In  general,  then,  the  moment  of  a  force-couple  relative  to  any  point  in  its  plane  is  constant 
and  equal  to  the  product  Fp  of  the  arm  p  by  one  of  the  forces. 

Line  Representative  of  a  Force-Couple.  —  The  moment  of  a  for-ce-couple  is  of  course 
represented  by  a  straight  line,  just  like  the  moment  of  a  force  in  general  (page  181),  and 
when  we  speak  of  the  "direction"  of  a  couple  we  mean  the  direction  of  its  line 
representative. 

Thus  the  line  AB  represents  the  magnitude/^  of  a  couple  F,  F,  with  lever-arm/,  and 
the  direction  of  the  couple  is  that  of  the  arrow  at  B,  so 
that  looking  from  A  to  B  in  the  direction  of  the  arrow, 
the  rotation  is  seen  clockwise.  The  plane  of  the  couple 
is  at  right  angles  to  AB.  The  line  representative  may 
be  drawn  at  right  angles  to  this  plane  through  anv  point 
of  the  plane  we  please,  since  the  moment  is  the  same  at  all  points, 

184 


CHAP.  IV.]  EFFECT  OF  A   FORCE-COUPLE   ON  A   RIGID  BODY.  185 

Resolution  and  Composition  of  Couples.  —  We  have  then  for  couples  the  same  princi- 
ples as  for  moments  in  general  (page  88).  Thus  We  have  component  and  resultant  couples, 
and  the  triangle  and  polygon  for  couples,  just  as  for  displacement  (page  54),  velocity  (page 
66),  acceleration  (page  76)  and  force  (page  176). 

The  following  corollaries  are  at  once  evident  : 

COR.  i.  —  A  couple  can  be  turned  in  its  own  plane  so  that  the  forces  have  any  desired 
direction,  or  it  can  be  shifted  to  any  position  in  its  own  plane.  For  so  long  as  the  plane 
and  the  arm  and  forces  are  unchanged  the  moment  is  unchanged,  and  the  line  representative 
has  the  same  magnitude  and  direction  and  can  pass  through  any  point  of  the  plane. 

COR.  2.  —  All  couples  whose  planes  are  parallel  and  moments  equal  are  equivalent.  For 
the  line  representative  is  the  same  for  all. 

COR.  3.  —  Any  couple  can  be  replaced  by  another  in  the  same  plane  of  the  same  direction 
and  moment  and  having  any  desired  arm  or  any  desired  force.  For  if  the  plane  and 
moment  are  unchanged,  the  line  representative  is  unchanged. 

COR.  4.  —  Any  number  of  couples  in  the  same  plane  or  in  parallel  planes  can  be  reduced  to 
a  single  resultant  couple  in  that  plane  or  a  parallel  plane  whose  moment  is  the  algebraic 
sum  of  the  moments  of  the  couples.  For  the  line  representatives  of  all  are  parallel  and  can 
all  be  taken  as  acting  at  the  same  point  of  the  plane.  The  resultant  is  therefore  the 
algebraic  sum. 

Resultant  of  a  Force-Couple.  —  A  force-couple  is  only  a  special  case  of  two  parallel 
forces  which  we  have  discussed  already  on  page  183. 

Thus  for  two  parallel  forces  Fl  and  F2  acting  at  points  Alt  A2  we  have  found  for  the 
resultant  in  general 


and  for  the  distance  of  the  point  of  application  C  of  the  resultant  from  Al  along  the  YmeA1A 


Hence  for  the  perpendicular  distance  from  Fl  to  the  resultant  we  have 

distance  =  -j=r  •  p, 

where/  is  the  lever-arm  or  perpendicular  distance  between  the  forces. 

Now,  for  a  couple,  Fl  and  F2  are  both  equal  to  F  and  opposite  in  direction.      Hence 
F  —  o  and 

Fp       M 

distance  =  —  =  —  =±00, 
o         o 

where  Mis  the  moment  of  the  couple.      That  is,  the  resultant  of  a  force-couple  is  a  force  oj 
zero  at  an  infinite  distance,  plus  or  minus  according  to  the  sign  of  the  moment. 

A.  couple  cannot,  therefore,  be  replaced,   nor  can    it  be  held  in  equilibrium  by  a  single 
force,  but  only  by  another  couple, 


1 86 


DYNAMICS.    GENERAL  PRINCIPLES. 


[CHAP.  IV. 


This  is  also  evident  from  the  fact  that  a  single  force  has  a  different  moment  at  different 
points,  the  moment  varying  as  the  lever  arm,  while  a  couple 
has  the  same  moment  for  all  points  in  its  plane.     It  cannot, 
then,  be  replaced  by  a  single  force. 

Effect  of  a  Force-Couple  on  a  Rigid  Body.  —  Let  a  rigid 
body  have  angular  acceleration  a  about  an  axis  through  its 
centre  of  mass  O.  Let  ml  and  ;«2  be  the  masses  of  two  parti- 
cles of  the  body  situated  in  a  line  passing  through  the  axis  of 
acceleration  and  at  right  angles  to  it.  Let  the  distance  of  ml 
be  r,,  and  of  w/2  be  r2,  and  let  the  centre  of  mass  of  the  two 
particles  be  on  the  axis  through  O,  so  that 


Since  all  particles  of  the  body  have  the  angular  acceleration  a  about  the  axis,  the  accel- 
eration of  ml  is  /,  =  r^,  and  of  m2  is  /2  =  r2a.      The  force  upon  m^  is  then 


and  the  force  upon  m2  is 


and  these  two  forces  are  parallel  and  opposite.      But,  since    W1r1  =  w2r2,  these  two   forces 
are  not  only  parallel  and  opposite,  but  equal.      They  therefore  constitute  a  couple. 

Now  the  entire  body  is   composed  of   such  pairs  of  particles,  and  each  pair  therefore 
constitutes  a  couple.     All  these  couples  are  in  parallel  planes,  and  all  have  the  same  direc- 
tion.     They  can,  then,  be  reduced  to  a  single  resultant  couple  having  the  same  direction, 
whose  moment  is  the  sum  of  the  moments  of  all  the  components  (COR.  4,  page  186).     Hence 
If  a  rigid  body  has  angular  acceleration  about  an  axis    through  its  centre  of  mass,  the 
resultant  is  a  force  couple  in  a  plane  at  right  angles  to  this  axis. 
Conversely, 

If  a  rigid  body  is  acted  upon  by  a  force  couple,  the  effect  is  to  cause  angular  acceleration 
about  an  axis  through  the  centre  of  mass  at  right  angles  to  the  plane  of  the  couple. 

Resultant  for  Non- Concurring  Forces  acting  on  a  Rigid  Body. — Let  a  force  F,  whose 
line  representative  is  O'a',  act  at  any  point  O'  of  a  rigid  body. 
Let  O  be  any  point  of  the  body.      If  at  the  point  O  we 
M  apply  two  equal  and  opposite  forces    Oa  =  F  and    Ob  =  F, 

both  parallel  and   equal  to  O'a'  =  F,  the  previous  motion  of 
„ i/Pp 


>a  the  body  is  evidently  not  affected. 

Hence  the  single  force  O'a'  =  F  can  be  replaced  by  the 

same  force,  Oa  —  F,  acting  at  any  given  point  O,  and  a  force  couple  O'a'  and  Ob.  Let  p 
be  the  arm  of  the  couple.  Then  the  moment  of  the  couple  is  M  —  Fp,  and  its  line  repre- 
sentative is  OM  at  right  angles  to  the  plane  of  the  couple. 

When  we  speak  of  the  direction  of  a  moment  or  force  couple  we  always  mean  the 
direction  of  its  line  representative. 

Hence  (compare  page  151)  we  have  the  following  principles: 

(a)  A  single  force  /"acting  at  any  point  of  a  rigid  body  can  be  replaced  by  an  equal 
and  parallel  force  .F  acting  at  any  given  point  O  and  a  couple  whose  moment  is  M  =  Fp  and 
whose  line  representative  through  O  is  at  right  angles  to  the  plane  of  the  couple. 


CHAP.  IV.]  DYNAMICS.     GENERAL  PRINCIPLES.  187 

(#)  Conversely,  the  resultant  of  a  couple  M  and  a  force  F  in  the  plane  of  the  couple  is 

M 
a  single  equal  and  parallel  force  in  that  plane  at  a  distance  /=-—.. 

(c]  If  any  number  of  forces  act  upon  a  rigid  body,  each  acting  at  a  different  point  and 
in  a  different  direction,  then  by  (a)  we  can  replace  each  one  by  an  equal  and  parallel  force 
acting  at  any  point  O,  and  a  couple  whose  line  representative  through  O  is  at  right  angles 
to  the  force. 

We  can  reduce  all  the  forces  at  O  by  the  polygon  of  forces  (page  176)  to  a  single 
resultant  force  jpat  O,  and  all  the  couples  at  O  by  the  polygon  of  moments  (page- 88)  to  a 
single  resultant  couple  M  whose  line  representative  is  not  necessarily  at  right  angles  to  F. 
Hence,  generally, 

Any  system  of  forces  acting  on  a  rigid  body  can  be  reduced  at  any  point  O  to  a  single 
resultant  force  F  and  a  single  resultant  couple  M  ivhose  line  representative  is  not  necessarily 
at  riglit  angles  to  F. 


CHAPTER    V. 


CENTRE   OF   PARALLEL   FORCES.     CENTRE   OF   MASS. 

Centre  of  Parallel  Forces.  —  Let  Flt  Ft,  F3,   etc.,   be  any  number  of  parallel  forces 

acting  at  the  points  Alt  A2,  A3,  etc.,  of  a  rigid  body 
and  given  by  the  co-ordinates  (xl  ,  yv  ,  ^),  (^2,  j/2,  ~2), 
etc.,  and  making  the  angles  a,  /?,  y  with  the  co- 
ordinate axes. 

Then  the  resultant  F  is  parallel  to  the  forces  and 
equal  to  their  algebraic  sum,  or 

F=Fl  +  Ft  +  F9+...  =  2F.     .     .     (i) 

In  taking  the  algebraic  sum  in  (i),  forces  acting 
in  one  direction  are  taken  as  positive,  in  the  other  di- 
rection as  negative. 

If  2F  is  not   zero,  we  have  a  single  resultant. 
It  remains  to  find  its  position. 
For  the  moments  M*  ,  My,  Mz  of  the  forces  about  the  axes  of  X,   Y,  Z  we  have 

Mx  •=.  2F  cos  y  .  y  —  ^F  cos  /?  .  z  =  cos  y2Fy  —  cos  fi2Fz, 

My  =  2F  cos  a  .  z  —  2F  cos  y  .  x  =  cos  a2Fz  —  cos  y2Fx,       ...     (2) 

M,  =  2F  cos  ft  ,  x  —  2F  cos  a  .  y  =  cos  fi^Fx  —  cos  a2Fy.  j 

Let  the  point  of  application  O'  of  the  resultant  be  given  by  the  co-ordinates  ~x,  y,  ~z. 
If  we  transfer  the  origin  O  to  this  point,  and  take  co-ordinate  axes  X',  V,  Z'  at  this  point 
parallel  to  X,  Y,  Z,  we  can  replace  x,  y,  z  in  equations  (2)  by  x  —  Ic,  y  —  ~y,  z  —  2,  and  we 
have  for  the  moments  of  the  forces  about  the  axes  X't  Y',  Z'  through  the  point  of  applica- 
tion O'  of  the  resultant 

Mx'  =  cos  y2F(y  -  y)  -  cos  ftZF(z  -  2),  j 

M,'  =  cos  a2F(s  -z)-  cos  y2F(x  -  *),  j.     ......     (3) 

M.'  =  cos  ft2F(x  -  £)  -  cos  a2F(y  -  y).  j 

But  the  moment  of  the  resultant  is  equal  to  the  algebraic  sum  of  the  moments  of  the 
components,  and  for  the  point  of  application  of  the  resultant  its  moment  is  zero.      Hence 
x,  y,  z  must  have  such  values  as  to  make  equations  (3)  zero. 
We  have  then 


2F(x  -*)= 


=  o,     or     x  = 


=o,     or 


=  o,     or     z  -          . 


(4) 


1 88 


CHAP.  V.] 


PROPERTIES  OF  CENTRE  OF  MASS. 


189 


Equations  (4),  then,  give  the  position  of  the  point  of  application  for  the  resultant  of  a 
system  of  parallel  forces  for  any  assumed  origin  and  co-ordinate  axes. 

This  point  is  called  the  CENTRE  OF  PARALLEL  FORCES.  We  see  that  its  position  is 
independent  of  the  direction  of  the  forces  and  depends  only  upon  the  forces  and  their  points 
of  application. 

Properties  of  Centre  of  Mass. — When  a  body  has  motion  of  translation  all  particles  of 
the  body  move  in  parallel  paths  with  the  same  velocity,  in  the  same  direction,  at  the  same 
instant.  The  acceleration  of  every  particle  is  therefore  the  same  and  in  the  same  direction 
at  any  instant. 

Let  ml ,  m2,  m3,  etc. ,  be  the  masses  of  par- 
ticles of  a  translating  body,  each  one  therefore 
having  the  same  acceleration  /  in  the  same 
direction.  The  forces  on  the  particles  are  mlf, 
mzf>  m3/>  etc-»  and  these  forces  constitute  a 
system  of  parallel  forces.  Let  m  =  2m  be  the 
mass  of  the  entire  body,  and  F  the  resultant 
force. 

Then,  by  the  preceding  article,  we  have 


F=m/ 


F  =  m,f  +  mtf  +  mj  +  .   .  .    =  f2m  =  m/,     . 

and  for  the  point  of  application  O'  of  this  resultant  we  have  the  co-ordinates 

f2mz 


(0 


_ 
~~ 


or,  since /is  constant, 


_ 

m/ 


2  my 




z  =  ~ 


(2) 


But  we  have  seen,  page  21,  that  these  values  of  x ,  y,  ~z  give  the  position  of  the  centre 

of  mass  of  the  body. 

Hence  the  center  of  mass  of  a  body  coincides  with  the  point  of  application  of  the  resultant 
of  that  system  of  parallel  forces  which  acts  upon  all  the  particles  of  the  translating  body.  If 
m  is  the  mass  of  the  body  and  f  the  common  acceleration,  the  resultant  F  is  given  by  F  =  in/. 

Conversely,  if  a  force  F  act  at  the  centre  of  mass  of  a  body  of  mass  m,  it  will  cause  in 

every  particle  the  same  acceleration  f,  in  the  same  direction,  given  by  f  =  g  •    That  is,  it  will 

cause  acceleration  of  translation  only. 

We  have  seen,  page  188,  that  any  number  of  forces  acting  in  any  directions  on  a  body 
can  be  reduced  to  a  force  .F  acting  at  any  point  and  a  couple.  Take  this  point  as  the  centre 
of  mass.  Then  we  have  a  force  F  acting  at  the  centre  of  mass  and  a  couple.  But  we  have 
seen,  page  187,  that  the  effect  of  a  couple  is  to  cause  angular  acceleration  about  an  axis 
through  the  centre  of  mass.  It  does  not,  then,  affect  the  motion  of  the  centre  of  mass  itself. 
We  have  also  just  seen  that  the  effect  of  the  force  F  acting  at  the  centre  of  mass  is  to  cause 

—      F 
in  every  particle  an  acceleration  /in  the  same  direction  given  by/=  -^.-    Since,  then,  the 

acceleration  of  the  centre  of  mass  is  not  affected  by  the  couple,  we  have  for  the  acceleration 


190  DYNAMICS.     GENERAL   PRINCIPLES.  [CHAP.  V. 

_  p 
f  of  the  centre  of  mass  for  any  number  of  forces  in  any  directions,  =/  •= .     Hence  F  —  m/, 

where  (page  188)  F  is  the  resultant  of  all  the  forces  acting  upon  the  body,  each  one  con- 
sidered  as  transferred  to  the  centre  of  mass,  without  change  in  magnitude  or  direction. 

Hence  the  acceleration  f  of  the  centre  of  mass  of  a  body  is  the  same  as  if  all  the  forces 
acting  on  the  body  were  applied  to  the  entire  mass  m  concentrated  there. 

It  is  this  property  which  makes  the  centre  of  mass  of  such  importance  in  Mechanics. 
So  far  as  the  motion  of  the  centre  of  mass  of  a  body  is  concerned,  we  can  always  consider  it 
as  a  particle  of  mass  equal  to  the  mass  of  the  body  and  acted  upon  by  all  the  forces  which 
act  upon  the  body,  unchanged  in  magnitude  and  direction. 

We  have  also  the  following  properties  of  the  centre  of  mass : 

1.  The  attraction  of  the  earth  for  a  body  whose  longest  dimension  is  insignificant  com- 
pared to  the  earth's  radius  is  practically  a  parallel  force  mg  on  every  particle  of  the  body  of 
mass  m.     The  entire  weight  of  the  body,  mg.  acts,  in  such  case,  practically  at  the  centre  of 
mass  for  all  positions  of  the  body,  and  a  body  acted  upon  by  its  weight  only  has  motion  of 
translation. 

Hence  the  centre  of  mass  is  often  called  erroneously  the  "  centre  of  gravity  "  (page  207). 

2.  If  a  rigid  body  at  rest  is  supported  at  its  centre  of  mass  and  is  acted  upon  by  gravity 
only,  it  will  remain  at  rest  in  all  positions. 


CHAPTER   VI. 

NON-CONCURRING  FORCES  IN  GENERAL.     ANALYTIC  EQUATIONS. 

Resultant  for  Non-Concurring  Forces. — We  have  seen  (page  188)  that,  generally,  any 
system  of  forces  acting  on  a  rigid  body  can  be  reduced  at  any  point  O  to  a  single  resultant 
force  F  and  a  single  resultant  couple  M,  whose  line  representative  is  not  necessarily  at  right 
angles  to  F. 

The  force  F  does  not  change  in  magnitude  or  direction  no  matter  where  the  point  O  is 
taken.  '  The  couple  M,  however,  changes  in  magnitude  and  direction  with  the  position  of  the 
point  O,  The  line  representative  of  the  couple  M  at  the  centre  of  mass  is  the  spontaneous 
axis  of  angular  acceleration  (page  152). 

Since  M  is  not  necessarily  at  right  angles  to  F,  we  can  resolve  it  at  any  point  O  into  a 
component  Ma  along  F  and  a  component  Mn  normal  to  F.  But  by  (fr),  page  188,  we  can 

reduce  F  and  Mn  to  an  equal  and  parallel  force  at  a  distance  p  —  ^.     Since  the  couple  Ma 

r 

at  O  can  be  replaced  by  the  same  couple  along  F  (page  186),  we  have  a  force  .Fand  a  couple 
Ma  whose  line  representative  coincides  with  F.      Hence,  generally, 

Any  system  vf  forces  acting  on  a  rigid  body  can  be  reduced  at  any  point  O  to  a  resultant 
force  F,  a  couple  Ma  along  F  and  a  couple  Mn  normal  to  F.  Or  to  a  force  F  at  a  distance 

M 
p  =  -=?  and  a  couple  Ma  along  F  at  this  distance. 

Again,  since  Mis  not  necessarily  at  right  angles  to  F,  we  can  resolve  F  at  any  point  O 
into  a  component  Fa  along  M  at  that  point  and  a  component  Fn  normal  to  M  at  that  point. 
But  by  (&),  page  188,  we  can  reduce  Fn  and  M  to  an  equal  and  parallel  force  Fn  at  a  distance 

M 
p  =  -p.      Hence,  generally, 

Any  system  of  forces  acting  on  a  rigid  body  can  be  reduced  at  any  point  O  to  a  couple  M, 
a  force  Fa  along  M  and  a  force  Fn  normal  to  M.  Or  to  a  force  Fa  at  O  and  a  normal  force 

Fn  at  a  distance  p  =  — . 

ft 

As  M  changes  in  magnitude  and  direction  with  the  point  O,  Fa  and  FM  will  change 
also.  But  the  resultant  F  is  unchanged  in  magnitude  and  direction  no  matter  where  O  is 
taken. 

'The  line  representative  of  M  at  the  centre  of  mass  is  the  spontaneous  axis  of  angular 
acceleration,  and  the  instantaneous  axis  is  parallel  to  it  (page  150). 

Effect  of  any  System  of  Forces  acting  on  a  Rigid  Body. — Since  we  can  take  the  point 
O  where  we  please,  let  us  always  take  it  at  the  centre  of  mass, 

We  have  seen  (page  190)  that  a  force  acting  at  the  centre  of  mass  of  a  rigid  body  causes 

acceleration  of  translation  /of  the  body  in  the  direction  of  the  force.      Also  (page  187)  that 

191 


I92  DYN/1MICS.    GENERAL  PRINCIPLES.  [CHAP  VI. 

a  force-couple  acting  on  a  rigid  body  causes  angular  acceleration  a  of  the  body  about  the 
line  representative  through  the  centre  of  mass  as  axis. 

Therefore  the  force  Fm  along  M  at  the  centre  of  mass  causes  acceleration  of  translation 
fa  of  the  body  along  the  spontaneous  axis  of  angular  acceleration.  The  normal  force  Fn  at 
the  centre  of  mass  causes  acceleration  of  translation  /„  of  this  axis.  The  couple  M  at 
the  centre  of  mass  causes  angular  acceleration  at  about  the  spontaneous  axis  of  accelera- 
tion. 

But  we  have  seen  (page  151)  that  fa,f«  and  a  reduce  to  fa  and  a  about  the  instan- 
taneous axis  of  acceleration,  or  to  a  screw-twist.  Hence,  generally, 

The  effect  of  any  system  of  forces  acting  on  a  rigid  body  at  any  instant  is  to  cause  a  screw- 
twist ',  or  angular  acceleration  about  tlie  instantaneous  axis  and  acceleration  of  translation  along 
this  axis. 

Wrench.  Screw-Wrench. — A  force-couple  acting  on  a  rigid  body  we  call  a  wrench. 
It  causes  angular  acceleration  about  its  line  representative  through  the  centre  of  mass  (page 
187).  The  line  representative  is  the  axis  of  the  wrench  and  may  be  taken  anywhere 
parallel  to  itself  (page  186). 

A  force-couple  or  wrench,  and  a  force  parallel  to  the  axis  of  the  wrench,  we  call  a 
scr  civ -wrench. 

The  preceding  principles  may  then  be  expressed  as  follows: 

Any  system  of  forces  acting  on  a  rigid  body  reduces  in  general  to  a  resultant  force  Fat 
the  centre  of  mass,  and  a  resultant  couple  or  wrench  M  with  axis  at  the  centre  of 
mass. 

Any  system  of  forces  acting  on  a  rigid    body  reduces  in  general  to  a  screw-wrench  Ma 

about  an  axis  coinciding  with  F  at  a  distance  from  the  centre   of  mass  given  by  />  =  -^, 

where  Mn  is  the  component  of  M  normal  to  F. 

Any  system  of  forces  acting  on  a  rigid  body  reduces  in  general  to  a  screw-wrench  M 
with  axis  at  the  centre  of  mass,  a  force  Fa  coinciding  with  this  axis  and  a  force  Fn  normal 
to  it.  Or  to  a  force  Fa  at  the  centre  of  mass,  and  a  normal  force  Fn  at  a  distance  from  the 

M. 

centre  of  mass  given  by/>  =    ~ 

^ '  H 

The  effect  of  any  system  of  forces  acting  on  a  rigid  body  is,  generally,  to  cause  a  screw- 
twist  about  the  instantaneous  axis  of  acceleration. 

This  axis  is  parallel  to  the  spontaneous  axis  of  acceleration,  that  is  to  M  at  the  centre 
of  mass. 

Dynamic  Components  of  Motion. — We  have  just  seen  (page  191)  that  any  number  of 
forces  acting  upon  a  rigid  body  reduces  to  a  single  resultant  force  F  at  the  centre  of  mass  and  a 
resultant  couple  M  at  the  centre  of  mass.  Let  us  take  co-ordinate  axes  through  the  centre 
of  mass  O,  and  let  the  components  of  Fbe  FMt  Fr,  Fx ,  and  the  components  of  M  be  Mx, 
Myt  M  . 

The  motion  of  the  body  under  the  action  of  forces  is  then  known  if  these  six  quantities 
are  known.  These  six  quantities 

Fmt  Fy,  F,,  Mx,  M,,  M,, 

with  reference   to  'co-ordinate   axes  through   the   centre  of  mass,  are  therefore  called  the 
dynamic  COMPONENTS  OF  MOTION  of  the  body. 


CHAP.  VI.] 


GENERAL  ANALYTIC  EQUATIONS. 


193 


General  Analytic  Equations. — Let  any  number  of  forces  Flt  Fz,  F^  act  at  points  A19 


Az,  A3  of  a  rigid  body  given  by  the  co-ordinates^,  y^  ^), 
0*2 »  yv  "2)'  e^c*  -^e^  ^i  make  with  the  co-ordinate  axes 
the  angles  (orp  /?x,  y^,  F2  the  angles  (av  >52,  y^,  etc.  Take 
the  origin  O  at  the  centre  of  mass. 

We  can  replace  each  force  by  an  equal  parallel  force 
in  the  same  direction  at  O  and  a  couple  with  axis  through 
0  (page  187).  All  the  forces  can  then  be  reduced  to  a 
single  resultant  F,  and  all  the  couples  to  a  single  couple 
whose  moment  is  M. 

RESULTANT  FORCE  AT  CENTRE  OF  MASS. — We  have 
for  the  algebraic  sum  of  all  the  components  along  the  co- 
ordinate axes  through  the  centre  of  mass  O 


Fx  =  Fl  cos  al  -J-  F2  cos  ar2  -|-  .  .  .  =  2Fcos  a, 

Fy  =  Fl  cos  A +  ^2  cos  £2  +  •  -  •  =2Fcos/3, 

Fz  =  Fl  cos  yl  4-  F2  cos  y2  -|-  .  .  .  =  2Fcos  y.  ^ 

The  resultant  force  is  then 


(2) 


The  line  representative  passes  through  the  centre  of  mass  O,  and  its  direction  cosines  are 

F. 


Fx  Fy 

COS  Of  =  —  t  COS  p  —  -ff- , 


(3) 


These  equations  give  in  any  case  the  components  of  motion  Fx,  Fy,  Ftt  and  the 
magnitude  and  direction  of  the  resultant  force  .pof  the  screw-wrench.  They  hold  good, 
evidently,  no  matter  where  the  origin  is  taken,  whether  at  the  centre  of  mass  or  not. 

RESULTANT  COUPLE  OR  WRENCH  AT  CENTRE  OF  MASS. — For  the  moments  about  the 
co-ordinate  axes  through  the  centre  of  mass  we  have 


Mx  =  2F  cos  y  .y  —  2F  cos  /? .  z, 
My  =  2F  cos  a  .  z  —  2F  cos  y  .  x, 
M,  =  2F  cos  /3 .  x  —  2F  cos  a  .  y. 

The  moment  of  the  resultant  couple  M  at  the  centre  of  mass  is  then 


(4) 


(5) 


Its  line  representative  passes  through  the  centre  of  mass  O,  and  its  direction  cosines  are 


cos  «  = 


(6) 


These    equations  give  in  any   case   the   components  of  motion  Mx,   M  ,   Mt,  and  the 
magnitude  and  direction  of  the  resultant  moment  M  at  the  centre  of  mass  O.      If  we  take  the 


i94 


DYNAMICS.    GENERAL  PRINCIPLES.  [£HAI'.  VI- 


origin  at  any  other  point,  the  moment  M  changes  in  direction  and  magnitude.  These 
equations  hold  good,  therefore,  only  for  origin  at  the  centre  of  mass. 

RESULTANT  COUPLE  OR  WRENCH  AT  ANY  POINT. — Instead  of  taking  the  origin  at  the 
centre  of  mass  O,  let  us  take  it  at  any  point  O'  not  at  the  centre  of  mass.  Take  co-ordinate 
axes  O 'X't  O'Y't  O'Z'  at  the  origin  O'  parallel  to  OX,  OY,  OZ at  the  centre  of  mass  O  (see 
figure,  page  193).  Let  the  co-ordinates  of  the  centre  of  mass  O  be  ~x,  y,  ~z.  Then  equations 
(4)  still  hold  if  we  put  x  -\-  x,  y  -(-  y,  ~z  -\-  z  in  place  of  x,  y,  s. 

We  have,  then,  for  the  moments  about  the  co-ordinate  axes  at  any  point  O' 


'  =  M,  +  FJ>- 


M:  =  M.+  F~X  -  Fxy. 

The  moment  of  the  resultant  couple  M'  is  then 

M'  = 


(7) 


Its  line  representative  passes  through  the  origin  O'  ,  and  its  direction  cosines  are 


These  equations  give  the  components  of  motion  Mx',  My'  ',  Mt'  and  the  magnitude  and 
direction  of  the  resultant  moment  M'  for  any  point  O'.  If  we  take  the  origin  at  the  centre 
of  mass,  they  reduce  to  equations  (4),  (5)  and  (6). 

MOMENT  Ma  AT  CENTRE  OF  MASS  ALONG  THE  RESULTANT  FORCE.  —  Let  the  compo- 
nent of  the  moment  M  at  the  centre  of  mass  along  the  line  representative  of  the  resultant 
force  F  be  Ma.  Then  we  have 

Ma  =  Mx  cos  a-\-  My  cos  ft  -f  Mt  cos  y, 
where  cos  a,  cos  ft,  cos  y  are  given  by  equations  (3).     We  have  then 


The  resultant  force  F  is  independent  of  where  we  take  the  origin,  and  the  moment  J/, 
at  the  centre  of  mass  remains  the  same  no  matter  where  we  take  the  origin.  Equation  (8) 
therefore  holds  good  no  matter  where  we  take  the  origin,  whether  at  the  centre  of  mass 
or  not. 

MOMENT  Mn  AT  CENTRE  OF  MASS  NORMAL  TO  THE  RESULTANT  FORCE.  —  Let  the 
component  of  the  moment  M  at  the  centre  of  mass  normal  to  the  line  representative  of  the 
resultant  force  F  be  ATn.  The  components  of  Ma  along  the  co-ordinate  axes  through  the 
centre  of  mass  O  are 

Ma  cos  a,          Ma  cos  /3,          Ma  cos  y, 


CHAP.  VI.] 


FORCE  Fa  AT  CENTRE  OF  MASS. 


195 


where  cos  or,  cos/3,  cos  y  are  given  by  equations  (3).  .If  we  subtract  these  components  from 
the  components  Mx,  My,  Mz  of  the  resultant  moment  M  at  the  centre  of  mass,  we  have  the 
components  of  the  normal  moment  M»  at  the  centre  of  mass  O 


Mnx  =  MX- 


Mny  =My- 


cos 


a  cos 


M          M*F* 

=.  Mx „--- 


Mnz  =  Mz  -  Ma  cos  y  =  Mz- 


(9) 


FORCE  Fa  AT  CENTRE  OF  MASS  ALONG  M.  —  Let  the  component  of  the  force  F  along 
M  at  the  centre  of  mass  be  Fa.      Then  we  have 

Fa  =  Fx  cos  a.  -f-  Fy  cos  ft  -}-  F,  cos  y, 
where  cos  a,  cos  ft,  cos  y  are  given  by  equations  (6).      We  have  then 


The  resultant  force  F  is  independent  of  where  we  take  the  origin,  and  the  moment  M 
at  the  centre  of  mass  remains  the  same  no  matter  where  we  take  the  origin.  Equation  (10) 
therefore  holds  good  no  matter  where  we  take  the  origin,  whether  at  the  centre  of  mass 
or  not. 

FORCE  Fn  AT  CENTRE  OF  MASS  NORMAL  TO  M.  —  Let  the  component  of  the  force  F 
normal  to  the  axis  of  M  at  the  centre  of  mass  be  FH.  The  components  of  Fa  along  the 
co-ordinate  axes  through  the  centre  of  mass  O  are 


Fa  cos  or,          Fa  cos 


Fa  cos  y, 


where  cos  a,  cos  ft,  cos  y  are  given  by  equations  (6).      If  we  subtract  these  components  from 
the  components  Fx  ,  Fy  ,  Fz  of  the  force  F,  we  have  the  components  of  Fni 


,  =  F,  -  Fa  cos  y  —  Fz  - 


M 


(n) 


POSITION  OF  RESULTANT  FORCE  F  OF  THE  SCREW-WRENCH  Ma.—  We  have  from  (8) 
the  moment  of  the  screw-wrench  Ma,  and  from  (2)  the  force  /''along  its  axis,  and  from  (3) 
the  direction  of  the  axis.  It  remains  only  to  find  the  position  of  F  relative  to  the  centre  of 
mass. 

The  moment  Mn  at  the  centre  of  mass  is  equal  to  Fp,  where/  is  the  distance  of  F  from 
the  centre  of  mass  O.  Let  x,  y,  z  be  the  co-ordinates  of  any  point  on  the  line  representative 
of  F.  Then  we  have 


—  Fyz  =  Mnx, 


—  Fx  =  Mn 


—  Fxy  =Mnz. 


(12) 


196 


DYNAMICS.    GENERAL  PRINCIPLES. 


[CHAP.  VI. 


These  are  the  equations  of  the  projection  of  F  on  the  three  co-ordinate  planes. 
Let  .r,,  yv  sl  be  the  intercepts  on  the  co-ordinate  axes  of  these  projections.      Then  we 
have  from  equations  (12),  making^  =  o  and  s  =  o  in  the  last  two, 


_M_Hy  _M^ 
~     ~       "" 


making  x  —  o  and  z  =  o  in  the  first  and  last, 

x 

making^  =  o  and  x  =  o  in  the  first  two, 


03) 


^_^ 
F.        F. 


Let  the  distance  of  the  force  F  from  the  centre  of  mass  O  be  /,  and  let  px,  py,  pt  be  its 
projections  on  the  co-ordinate  axes.  Let  the  intersection  of/ 
with  F  be  O',  so  that  OO'  =  /. 

Let  us  consider  the  projection  of  F  on  the  plane  KZ.      In 
the  figure  A  =  Ob  and  py  =  Oc .     We  have  then 


where  £,  and  j/,  are  the  intercepts  Oa  and  Od  given  by  equations  (13). 
We  have  also  the  distance 


ab  =^. 


Hence 


Substituting  the  value  of  py,  we  obtain 


Substituting  the  values  of  ^r,  and  _y,  from  equations  (13),  we  have 


In  the  same  way  we  can  find  pm  and  /,  ,  /,  and  py  on  the  other  two  co-ordinate  planes. 
We  thus  have 


*  "-  /7»_i_  /ra  '      A  =  ~ 

»        *f     i    *• 

A  =  - 


•     •     .     (14) 


CHAP.  VI.]  COMPONENTS  OF  MOTION.  197 

The  screw-wrench  Ma  is  thus  completely  determined  if  the  components  of  motion  Fx  , 
Fy ,  Fz ,  Mx ,  My ,  Mz  are  given.  From  (2)  and  (3)  we  have  the  magnitude  and  direction  of 
the  resultant  force  Ft  and  from  (14)  its  position.  From  (8)  we  have  the  moment  Ma. 

COMPONENTS  OF  MOTION. — Suppose,  on  the  other  hand,  the  screw-wrench  Ma  is 
given,  that  is,  we  have  Ma  and  F  and  the  position  and  direction  of  /''given. 

From  equations  (12)  and  (3)  we  have 

Mx  —  Ma  cos  a  —  F(ps  cos  /3  — /„  cos  y},  Fx  =  F  cos  a; 
My  =  Ma  cos  /3  —  F(  px  cos  y  —  pz  cos  a),  Fy  —  F  cos  fi ; 
Mz  =  Ma  cos  Y  —  F(py  cos  a  —  yx  cos  /?),  F2  =  F  cos  y. 

If,  then,  the  screw-wrench  Ma  is  given,  we  have  from  (15)  the  components  of  motion. 

SCREW-WRENCH  M. — The  screw-wrench  M  is  already  completely  determined.  Its 
axis  passes  through  the  centre  of  mass,  and  its  direction  is  given  by  equations  (6),  its 
moment  by  (5),  and  the  force  Fa  along  its  axis  by  (10). 

POSITION  OF  THE  FORCE  Fn. — We  have  seen  (page  193)  that  the  force  system  reduces 
to  a  force  Fa  along  M  at  the  centre  of  mass,  or  the  spontaneous  axis  of  angular  acceleration, 

M 

and  a  force  F*  normal  to  this  axis,  at  a  distance  p  —  —. 

For  any  point  on  the  line  representative  of  Fn  given  by  the  co-ordinates  x,  y,  z  we 
have  then 

Fnzy  -  Fnyz  =  Mx,         Fnxz  -  Fnzx  =  My,          Fnyx  -  Fnxy  =  Mz. 

These  are  the  equations  of  the  projections  of  FM  on  the  these  co-ordinate  planes. 
Let  xv,  y^  zv  be  the  intercepts  on  the  co-ordinate  axes  of   these  projections.     Then 
we  have,  just  as  on  page  196, 


Proceeding,  then,  just  as  on  page  196,  we  have  for  the  co-ordinates/,,  py,  p*  giving  the 
position  of  Fn 

FnzMy  FnyMz 

P*  7T2      i     c*2  rrz     i      772    » 


WT  <•*  ^*-r^      ".-r^    , 

^i  ^>,  ^    °6) 


Equations  (16)  give  the  position  of  Fn. 

The   values  of   Mx ,   My ,  Mz  are  given  by  equations  (4) ;    the  values  of  M  and  Fa  by 
equations  (5)  and  (10). 

THE  INVARIANT.— From  (8)  and(io)  we  have 


FMa  =  F<tM=FxMx-\-  FyMv+  FZMZ.   .     .     .     .     .     .     .     (17) 

Whatever  point  we  take  for  origin  the  moments  Ma  and  M  at  the  centre  of  mass  remain 
unchanged,  and  F  does  not  change  for  any  origin. 


198  DYNAMICS.    GENERAL  PRINCIPLES.  [CHAP.  VI. 

The  quantity 


remains  the  same,  therefore,  no   matter  where  we   take  the  origin,   and  we   can  write  in 
general 

...     r   .....    (18) 


where    F,,  Fy,  Ft,  and  F  are  given  by  equations  (i)  and  (2);  Mx,  M'y,  Aft  by.  equations 
(7);  At  and  Afa  by  equations  (5)  and  (8)  ;  and  Fa  by  equation  (10). 
Since  the  quantity 


is  constant  no  matter  where  the  origin  is  taken,  it  is  called  the  invariant  for  non-concurring 
forces. 

If  the  invariant  is  not  zero,  the  force  system  reduces  to  the  screw-wrench  Mlt  and  F 
given  by  (8),  (2),  (3)  and  (14).  Or  to  two  forces  Fa  and  Fn  given  by  (10),  (6),  (u)  and 
(16). 

If  the  invariant  is  zero  and  F  and  M  are  not,  we  have  Ma  =  o  and  Fa  =  o,  and  F  is  at 
right  angles  to  Mat  the  centre  of  mass.  In  this  case  the  force  system  reduces  to  a  single 
force  F  =  FH  at  a  point  given  by  equations  (16). 

If  the  invariant  is  zero  but  Ma  is'  not,  then  F  =  o.  In  this  case  the  force  system 
reduces  to  a  resultant  couple  or  wrench  M  at  the  centre  of  mass. 

If  the  invariant  is  zero  but  Fa  is  not,  then  M  =  o  and  the  force  system  reduces  to  a 
resultant  force  at  the  centre  of  mass. 

Thus,  for  example,  let  all  the  forces  be  coplanar.  Take  the  plane  as  that  of  X'  V  . 
Then  F,  —  O,  Mx  —  o,  My  =  o  and  the  invariant  is  zero.  Since  F  and  M  are  not  zero,  we 
have  Fat  right  angles  to  M  at  the  centre  of  mass.  If  F  is  zero,  we  have  then  a  resultant 
moment  only.  If  M  is  zero,  we  have  a  resultant  force  only  If  neither  F  nor  M  are  zero, 
\ve  have  a  single  force  F  only  at  a  point  given  by  equations  (16). 

Hence  a  system  of  coplanar  forces  must  reduce  to  either  a  single  force  or  a  single 
couple.  We  cannot  have  both. 

Let  all  the  forces  be  parallel.  Take  them  all  parallel  to  the  axis  of  27.  Then  Fx  =  o, 
Fy  =  o,  ^^t  =  o  and  the  invariant  is  zero.  Since  F  and  Mare  not  zero,  we  have  F  at  right 
angles  to  M  at  the  centre  of  mass,  and  the  force  system  reduces  to  a  single  force  F  only  at 
a  point  given  by  equations  (16).  If  F  is  zero,  we  have  a  resultant  moment  only.  If  M  is 
zero,  we  have  a  resultant  force  only. 

Hence  a  system  of  parallel  forces  must  reduce  to  either  a  single  force  or  a  single  couple. 
We  cannot  have  both.  If  the  parallel  forces  are  all  in  the  same  direction,  F  cannot  be 
zero  and  we  must  have  a  resultant  force  only. 

Examples.  —  (i)  Find  the  resultant  for  a  system  of  parallel  coplanar  forces  given  by 
F,  =  +    33  Ibs..    x,  =  +  25  ft..    _y,  =  +  13  ft.; 

Ft  =   +     20    "  X*  =r  -  10    "        _y,  =  —  I  5    " 

Ft  =  -    35    "        .r,  =  +  1  5  "      /,  =  -  27  " 

F4  =  —    72    "       x<  =  —  31    "     _y«  =  +  17  " 

Ft  =  +  120    "        jn  =  +  23   "     yt  =  -  19  " 
ANS.  F=  +  66  Ibs.,  p,  =  +  77.1  5  ft.,  p  =  -  36.82  ft. 
If  the  forces  are  parallel  to  the  axis  of  }',  Mz  =  +  5091.9  Ib.-ft. 
If  the  forces  are  parallel  to  the  axis  of  X,  Mz  =  +  2430.  12  Ih.-ft. 

If  we  look  along  the  line  representative  of  the  moment  towards  the  origin,  the  rotation  is  seen  counter- 
clockwise. 


CHAP.  VI.]  EXAMPLES.  199 

(2)  Find  the  resultant  for  the  parallel-force  system  given  by 

Fi  =  4    60  Ibs.,  x\  —  o,  y\  =  o,  Zi  =«  o; 

^2=+    70    "  ;r2  =  +  ift.,  j.,  =  4  2  ft.,  *a=  +  3ft.; 

^3  =  -    90    "  *3  =  +  2  "  _>/3  =  +  3  "  23  =  4  4  " 

/%  =  —  1 5°    "  -r4  =  +  3  "  y*  =  +  4  "  z*  =  +  5  " 

/^6  =  +  200    "  xs,  =  +  4  "  y&  —  +  5  "  2-6  =+  6  " 

ANS.  F  =  +  90  Ibs.,    px  =  +  2f  ft.,    A  =  +  3  ft.,    />*  =  4  3i  ft. 
If  the  forces  are  parallel  to  the  axis  of  Y,  we  have 

Mx  =  +  31 5  lb.-ft.,        Mz  =  +  240  lb.-ft.,        Mr  =  396  Ib.-ft. 
The  line  representative  making  the  angles  with  the  axes  of  X,  Y,  Z  given  by 

+H°, 

or 

a  =  322°  41'  41",  /S  =  90°,  r  =  52°  4i'  4i'. 

If  we  look  along  the  line  representative  towards  the  origin,  the  rotation  is  seen  counter-clockwise. 

(3)  Let  a  rigid  body  be  acted  upon  by  the  coplanar  forces 

•F!  =  50  Ibs.,         F*  =  30  Ibs.,        F3  =  7o  Ibs.,        /\,  =  90  Ibs.,        Fs  =  120  Ibs. 
icting  at  the  points  given  by 

Xl  =  4-  5  ft.,  j,  =  +  10  ft.  ;  x*  =  +  9  ft.,  ^2  =-fi2  ft.; 
jr.  =  4-  17  ft.,  j3  =  4-  14  ft.  ;  :r«  =  4-  20  ft.,  j4  =  4  13  ft. ; 
jr6=  +  15  ft.,j6=  4  8  ft. 

Let  the  forces  make  angles  with  the  axes  of  X  and  Y  given  by 

«i  =    70°,  /?!  =    20° ;     a,  =  60°,  ft*  =  1 50° ;     a3  =  120°,  /&,  =30° ; 
«4  =  1 50°,  /?4  =  120° ;     «6  =  90°,  /35  =      o°. 

Find  the  resultant,  etc. 

ANS.  We  have  for  the  components  parallel  to  the  axes  X  and  Y: 

Fx  =  50  cos  70°  4  30  cos  60°  —  70  cos  60°  —  90  cos  30°  =  —  80. 842  Ibs. ; 

Fy  =  50  cos  20°  —  30  cos  30°  4  1 20  4  70  cos  30°  —  90  cos  60°  =  4  1 56.626  Ibs. ; 

Fz  =  o. 

The  resultant  is  given  in  magnitude  by 


F  =  \/F*x  +  ^  =  176.259  Ibs., 
and  its  direction  cosines  by 


cos  a  = 


, 
F  176.259 

cos  ft  =  $  =  +  If-626.     or     ft  =  27°  18'  i». 
F  176.259 

We  have  from  equation  (4),  page   193, 
SFx  cos  ft  =  4  50  cos  20°  x  5  —  30  cos  30°  x  9  4  70  cos  30°  x  17  —  90  cos  60°  x  20  4  120  x  15 

=  4  1931.670  lb.-ft.  ; 
2Fycos  a  =  4  50  cos  70°  x  10  4  30  cos  60°  x  12  —  70  cos  60°  x  14  —  90  cos  30°  x  13  =  —  1152.245  lb.-ft. 

M*  =  o,          My  =  o,          M,  =  2F.v  cos  ft  —  2Fy  cos  a=  4  3083.915  lb.-ft. 
Since,  then,  equation  (18),  page  198, 

FXMX  4  FyMy  4  F,M,  =  o, 

is  satisfied,  the  forces  reduce  to  a  single  resultant  force. 

The  moment  of  this  resultant  force  relative  to  the  origin  is 


M  =  ^ Ml  +  My  +  Ml  =  Mz  =  +  3083.915  lb.-ft. 


200  DYNAMICS,    GENERAL  PRINCIPLES.  [C"AP-  V1- 

Its  lever-arm  is 


M  ==          ft< 
f       /.'  -    176.259 

The  equation  of  the  line  of  direction  of  the  resultant  is 

y  -  FjLx  _  ^  =  -  I.95J-  +  38.14. 
The  co-ordinates  of  the  point  of  application  of  the  resultant  are  given  from  equation  (10),  page  195  : 


(4) 


Find  the  resultant,  etc.,  for  the  force  system  acting  on  a  rigid  body  given  by 


Ft  =  $o  IDS., 

a,  =  60°, 

ftt  =  40°, 

^i  acute  ; 

F,  =  70   " 

at  =  65°, 

/»»  =  45°. 

^3  obtuse  ; 

F»  =  90    " 

«a  =  70°, 

ft3  =  50°, 

y3  acute  ; 

Ft  =   120  " 

««  =  75°. 

/*«  =  55°. 

y<  obtuse. 

Xi  =  0, 

J'l  =  O, 

^•1  =  0; 

JCt  =    +    I  ft., 

j,  =  +  4  ft., 

zt  =  +  7  ft.  ; 

x.  =  +  2  •• 

/»  =  +  5  " 

*3    =     +    8     " 

-r.  =  +  3  " 

/4     =      +     6     " 

z*  =  +  9  " 

ANS.  We  find  the  angles  y  bV  the  formula,  page  13. 

cos3  Y  —  —  cos  (a  +#)  cos  (a  —  ft). 

Then,  from  page  193,  we  have 

F,  =  +  116.423105.,        ^=  +  214.480^5.,         F,  =  -51.057  Ibs. 

Therefore  the  resultant  is  

F  =  ^Fl  +  Fl  +  F1=  +  249.325  Ibs., 

and  its  direction  cosineS  are  given  by 

Cosa  =  ^,         cos  ft  =  -p-,         co*Y  =  ~, 

or  «=  62°  9'  48",         ft  =  30°  39'  20",        Y  =  101°  49'. 

We  also  have  for  the  moments  from  equation  (4),  page  193, 

Mx  =  -  1838.604,         My  =  +928.947,         M,  =  -  86.903  Ib.-ft. 
The  resultant  moment  about  the  origin  is 


M=  4/yJ/|  +  M\  +  Ml  =  +  2061.789  Ib.-ft., 
and  the  direction  cosines  of  its  line  representative  are  given  by 

M~  M9  M* 


or  a  =  153°  5'  40",        ft  =  63°  14'  15",        Y  —  92°  24'  56". 

Looking  along  this  line  representative  towards  the  origin,  the  direction  of  rotation  is  seen  <roon».er. 
clockwise. 

The  equations  of  the  projection  of  the  resultant  on  the  co-ordinate  planes  are 

y  =  1.885^+0.746,         *=-  2.28*+  18.19.        *=  -0.238^-8.57. 
We  see  that  FXMX  +  F,M,  +  F.M, 

does  not  in  this  case  equal  zero.    Hence,  page  198,  the  forces  do  not  reduce  to  a  single  resulta\u  force,  bin. 
to  a  resultant  force  and  a  couple, 


CHAP.  VI.]  EXAMPLES.  201 

The  resultant  force  is,  as  already  found,  F  —  249.3^5  Ibs.,  and  its  angles  with  the  co-ordinate  axes  are 
as  already  found. 

The  co-ordinates  of  the  axis  of  the  couple  are  given  by  equation  (14),  page  196  : 

FyM,  -  FzMy  FZMX  -  FXM,  FxMy  -  FyMx 

Px  =  ~F7~^  -=+°-4  63  ft.,         py=-     *F*       —  =  +  1.673  ft..         P*=-     yFj?  JF-=+8.o8ft. 

The  resultant  couple  Ma  is  given  by  equation  (8),  page  194, 

X.  =  M'+Ff,+™.  =  _  4,.624  lb..ft. 

The  direction  cosines  of  its  line  representative  are  the  same  as  for  the  resultant  F,  and  looking  along 
this  line  representative  towards  the  origin  the  rotation  is  seen  counter-clockwise. 

(5)  In  the  preceding  example  find  what  the  co-ordinates  -r4,  yt,  z<  of  the  force  Ft  =  120  Ibs.  must  be  in 
order  that  all  the  forces  may  reduce  to  a  single  resultant. 

ANS.  We  evidently  have  Fx,  Fy,  Fz,  Fr  and  the  angles  a,  ft,  y  unchanged,  since  changing  the  point  of 
application  of  Ft,  without  changing  its  direction  or  magnitude  has  no  effect  on  the  magnitude  of  the  resultant 
or  its  direction. 
We  have  then 

Mx  =  —  659  571  —  93.262/4  —  68.  829.24,  "| 

My  =  +  369-629  +  31.059*4  +  93.262*4,   V      .     ....  1    ....    (l) 
Mz  =  —  107.036  +  68.829^4  —  3'-°59/4-  J 
We  have  as  the  equation  of  condition  for  a  single  resultant 


FXMX  +  FyMy  +  F,M,  =  o, 
or 

1  1  6.  423  Af*  +  214.48^/7  —  51.057^/3  =  o, 
or 

Af*  +   1.842^  —  0.4386^/2=0  .............      (2) 

From  (i)  we  obtain 

(Mx  +  659.571)31.059  +  (My  —369.629)68.829=  (A/,  +  107.036)93.262, 

or 

Mx  +  2.21  6My—  3.003^/2=  +  481.034  ............     (3) 

From  (2)  and  (3)  we  obtain 

0.374^-  2.564^  =  +  481-034. 

If  we  retain  for  My  its  value  in  the  preceding  example,  +  928.947  lb.-ft.,  we  shall  have 

M*=  —      52.108  lb.-ft., 
^=-1733.95 
If  we  substitute  these  values  in  (i),  we  obtain 

93.262/4  +  68.8295-4  =  +  1074.4, 
31.0595-4  +  93.262.*:,  =  +     559.308, 

68.829^-4—  31.059/4  =  +      54-934- 
Hence 

x,  =  -  0.3332-4  +    5.997, 

y^  —  —  0.7382-4  +   11.520. 
If  then  we  assume  2-4  =  o,  we  have 

*«  =  +  5'997,        J4  =  +  11.520. 

(6)    Using  the  values  of  the  preceding  example,  find  the  point  of  application  of  the  resultant. 
ANS.  We  have 

Fx  =  +  116.423  Ibs.,         Fy  =  +  214.480  Ibs.,        Ft  =  —  51.057  Ibs  ,        Fr  —  +  249.325  Ibs.; 
a  =  62°  9'  48",        (1  =  30°  39'  20",        Y  =  101°  49'; 


202  DYNAMICS.     GENERAL  PRINCIPLES.  [CHAP.  VI. 

Mx  =  -  1733.975  Ib.-ft.,          M,  =  +  928.947  lb.-ft.,        Mt  =  -  52.108  lb.-ft.,         Mr  =  +  1967.823  lb.-ft.; 

a  =151°  47',        ft  =  6i°49'  53".        Y  =  9'°  3»'  3"- 

The  co-ordinates/,,  p1t  p,  of  the  point  of  application  of  the  resultant  are  given  by 
-  1733-975  =  F*P,  ~  Fyfi*  =  ~    5'-°57j'  -  214.480*. 
+    928.947  =  Fxpt  -  Ftpx  =  +  116.4232:  +     51.057*. 

—      52.108  =  Fypx  —  Fxpy  =       214.480.1:  —  116.423^. 
Hence  we  obtain 

px  =  —  2.2802,  +  18.194, 

p,  =  —4.2008,  +  33.961. 
If  we  assume  p,  =  o,  we  have  then 

px  =  +  18.194  ft.,        p,  —  +  33.961  ft. 

If  we  introduce,  then,  a  fifth  force,  Ft  =  +  249.325  Ibs.,  whose  direction  makes  with  the  axes  the  angles 
at  =  117°  50'  12",         /35  =  149°  20'  40",         yt  =  78°  ii', 

acting  at  a  point  whose  co-ordinates  are/*  =  +  18.194 .ft.  and  py  =  33.961  ft.,  pz  —  o,  we  have  a  system  of 
forces  in  equilibrium. 

(7)  Find  the  resultant,  etc.,  for  the  parallel-force  system  given  by 

j,  =  o,  zi  =  o; 

y*  =  +  2  ft.,     *,  =  +  3  ft.; 

_y,  =  +  3    "      .sr,  =   +  4  « 

^r4    =    +   4      "        2-4    =    4.  -5    « 

yb  =   +  5    "      zt  =  +  6  " 
ANS.  F  =  2F  =  +  90  Ibs.;  px  =  =  +  2|  ft.,        p,  =  =  +  3  ft.,        pK  =       £  =  +  $  ft. 


ft  =  +  60  Ibs.  ; 

x\  =  o. 

Ft  =  +  70  " 

x,  =  +  I  ft.. 

Ft  =  —  90  " 

Xt  =  +  2  " 

Ft  =  -  150  •« 

*4  =  +  3  " 

Ft  =  +  200  " 

*6  =  +  4  " 

CHAPTER  VII. 

FORCE  OF  GRAVITATION.  CENTRE  OF  GRAVITY. 

Force  of  Gravitation. — The  "law  of  gravitation,"  as  formulated  by  Newton,  asserts 
that  every  particle  of  matter  attracts  every  other  particle  with  a  force  which  acts  in  the 
straight  line  joining  the  particles,  and  whose  magnitude  is  directly  proportional  to  the  product 
of  the  masses  of  the  particles,  and  inversely  proportional  to  the  square  of  the  distance  between 
them. 

If,  then,  ///j  and  /«,  are  the  masses  of  two  particles,  and  r  is  the  distance  between  them, 
the  mutual  force  of  attraction  F  is  given  by 


where   k   is   a   constant    to.be    determined   by  experiment,  and    is    called    the    constant   of 
gravitation. 

For  absolute  accuracy  and  universal  generality,  as  well  as  for  far-reaching  consequences, 
this  statement  is  without  parallel  in  the  history  of  science.  The  facts  that  by  means  of  it 
the  motions  of  all  the  bodies  of  the  solar  system  are  completely  explained;  that  their  past 
and  future  positions  can  be  told;  that  the  existence  of  Neptune  was  deduced  from  the 
assumption  that  certain  disturbances  in  the  motion  of  Uranus  were  due  to  the  attraction  of 
an  unknown  planet  according  to  this  law,,  all  go  to  prove  that  the  law  holds  with  absolute 
accuracy,  so  far  as  the  action  upon  each  other  of  large  masses  separated  by  distances  which 
are  great  compared  ivith  their  linear  dimensions  is  concerned. 

The  enunciation  of  the  law  expressly  confines  it  to  such  cases,  since  only  when  the 
linear  dimensions  of  the  attracting  bodies  are  insignificant  compared  to  the  distance  between 
them  can  we  conceive  them  as  "  particles." 

We  shall,  however,  show  in  the  next  article  that  if  bodies  are  homogeneous  and  spheri- 
cal, this  limitation  may  be  removed  and  the  "distance  between  them"  is  the  distance 
between  their  centres  of  mass. 

Attraction  of  a  Homogeneous  Shell  or  Sphere. — Let  the  circle  ACA' ,  with  centre  at 

C,  represent  a  uniform  thin  homogeneous  spherical 
shell  whose  surface  density  is  ft.'  Suppose  a  par- 
ticle at  P  whose  mass  is  m.  Join  C  and  P.  Take 
any  point  A  of  the  shell  and  draw  CA  and  A  P. 
Let  APmake  the  angle  6  with  CP,  and  draw  a  line 
AB  through  A,  making  the  same  angle  0  with  CA. 
Then  in  the  two  triangles1  CAB  and  CA?  v\e 
have  the  side  CA  and  the  angle  at  C  common  to  both,  and  the  angles  at  A  and  P  ( OHM.!  c>y 
construction.  These  triangles  are  therefore  similar  and  we  have 

AB  _  CA 
AP        CP. 


204  DYNAMICS.    GENERAL  PRINCIPLES.  LCHAP-  VII« 

Now  let  As  represent  any  small  elementary  area  of  the  spherical  surface,  and  An  its 
projection  normal  to  AB. 

Let  oo  square  radians  (page  7)  denote  the  conical  angle  subtended  at  B  by  An.     Then 

Al?  .  co 
the  area  denoted  by  An  is  equal  to  Alf  .  co,  and  the  area  denoted  by  As  is  equal  to     CQS  ^   , 

since  the  angle  nAs  =  BAC=Bt  and  the  angle  snA  is  a  right  angle. 

The  mass  of  the  elementary  area  denoted  by  As  is  then  —    —  '——  ,  and  the  attraction  of 
this  mass  for  the  particle  of  mass  m  at  P  is,  by  Newton's  law, 

m.  dA£?.  GO 


and  acts  in  the  line  AP. 

If  we  draw.4/4'  perpendicular  to  CP,  we  have  evidently  the  same  attraction  between 
the  equal  elementary  mass  at  A'  and  the  particle  of  mass  m  at  P  acting  in  the  line  A'  P. 

We  can  resolve  each  of  these  equal  forces  into  a  component  along  the  line  CP  and  at 
right  angles  to  CPat  P.  Since  the  angles  A  PC  and  A'  PC  are  each  equal  to  B,  the  two  com- 
ponents at  right  angles  to  CP  at  Pare  equal  and  opposite  and  therefore  produce  no  effect  upon 
P.  The  resultant  attraction  of  the  two  elements  at  A  and  A'  upon  the  particle  of  mass  m  at 
P  acts  then  in  the  line  CP  and  is  equal  to 

m  .  8AJ?.  GO  , 


or,  since  -r-n  —  -^  »  tne  resultant  attraction  is 
At          C.J 

in  .  fiCA   . 


But  CA   .  co  is  the  area  of  the  elementary  area  at  A  or  A',  and  2k  ~^z  is  constant  for  all 

pairs  of  elements  A  and  A'  .     The  total  attraction  of  the  shell  for  the  particle  of  mass  m  at 
Pacts,  then,  in  the  line  CP  and  is  equal  to 


where  the  summation  is  to  be  taken   for  an  entire  hemisphere.      But  2CA  .  co  for  a  hemi- 


sphere is  2nCA*,  and  hence  the  attraction  is  equal  to 


4rr6CAz  .  m         w/m 


where  m  =  4*6  C  A   is  the  total  mass  of  the  spherical  shell. 

We  see,  then,  that  the  spherical  shell  attracts  a  particle  of  mass  m  at  any  outside  point 
P,  just  as  if  its  entire  mass  were  condensed  at  the  centre  of  the  shell. 

If  instead  of  a  homogeneous  spherical  shell  we  have  a  solid  homogeneous  sphere,  we 
may  consider  it  as  composed  of  an  indefinite  number  of  concentric  homogeneous  spherical 
shells,  each  of  which  attracts  the  mass  at  Pas  if  its  entire  mass  were  condensed  at  its  centre. 


CHAP.  VII.  ]  ATTRACTION  OF  A  HOMOGENEOUS  SHELL   OR  SPHERE.  205 

Hence  the  attraction  of  a  homogeneous  spherical  shell  or  of  a  homogeneous  sphere  upon  a 
particle  at  any  outside  point  is  the  same  as  if  the  entire  mass  of  the  shell  or  sphere  were  con  - 
densed  in  a  point  at  the  centre. 

We  can  therefore  consider  a  homogeneous  shell  or  sphere  as  a  particle  of  equal  mass  at 
the  centre,  so  far  as  its  attraction  upon  an  outside  particle  is  concerned. 

COR.  —  If  the  sphere  is  not  homogeneous,  but  the  density  of  every  point  at  the  same 
distance  from  the  centre  is  the  same,  we  may  still  consider  the  sphere  as  composed  of  homo- 
geneous spherical  concentric  shells,  each  one  of  which  attracts  an  outside  mass  as  if  its  entire 
mass  were  condensed  at  the  centre.  Hence  the  same  holds  true  for  the  sphere. 

Value  of  Constant  of  Gravitation.  —  We  have  seen  that  Newton's  law  is  expressed  by 


where  the  constant  k  must  be  determined  by  experiment.  This  determination  we  are  now 
able  to  make. 

Thus  for  a  body  of  mass  m^  at  any  point  on  the  earth's  surface  where  the  acceleration 
is  g,  we  know  that  the  force  of  gravity  is  F  =  mlg  poundals.  Let  mQ  be  the  mass  of  the 
earth,  and  let  rQ  be  the  radius  of  the  earth  at  the  place  of  experiment. 

If  we  consider  the  earth  as  a  sphere  whose  density  is  either  constant  or  the  same  at  all 
points  at  equal  distances  from  the  centre,  then,  as  we  have  just  seen,  we  may  consider  it  as  a 
particle  of  equal  mass  at  the  centre  of  mass  so  far  as  its  attraction  upon  any  outside  particle 
is  concerned,  and  the  centre  of  mass  is  the  centre  of  figure. 

The  earth  is  not  strictly  spherical,  but  its  deviation  from  sphericity  is  very  slight.  Also, 
the  density  is  not  constant  nor  strictly  the  same  at  points  equidistant  from  the  centre. 
But  the  small  distance  between  the  centre  of  mass  and  that  point  at  which  in  any  case  of 
attraction  we  may  consider  its  mass  condensed  is  insignificant  compared  to  its  radius.  So 
far  as  its  attraction  for  any  outside  particle  is  concerned,  we  may  then  still  consider  it  as  a 
particle  of  equal  mass  at  the  centre  of  mass,  and  the  centre  of  mass  as  the  centre  of  figure. 

Also,  since  the  dimensions  of  any  body  with  which  we  experiment  at  the  earth's  surface  are 
insignificant  compared  to  the  radius,  we  may  consider  any  such  body  as  a  particle.  Hence  in 
equation  (i)  we  can  take  r  as  the  radius  of  the  earth  rQ,  and  m2  as  the  mass  ;«0  of  the  earth,  and 
we  then  have 


Hence  we  have 

*=*£•     ...........    (*) 

"*0 

Substituting  this  value  of  /fin  (i),  we  have 


Equation  (3)  gives  the  force  of  attraction  F  between  two  masses  ml  and  mz  at  any  dis- 
tance r,  in  absolute  units.  Thus  if  m0  ,  m^,  m2  are  taken  in  Ibs.,  r0  and  r  in  feet,  and  g  in 
ft.-per-sec.  per  sec.,  equation  (3)  gives  Fin  poundals. 

If  we  take  mass  in  grams  and  distance  in  centimeters,  we  have  F  in  dynes. 


zo6  DYNAMICS.    GENERAL  PRINCIPLES.  [CHAP.  VII. 

If  we  divide  by^-,  we  have  in  gravitation  units 

f=jyb-r~ .  ;  .  (4) 

If  M  is  the  mass  of  the  sun  and  m  the  mass  of  a  planet,  we  have,  from  (3),  for  the 
mutual  force  of  attraction 

.,        Mm    r* 
F=~^~  -m/' 

Since  force  equals  mass  X  acceleration,  if  we  divide  this  by  M,  we  have  for  the  acelera- 
tion  SO  of  the  sun  relative  to  a  fixed  point  O  s  :*u 


If  we  divide  by  ;«,  we  have  for  the  acceleration  PO  of  the  planet  relative  to  „ 

a  fixed  point  O 

M_     rj_ 

*2       •       M  S 

I  /'*fl 

opposite  in  direction  to'SO.     We  have  then  for  the  acceleration  PS  of  the  planet  relative  to 
the  sun  PO  +  OS,  or  s*~  ~~O  P 

M+  m     r_l 
r>       '  m/' 

At  a  distance  r0  we  have,  making  r  —  r0  ,  the  acceleration /0  of  the  planet  relative  to 
the  sun 

jyi  — (—  m  i  \ 

/0  =  —       —  .  g. (5) 

m0 

This  is  the  value  of  f0  to  be  used  in  our  equations  for  planetary  motion,  page  123. 

Astronomical  Unit  of  Mass. — That  mass  which  attracts  an  equal  mass  at  unit  distance 
with  unit  force  is  called  the  ASTRONOMICAL  UNIT  OF  MASS. 

Let  [F]  denote  the  unit  of  force,  and  [£]  the  unit  of  length.  Then,  from  equation  (3), 
we  have  for  the  astronomical  unit  of  mass  [»/] 


r/n      M'      r"  v 

MZj-.i* 

Hence  we  obtain  for  the  astronomical  unit  of  mass  [#/] 


Equation  (6)  gives  by  definition  the  astronomical  unit  of  mass. 

The  mass  w0  of  the  earth  is  about  I  1920  X  io21  Ibs.  The  mean  radius  of  the  earth  r0 
is  about  21  X  10*  ft.  Taking  these  values  and  £•  =  32  ft.-per-sec.  per  sec.,  we  have,  from 
(6),  for  the  astronomical  unit  of  mass 

[m]  =  29063  Ibs. 

Taking  w0  equal  to  6.14  X  io27  grams,  r0  equal  to  6.37  X  io8  cm.  and^-=  981  cm.-per- 
sec.  per  sec.,  we  have 

\nt]  =  3928  grams. 


CHAP.  VII. J  ASTRONOMICAL   UNIT  OF  MASS.  207 

If  we  adopt  the  astronomical  unit  of  mass,  we  can  then  write  simply  the  numeric  equation 

m^ 
~^' 

where  ml  and  m.2  are  the  number  of  astronomical  units  of  mass  in  the  two  particles,  r  the 
number  of  units  of  length  in  the  distance  between  them,  and  F the  number  of  absolute  units 
of  force  in  the  attraction. 

Centre  of  Gravity. — When  a  body  attracts  and  is  attracted  by  all  external  bodies, 
whatever  their  distance  and  position,  as  though  its  mass  were  condensed  in  a  single  point 
fixed  relatively  to  the  body,  that  point  is  properly  called  the  CENTRE  OF  GRAVITY  (page  23), 
A  body  which  has  a  centre  of  gravity  is  said  to  be  centrobaric  or  barycentric.  In  gen- 
eral bodies  are  not  centrobaric  if  the  law  of  attraction  follows  Newton's  law,  that  is,  if  the 
force  is  inversely  as  the  square  of  the  distance. 

As  we  have  seen,  page  205,  a  homogeneous  spherical  shell  or  a  homogeneous  sphere  is 
centrobaric  and  the  centre  of  gravity  is  at  the  centre  of  mass.  So  also  for  a  non-homo- 
geneous sphere  whose  density  is  the  same  at  all  equidistant  points  from  the  centre. 

In  general  if  a  body  has  a  centre  of  gravity,  it  must  always  coincide  with  the  centre  of 
mass,  because  the  attraction  of  an  infinitely  distant  body  upon  it  constitutes  a  system  of 
parallel  particle  forces,  and  the  centre  for  such  a  system  is  (see  page  190)  at  the  centre  of 
mass. 

All  bodies,  then,  have  a  centre  of  mass,  but,  as  we  have  seen,  only  a  few  bodies  have  a 
centre  of  gravity. 

Examples. — (i)  Show  that  the  attraction  of  a  thin  spherical  shell  of  uniform  thickness  and  density  upon  a 
particle  inside  is  zero.  -"• 

ANS.  Let  P  be  the  particle  of  mass  m\.  Take  any  point  A  on  the  spherical  surface.  Join  AP  and 
produce  to  A'.  If  from  all  points  of  a  small  element  of  the  surface  at  A  lines  be 
drawn  through  P,  they  will  mark  off  a  corresponding  element  at  A'.  Both  these  ele- 
ments subtend  the  same  conical  angle  (page  7),  oo  square  radians.  The  area  of  the 
element  at  A  is  then  AP*.  <a  (page  7),  and  the  area  of  the  element  at  A'  is  A'P*.oo. 
If  8  is  the  uniform  surface  density,  the  mass  of  the  element  at  A  is  m*=  SAP  .  oo  and 
the  mass  of  the  element  at  A'  is  ;«a'  =  SA'P*.  <».  The  attraction  of  the  element  at  A 
for  a  particle  of  mass  m\  at  P  is  then  (page  203) 


and  acts  in  the  line  PA.     The  attraction^!  the  element  at  A'  for  the  particle  of  mass  m  at  P  is 

— ' =  KmiSoo 

A'P* 

and  acts  in  the  line  PA'.  The  resultant  attraction  upon  the  particle  at  P  of  the  pair  of  elements  at  A 
and  A'  is  then  zero.  The  whole  shell  consists  of  such  pairs  of  elements.  Hence  the  resultant  attraction  of 
the  shell  on  a  particle  at  P  is  zero. 

(2)  Show  that  the  attraction  of  a  homogeneous  sphere  on  a  particle  it'ithin  it  is  directly  proportional  to  its 
distance  from  the  centre. 

ANS.  Let  P  be  a  particle  of  mass  m\  situated  within  a  homogeneous  sphere  at  any  distance  PC  from 
the  centre  C.     Then,  from  the  preceding  example,  we  know  that  the  attraction  upon 
the  particle  at  P  due  to  the  shell  outside  of  the  sphere  whose  radius  is  PC  is  zero. 
The  attraction  upon  the  particle  of  mass  m\  at  P  is  then  due  to  the  attraction  of  the 

sphere  whose  radius  is  PC.     The  volume  of  the  sphere  is  —it  PC3.     If  8  is  the  uniform 

density,  the  mass  of  this  sphere  is  —  Sir/'C3      Its  attraction  for  a  particle  of  mass  mi 
at  P  is  (page  205)  the  same  as  if  the  entire  mass  of  the  sphere  were  condensed  at  the 


centre,  or  (page  203)  6mi^-==^ —  =  km\ .  ~8it .  PC. 


208  DYNAMICS.     GENERAL   PRINCIPLES.  [CHAP.  VII. 

The  attraction  is  therefore  directly  proportional  to  the  distance  PC  of  the  particle  from  the  centre. 

(3)  Assuming  tlit  earth  to  be  a  homogeneous  sphere,  compare  its  attraction  on  a  given  mass  at  a  distance 
from  its  centre  equal  to  one  half  its  radius,  with  the  attraction  when  the  given  mass  is  at  a  distance  equal  to 
twict  tht  radius, 

ANS.  2  to  i. 

(4)  Find  in  dynes  the  attraction  of  two  homogeneous  spheres,  each  of  100  kilograms  mass,  with  their  centres 

/  metre  apart. 

ANS.  0.0648  dynes  nearly. 

(5)  How  far  would  a  body  fall  toward  the  earth  in  one  second  from  a  point  at  a  distance  from  the  earth's 
surface  equal  to  the  radius  of  the  earth  ? 

ANS.  The  acceleration   is  inversely  as   the   square  of   the   distance.     We   have  then  g  :g  ::  r*  :  4^'.  of 

gi  _    '         Tnat  jSf  the  acceleration  is  one  fourth  of  the  acceleration  at  the  surface. 

The  distance  is  then  s  =  ^g1?,  or,  taking^  =  32  ft.-per-sec.  per  sec.  and  /  =  i,  s  =  4  ft. 

(6)  The  moon's  mass  is  136  x  /o"  Ibs.  ;  the  moons  radius,  3.70  x  io*ft.;  the  mass  of  the  earth,  11920 
x  /0"  Ibs.  ;  the  radius  of  the  earth,  21  x  10*  ft.     Find  how  far  a  stone  at  the  moon's  surface  would  fall  in  ./ 

second,  the  attraction  of  the  earth  being  neglected. 

ANS.  If  M  is  the  mass  of  the  moon  and  m  that  of  the  stone,  the  force  of  attraction,  if  X  is  the  radius  or 
the  moon,  is,  from  equation  (3),  page  205, 


The  acceleration  of  the  stone  at  the  moons's  surface  is  then 


The  distance,  then,  is  -g'f.  or,   taking  /  =  i  sec.,  s  —  2.5  ft. 

(7)  Suppose  the  earth  to  contract  until  its  diameter  is  6000  miles,  what  wovld  be  the  effect  on  the  -weight  of 
an  inhabitant  ?     The  diameter  of  the  earth  to  be  taken  at  8000  miles. 

ANS.  Increased  in  the  ratio  of  16  to  9. 

(8)  If  the  mass  of  the  sun  is  300,000  times  the  mass  of  the  earth,  and  its  radius  is  zoo  times  the  radius  of 
the  earth,  find  the  attraction  at  the  surface  of  the  sun  of  a  mass  which  at  the  surface  of  the  earth  is  attracted 
by  the  force  of  one  pound  weight. 

ANS.  39^  poundals.  or  the  attraction  of  the  earth  for  30  Ibs. 

(9)  The  diameter  of  Jupiter  is  10  times  that  of  the  earth,  and  its  mass  300  times.     By  hou>  much  per  cent 
of  his  former  weight  would  the  weight  of  a  man  be  increased  by  being  removed  to  the  surface  of  Jupiter? 

ANS.  By  200  per  cent.  He  would  weigh  by  a  spring-balance  three  times  as  much  as  before.  The  same 
number  of  standard  pounds  would,  however,  balance  him  in  a  lever-balance.  The  standard  pound  at  Jupiter 
would  be  attracted  by  a  force  three  times  as  great  as  the  earth's  attraction  here.  The  lever-balance  weight 
which  gives  his  mass  is  unchanged. 

(10)  Find  the  acceleration  due  to  the  attraction  of  the  earth  at  the  distance  of  the  moon.     Assume  g  =  3* 
at  the  earth's  surface,  the  diameter  of  the  moon's  orbit  480,000  miles,  the  diameter  of  the  earth  8000  miles. 

ANS.  0.0089  ft.-per-sec.  per  sec. 


STATICS.     GENERAL   PRINCIPLES. 


CHAPTER   I. 

EQUILIBRIUM  OF  FORCES.     DETERMINATION   OF   MASS. 

Statics. — That  portion  of  Dynamics  which  treats  of  balanced  forces,  and  of  bodies  at 
rest  under  the  action  of  balanced  forces,  is  called  STATICS.  Statics  is  thus  a  special  case  of 
Dynamics  which  it  is  convenient  to  consider  separately. 

Equilibrium  of  Concurring  Forces. — Any  number  of  forces  acting  upon  a  point  or  par- 
ticle are  said  to  be  concurring  forces.  If  the  resultant  of  any  system  of  concurring  forces  is 
zero,  the  forces  are  said  to  be  in  equilibrium. 

The  effect  of  the  resultant  force  is  to  cause  acceleration  of  the  particle.  When  the 
forces  are  in  equilibrium,  then,  the  particle  has  no  acceleration,  and  if  the  particle  is  at  rest 
it  is  said  to  be  in  static  equilibrium. 

But  if  the  resultant  force  is  zero,  all  the  concurring  forces  must  evidently  reduce  to  two 
equal  and  opposite  forces,  or,  what  is  the  same  thing,  any  one  of  the  forces  must  be  equal 
and  opposite  to  the  resultant  of  all  the  others. 

Hence  the  algebraic  sum  of  the  components  of  all  the  forces  in  any  three  rectangular 
directions  must  be  zero. 

We  have  then,  from  equations  (i),  page  194,  for  the  conditions  of  equilibrium  of  con- 
curring forces 

Fx  =  2F  cos  a  =  o,         Fy  —  2F  cos  ft  =  o,          F,  =  2F  cos  7  =  0. 

We  obtain,  then,  the  following  obvious  results  from  the  condition  for  equilibrium  of 
concurring  forces,  which  will  be  found  useful  in  special  cases : 

(1)  If  two  concurring  forces  are  in  equilibrium,  they  must  be  equal  in  magnitude  and 
opposite  in  direction. 

(2)  If  three  concurring  forces  are  in  equilibrium,  they  must  all  act  in  the  same  plane. 
For  the  resultant  of  any  two  must  act  in  their  plane  and  be  equal  and  opposite  to  the  third. 

(3)  If  three  concurring  forces  are  represented  in  magnitude  and  direction  by  the  sides 
of    a    triangle    taken    the   same   way   round,  the   resultant    is    zero    and    the    forces    are   in 
equilibrium. 

(4)  Hence,  if  three  concurring  forces  are  in  equilibrium,  each  one  is  proportional  to  the 
sine  of  the  angle  between  the  other  two. 

(5)  If  three  concurring  forces  are  in  equilibrium  and  their  directions  are  represented  by 
the  sides  of  a  triangle  taken  the  same  way  round,  their  magnitudes  will  also  be  represented 
by  the  sides  of  that  triangle,  and  vice  versa. 

209 


210  STATICS.    GENERAL  PRINCIPLES.  \pUf.  1. 

(6)  If    any    number  of  concurring  coplanar  forces  are  represented  in   magnitude  am! 
direction  by  the  sides  of  a  plane  closed  polygon   taken   the   same   way  round,  they  are  in 
equilibrium.      If  their  magnitudes  are  given  by  the  sides  of  the  polygon,  their  directions  are 
also  given  by  the  directions  of  the  sides. 

Hut  if  the  directions  only  of  the  forces  are  given  by  the  sides  of  the  plane  polygon,  it 
does  not  follow  that  the  sides  of  this  polygon  represent  the  magnitudes,  because  any 
number  of  plane  polygons  with  parallel  sides  may  be  drawn,  the  magnitudes  of  the  sides 
varying. 

(7)  If  three  concurring  forces  in  different  planes  are  represented  by  the  three  edges  of 
a  parallelopipcdon,  the  diagonal  taken  the  opposite  way  round  will  represent  the  resultant 
in  direction  and  magnitude.      This  is  called  the  para  llelopipedon  of  forces. 

Examples.— (i)  Find  the  resultant  of  forces  of  7,  /.  /,  j  units,  represented  by  lines  drawn  from  one  angle  of 
a  regular  pentagon  towards  the  other  angles  taken  in  order. 
ANS.     f/74   units. 

(2)  P  andQ  are  two  component  forces  at  right  angles,  whose  resultant  is  R.     S  is  the  resultant  of  R  and 
P.     If  Q.  =  aP,  what  is  Sf 

ANS.  S  =  2/V7. 

(3)  Component  forces  P,  Q.  K  are  represented  in  direction  by  the  sides  of  an  equilateral  triangle  taken  the 
same  way  round.     Find  the  magnitude  of  the  resultant. 

ANS.  v7*  +  <£  +  A'8  -  QR  -PR-  PQ. 

(4)  A  weight  of  10  tons  is  hanging  by  a  chain  20  feet  long.      Find  how  much  the  tension  in  the  chain  is 
increased  by  the  weight  being  pulled  out  by  a  horizontal  force  to  a  distance  of  12  feet  from  the  vertical. 

ANS.    By  2.5  tons. 

(5)  A  weight  of  4  pounds  is  suspended  by  a  string,  and  is  acted  upon  by  a  horizontal  force.      If  in  the 
position  of  equilibrium  the  tension  of  the  string  is  j  pounds,  what  is  the  horizontal  force  ? 

ANS.  3  Ibs. 

(6)  A  mass  of  10  Ibs.  is  supported  by  strings  of  lengths  3  and  4  feet  attached  to  two  points  in  the  ceiling  5 
feet  apart.      W 'hat  is  the  tension  of  each  string  ? 

ANS.  8  Ibs.  and  6  Ibs. 

(7)  A  particle  is  actfd  on  by  a  force  whose  magnitude  is  unknown,  but  whose  direction  makes  an  angle  of 
60"  with  the  horizon.      The  horizontal  component  of  the  force  is  1.33  dynes.     Determine  the  total  force  and  its 
vertical  component. 

ANS.  2.7  dynes  and  2.34  dynes. 

(8)  Three  forces  proportional  to  /,  2,  j.  act  on  a  point.      The  angle  between  the  first  and  second  is  60", 
between  the  second  and  third  jo".     Find  Hie  angle  which  the  resultant  makes  with  the  first. 

ANS.    About  67°. 

(9)  Three  cords  are  tied  together  at  a  point.      One  is  pulled  in  a  northerly  direction  with  a  force  of  6 
pounds,  and  another  in  an  easterly  direction  with  a  force  of  8  pounds.       With  what  force  must  the  third  be 
pulled  in  order  to  keep  the  whole  at  rest  f 

ANS.  10  pounds,  at  an  angle  with  the  horizon  whose  tang  =  -. 

Static  —  Molar  —  Dynamic  and  Molecular  Equilibrium. — Forces  acting  at  different 
points  of  a  body  are  said  to  be  non-concurring  forces. 

We  have  seen  (page  188)  that  any  number  of  non-concurring  forces  acting  upon  a  rigid 
body  can  be  reduced  to  a  resultant  force  and  a  resultant  couple  at  the  centre  of  mass. 

The  effect  of  the  resultant  force  (page  190)  is  to  cause  linear  acceleration  of  translation 
of  the  body.  The  effect  of  the  resultant  couple  (page  187)  is  to  cause  angular  acceleration 
of  the  body  about  an  axis  through  the  centre  of  mass. 

If  both  resultant  force  and  couple  are  zero  and  the  body  is  at  rest,  it  will  remain  at  rest 
and  then  it  is  said  to  be  in  static  equilibrium.  If  every  pcint  has  the  same  velocity  of  trans- 
lation, this  velocity  of  translation  will  b<_  uniform,  and  the  body  is  then  in  molecular  equi- 


CHAP.  I.]  EQUILIBRIUM  OF  NON-CONCURRING  FORCES.  211 

librium.  If  the  body  has  angular  velocity  about  an  axis  through  the  centre  of  mass,  this 
angular  velocity  is  unchanged  and  the  body  is  said  to  be  in  molar  equilibrium. 

If  only  the  resultant  force  is  zero,  but  not  the  couple,  the  centre  of  mass  is  either  at 
rest  or  moves  with  uniform  velocity,  but  the  angular  velocity  changes.  The  body  is  then 
said  to  be  in  dynamic  equilibrium. 

Equilibrium  of  Non-Concurring  Forces  —  When  both  the  resultant  force  and  couple 
are  zero  the  forces  are  in  equilibrium.  In  such  .case  all  the  forces  must  evidently  reduce  to 
two  equal  and  opposite  forces  in  the  same  straight  line,  or  what  is  the  same  thing,  any  one 
of  the  forces  must  be  equal  and  opposite  to  the  resultant  of  all  the  others,  and  lie  in  the 
same  straight  line  with  it. 

We  have,  then,  for  the  equilibrium  of  non-concurring  forces  two  necessary  conditions. 

ist.  TJie  algebraic  sum  of  the  components  of  all  the  forces  in  any  three  rectangular  direc- 
tions must  be  zero. 

From  equations  (i),  page  194,  we  have,  then, 

Fx  —  2F  cos  a  =  o,          Fy  =  2Fcos  ft  =  o,          Fz  —  2F  cos  y  =  O.     .     .     (l) 

When  these  conditions  are  complied  with,  there  is  no  resultant  force  on  the  centre  of  mass 
of  the  body,  and  any  one  force  is  equal  and  opposite  to  the  resultant  of  all  the  others  but 
does  not  necessarily  lie  in  the  same  straight  line  with  it. 

2d.  The  algebraic  sum  of  the  component  moments  in  any  three  rectangular  planes  must  be 
zero. 

From  equations  (4),  page  194,  we  have,  then, 


Mx  —  2Fcos  y.  y  —  2F  cos  ft  .  z  =  O, 
My  —  2Fcos  of  .  2  —  2F  cos  Y  •  x  —  °» 
Mz  •=  2Fcos  ft  .  x  —  2F  cos  a  .  y  =  o. 


(2) 


When  these  conditions  are  complied  with  there  is  no  resultant  moment  at  the  origin. 
But  it  does  not  necessarily  follow  that  there  is  no  moment  at  any  other  point  unless  the 
resultant  force  is  also  zero. 

When,  then,  both  equations  (i)  and  (2)  are  satisfied  the  forces  are  in  equilibrium,  and 
the  body,  if  at  rest,  remains  at  rest  and  is  in  static  equilibrium.  If  every  point  has  the  same 
velocity  of  translation,  it  is  in  molecular  equilibrium.  If  the  body  has  angular  velocity  about 
an  axis  through  the  centre  of  mass,  it  is  in  molar  equilibrium. 

If  only  equations  (i)  are  satisfied  but  not  (2),  the  forces  reduce  to  a  couple.  The  body 
is  then  in  dynamic  equilibrium. 

If  only  equations  (2)  are  satisfied  but  not  (i),  we  have  a  single  resultant  force  passing 
through  the  origin. 

COR.  i.  —  If  the  forces  are  all  co-planar,  let  XFbe  their  plane.  Then  z  =  o,  y  =  90°, 
and  we  have 

Fx  =  2F  cos  <*  =  o,         Fy=2Fcos/3  .......     (0 

Mz  =  SFcosfi  .  x—  SFcosa  .y  =o    .......     (2) 

That  is 

ist.  The  algebraic  sum  of  the  components  of  all  the  forces  in  any  two  rectangular  direc- 
tions in  the  plane  of  Ihe  forces  must  be  zero. 

2d.  The  algebraic  sum  of  the  moments  of  the  forces  about  any  point  in  this  plane  must  be 
zero. 


212  STATICS.    GENERAL  PRINCIPLES.  [CtiAP.  1. 

If  the  first  condition  only  is  satisfied,  we  have  a  couple. 

If  the  second  only  is  satisfied,  there  is  a  single  resultant  passing  through  the  origin. 

COR.  2.  —  If  three  forces  acting  at  different  points  of  a  rigid  body  are  in  equilibrium, 
their  lines  of  direction  must  intersect  in  a  common  point  and  the  forces  must  be  coplcinar. 

For  the  resultant  of  any  two  must  pass  through  their  point  of  intersection  and  lie  in 
their  plane.  The  third  must  be  equal  and  opposite  to  this  resultant  and  lie  in  the  same 
straight  line. 

COR.  3.  —  If  the  forces  are  parallel,  take  their  common  direction  parallel  to  the  axis  of 
y.  Then  cos  a  =  o,  cos  y  =  o,  cos  ft  =  I  ,  Fx  =  o,  Ft  =  o,  Fy  =  2F,  and  we  have 


(i) 

(2) 


2Fy  =  o.  ) 

That  is: 

ist.    The  algebraic  sum  of  the  forces  must  be  zero. 

2d.    The  algebraic  sum  of  the  component  moments  of  the  forces  in  any  two  rectangular 
planes  must  be  zero. 

If  the  first  condition  only  is  satisfied,  we  have  a  couple. 

If  the  second  condition  only  is  satisfied,  the  resultant  passes  through  the  origin  and 
coincides  with  the  axis  of  Y. 

Determination  of  Mass  by  the  Balance.  —  Let  two  masses  m,  and  in~2  be  suspended 
from  the  arms  of  an  equal-armed  balance,  the  fulcrum  F 
being  in  a  vertical  through  the  centre  of  mass  O  of  the 
balance.  Let  the  balance  before  the  masses  nij  and  m.2  are 
applied  be  horizontal  and  at  rest.  Then  the  balance  is  in 
static  equilibrium,  and  its  weight  W  acting  at  the  centre  of 
mass  O  must  be  equal  and  opposite  to  the  upward  pressure 
of  the  fulcrum  and  in  the  same  line. 

Now  let  the  masses  nlj  and  m2  be  suspended,  the  length 

of  arm  AC  =  BC  being  /,  and  suppose  the  balance  still  remains  horizontal  and  at  rest.      It 
is  then  still  in  static  equilibruim  and  the  acting  forces  are  in  equilibrium.      These  forces  are 
the  weight  W  of  the  balance,  the  weights  m^  and  m^g  of  the  masses,  and  the  upward 
reaction  R  of  the  fulcrum. 
We  have,  then  : 
ist.   Algebraic  sum  of  the  forces  equal  to  zero,  or 


R  —  W—  mlg  —  m2£-=  o,     or    R  = 
2d.   Algebraic  sum  of  the  moments  about  any  point  zero.     Take  F  as  the  point.      Then 

m^  .  /  —  rx^gl  —  o,     or    m,  =  m2. 
Hence  for  equilibrium  the  masses  must  be  equal.      (See  page  173.) 

Examples.—  (i)  A  circular  table  -weighing  m  Ibs.  has  three  equal  legs  at  equidistant  points  on  its  circum- 
ference. The  table  is  placed  on  a  level  floor.  Neglecting  the  mass  of  the  legs  and  considering  table  and  floor 
and  legs  rigid,  find  the  mass  which,  placed  anywhere  on  the  table,  will  just  bring  it  to  the  point  of  over- 
turning. 

ANS.   m  Ibs. 


CHAP.   I.] 


EXAMPLES. 


213 


(2)  If  the  table  has  four  legs  at  equidistant  points  on  the  circumference,  find  the  mass  that  will  just  bring 
it  to  the  point  of  overturning. 

ANS.  2.4  m. 

(3)  The  centre  of  mass  of  a  ladder  weighing  50  Ibs.  is  12  ft.  from  one  end,  which  is  fixed.      What  force 
must  a  man  apply  at  a  distance  of  b  ft.  from  this  end  to  just  raise  the  ladder? 

ANS.   iop  Ibs. 

(4)  A  horizontal  beam  of  length  I  is  supported  at  its  ends.     It  is  acted  upon  by  vertical  downward  force: 
Fi ,  Fi ,  F3  acting  at  points  of  application  A\ ,  A* ,  A-j ,  dividing  the  beam  into  segments  b,  c,  d,  e.     Find  the 
resultant  pressures  R\ ,  R*  at  the  right  and  left  supports,  neglecting  the  weight  of  the  beam. 

ANS.  A*i  and  Ri  act  upwards,  and  are  given  by 

A5)  =  — 


(5)  A  mass  of  6  Ibs.  hangs  on  the  arm  of  a  safety-valve  at  a  distance  of  18  inches  from  the  fulcrum.     The 
valve-spindle  is  attached  at  i  inch  from  the  fulcrum.     Disregarding  friction  and  the  weight  of  the  arm,  fend 
the  steam-pressure  for  static  equilibrium. 

ANS.  108  pounds. 

(6)  In  a  wheel  and  axle  the  radius  of  the  axle  is  r,  and  of  the  wheel  R.    A  mass  Q  hangs  by  a  rope  wouna 
round  the  axle.     Find  the  force  P  acting  tangent  to  the  wheel  in  order  to  hold  Q  suspended,  disregarding 

friction. 

ANS.  P  =  £. 

A 

(7)  A  shopkeeper  has  correct  weights  but  an  untrue  balance,  one  arm  of  which  is  a  and  the  other  b.     He 
serves  out  to  each  of  two  customers,  according  to  his  balance,  m  Ibs.  of  a  commodity,  using  first  one  scale-pan  ana 
then  the  other.     Does  he  gain  or  lose,  and  how  much  f 

(a  - 


ANS.  Loses  m 


-, 
ab 


., 
Ibs. 


(8)  The  arms  of  a  balance  are  unequal,  and  one  of  the  scale-pans  is  loaded.  A  body  the  true  mass  of  which 
is  m  Ibs.  appears  when  placed  in  the  loaded  pan  to  weigh  ;«,  Ibs.,  and  when  placed  in  the  other  m,  Ibs.  Find 
the  ratio  of  the  arms  and  the  mass  with  which  the  pan  is  loaded. 

mi  —  m 


ANS.  Ratio  of  arms  = 


mass  = 


H-P 


m\  —  m 

(9)  A  balance  is  in  equilibrium  and  unloaded.     A  body  in  one  scale-pan  is  found  to  balance  m\  Ibs.     In  the 
other  scale-pan  it  balances  m*  Ids.     Find  the  true  mass. 

ANS.    ymitnt. 

( i  o)  Find  the  condition  for  static  equilibrium  for  a  screw,  neglecting  friction. 

ANS.  Let  +  P,—  P  be  the  couple  applied  at  the  top  with  lever-arm 
ac  =  /;  let  ab  =.  r  be  the  radius  of  the  screw,  and  p  the  pitch  or  distance 
between  the  threads,  so  that  if  the  screw  be  developed  we  have  an  in- 
clined plane,  which  has  a  rise  p  for  a  base  iitr.  Let  a  be  the  inclination 
of  the  screw  to  the  horizontal,  so  that 

P 
tan  a  —  -£- 

2nr  ' 

Let  Q  be  the  mass  supported.  We  can  resolve  Q  into  a  normal  to  the 
thread  — — — and  a  horizontal  force  Q  tan  a.  The  normal  is  balanced 

by  the  upward  normal  pressure  A7  on  the  threads.  The  horizontal  com- 
ponents at  every  two  diametrically  opposite  points  of  the  thread  are 
equal  and  opposite.  We  have  then  a  couple  Q  tan  a  x  r  balanced  by 
PI,  or 


gtan  ax  r=  PI;  .-.  ^  =  ^7-1 
Inserting  the  value  for  tan  a, 


e  =  8t. 


214 


STATICS.    GENERAL  PRINCIPLES. 


[CHAP.  I. 


(ii)    The  differential  screw  consists  of  a  screw  n  d  which  works  in  a  fixed  nut  CC.     The  screw  is  hollow 
and  has  a  thread  cut  inside  in  which  a  scre^u  it '  e  works.     This  screw  de  is  prevented  from  turning  by  the  rod 
_,          /•"/•',  whose  ends  slide  in  vertical  grooves.      Find  the  condition  for  static 
equilibrium,  neglecting  friction. 

ANS.  Let  pi  and  />„  be  the  pitch  of  the  screws,  and  a  the  common 
inclination  of  the  threads,  and  r\  and  rt  the  radii  of  the  screws.  When 
the  arm  ac  =  I  turns  through  lit  radians  screw  ad  rises  a  distance  p\ 
and  screw  de  falls  a  distance  pt. 

We  have  then  just  as  before,  in  the  preceding  example, 

Q  tan  a  x  r,  —  Q  tan  a  x  rt  =  PI, 
Hence 


Q(ri  -  r,)  tan  a 

I 


or,  since  tan  a  = 


P* 


If  pt  =o,  we  have  the  simple  screw.       By  making  p\  and  pi  nearly 
equal,  we  can  have  P  very  small  compared  to  Q. 

(12)  Let  the  force  P  act  normally  upon  the  middle  of  the  back  of  an  isosceles  wedge.     Find  the  condition  of 
static  equilibrium,  neglecting  friction.  p 

ANS.  The  pressure  N  on  each  side  must  be  normal.     Let  a  be  the  angle  of  the 
wedge.    Then 


Hence 


—  =  o, 


N 


or     P  =  2./Vsin  — . 


If  a  is  very  small,  N  may  be  made  very  great. 

(  1  3)  In  a  wheel  and  axle  the  radius  of  the  wheel  is  CA  =  a,  and  of  the  axle  CB 
=  b  Find  the  conditions  for  static  equilibrium  when  a  mass  P  hung  from  the 
wheel  balances  a  mass  Q  hung  from  the  a.rle,  neglfcting  friction  and  rigidity  of  ropes. 

ANS.  The  weight  of  /"is  Pg\  of  Q,  Qg  poundals.  Let  1\  be  the  upward 
pressure  of  the  journal  bearings.  Then 


LZ1P 


R  —  Pg  —Qg  =  o,     or, 

Taking  moments  about  the  centre  C,  we  have 

Qgb  —  Pga  =o,     or     Q 


=  (Pg  +<2.f.r|  poundals,     or,     in  gravitation  units, 
'/?  =  (>  +  0. 


CHAPTER   II. 


WORK.     VIRTUAL  WORK. 


FIG.  (i). 


FIG.  (2). 


Work. — The  product  of  a  uniform  force  by  the  component  displacement  along  the  line 
of  the  force  of  its  point  of  application,  is  called  the  WORK  of  the  force. 

Thus  let  F  be  a  force  acting  at  the  point  Pl ,  and  let  the  displace- 
ment be  d  =  P-JPy  Let  F  be  uniform  during  displacement,  and  let 
the  projection  Pvn  along  the  line  of  the  force  of  the  displacement  be 
p  =  d  cos  6,  where  0  is  the  angle  of  the  displacement  with  F. 

Then  the  work  of  F  is 

±  Fp  —  ±  Fd  cos  0. 

If  the  projection  p  is  in  the  direction  of  the  force,  the  work  ir.  positive  and  the  force  is 
said  to  do  work.  If  the  projection  p  is  opposite  to  the  direction  of  the  force,  the  work  is 
negative  and  work  is  done  against  the  force. 

Now  F  cos  d  is  the  projection  of  the  force  along  the  line  of  the  displacement.  We  can 
therefore  define  work  generally  as  the  product  of  a  uniform  force  by  the  component  displace- 
ment along  the  line  of  the  force  (±  F .  d  cos  0),  or  the  product  of  the  displacement  by  the  com- 
ponent force  along  the  line  of  the  displacement  (±  d  .  F  cos  #). 

Work  of  Resultant. — Let  Fr  Fv  Fy  etc.,  Fig    I,  be  any  number  of  forces  acting  on  a 

point  P  which  has  an  indefinitely  small  displacement 
ds  in  any  given  direction.  Then  during  this  dis- 
placement the  forces  remain  unchanged  in  magnitude 
and  direction. 

Let  F19  F2,  /3 ,  etc.,  Fig.  2,  be  the  line  repre- 
sentatives of  the  forces.  The  resultant  is  then  Fr 
given  in  magnitude  and  direction  by  the  closing  line 
of  the  force  polygon. 

The  projection  of  Fr  along  the  line  of  the  dis- 
placement ds  is  then  evidently  equal  to  the  algebraic  sum  of  the  projections  of  all  the  com- 
ponents. 

Hence  the , work  of  the  resultant  for  any  indefinitely  small  displacement,  is  equal  to  the 
algebraic  sum  of  the  works  of  the  components. 

If  the  forces  do  not  change  in  magnitude  or  direction  with  the  displacement,  then  the 
same  principle  will  hold  for  any  displacement  large  or  small. 

Work — Non- Concurring  Forces. — Let  a  rigid  body  acted  upon  by  any  system  of  non- 
concurring  forces  Fl  ,  F2 ,  F, ,  etc.  have  an  indefinitely  small  displacement.  Let  A  >  A  >  A  » 
etc.,  be  the  displacement  of  each  point  of  application  along  the  line  of  the  force  at  that  point. 

2*5 


216 


S7//7/CS.     GENERAL  PRINCIPLES. 


[CHAP.  II. 


Then  the  work,  no  matter  whether  the  displacement  is  one  of  translation  or  of  rotation  or  of 
both  combined,  is  given  by 

work  =  F,p,  -4-  Ftpt  +  F3/>3  +  .  .  .   =  2Fp.     .     .     .     (i) 

In  equation  (i)  each  term  is  to  be  taken  positive  or  negative  according  as  the  projections 
A  A'  A»  etc<»  are  m  t^lc  direction  of  the  corresponding  forces  or  in  the  opposite  direction. 
Let  us  suppose,  first,  that  the  displacement  is  one  of  translation  and  given  by  ds. 

Take  any  origin  O  and  co-ordinate  axes  X,   Y,  Z.      Let 

+  ^i»  ^2'  ^s»  etc-»  act  at  P°ints  PI*  PV  PV  etc-»  given  by  the 

\  co-ordinates  (*,,  j,,   2^,  (x2,  y.^  z2),  etc.      Let  the  direction 

id,  cosines  of  Flt  F2,  F3,  etc.,  be  (arlt  Pr  y^>  (a2,  /?2,  y2),  etc. 

Then  the  components  of  Fl  are  Fv  cos  alt  Fl   cos  /?,, 
-*  Fl  cos  yl\  of  F2,   F2  cos  cr2,   F.z  cos  fi2,  F2  cos  y2,  and  so 

on.      Also  the  components  in  the  direction  of  the  axes  are 
given  (page  194)  by 


r 


i 


ft* 


Fx  —  FI  cos  a1  -\-  F2  cos  «2  +  •  •  •  =  2F  cos  a, 

Fy  =  Fv  cos  /?,  -j-  F2  cos  /52  +  .   .  .  =  2F  cos  /?, 

Ff  =  ^,  cos  ^  -j-  F2  cos  ^2  -j-  .    .  .  =  ~2F  cos  y, 
and  the  moments  about  the  axes  are  given  (page  194)  by 

Mx  =  2F  cos  y  .  y  —  2F  cos  /?  .  z,  ~\ 
My  =  2F  cos  a  .  z  —  2F  cos  y  .  x, 
Mt  =  2F  cos  ft  .  x  —  2F  cos  a  .  y. 


(2) 


(3) 


Let  the  components  of  the  displacement  ds  be  dx,  dy,  ds. 

Since  the  work  of  any  force  is  equal  to  the  algebraic  sum  of  the  works  of  its  components, 
we  have  for  translation 


•      -      (4) 


Flpl  =  Fl  cos  «j  .  dx  -\-  FI  cos  /?! .  dy  -\-  Fl  cos  ^  .  ^, 
^2/2  =  F2  cos  a2 .  ^  -(-  /^  cos  ftz.  dy  -\-  F2  cos  ^2 .  ds, 

and  so  on.      By  addition,  we  have,  from  (i)  and  (2),  for  the  ivork  of  translation 

work  —  Fx.dx  +  Fy.  dy  -\-  Ft.dz (5) 

Now  let  us  suppose  the  displacement   is   one   of  rotation  due  to  an   indefinitely  small 
angular  displacement  dtt  about  an  axis  through  O.      Let  the  components  of  this  displacement 
along  the  axes  be  dOx,  dfty,  dVt. 
Then  we  have 

dx  —  z  .  dHy  —  y .  <fff,t 
dy  =  x  .  dftt  —  z  .  dttx, 
dz  =  y  .  dftx  —  x  .  dfiy. 

Substituting  these  values  in  (4)  and  adding,  we  have,  from  (i)  and  (3),  for  the  ^vork  of 
rotation 


CHAP.  II.] 


PRINCIPLE  OF  VIRTUAL   WORK. 


217 


For  translation  and  rotation  combined  we  have  then,  from  (5)  and  (6), 


M.    •     •     .     (7) 

Equations  (i)  and  (7)  are  then  equivalent.    Equation  (i)  is  general  and  includes  all  cases. 
Principle  of  Virtual  Work. — If  we  put  equation  (7)  equal  to  zero,  since  dx,  dy,  dz,  d6x, 
itfy,  dtfz  are  not  zero  for  any  displacement,  we  must  have 

Fx  =  o,         Fy  =  o,         F,  =  o, 
Mx  =  o,        My  —  o,        Mz  =  o. 

But  we  have  seen,  page  211,  that  these  are  the  conditions  for  equilibrium. 

Hence,  when  a  rigid  body  is  acted  on  by  any  system  of  non-concurring  forces  in  equilib- 
rium, the  algebraic  sum  of  the  works  of  all  the  forces  for  any  indefinitely  small  displacement 
either  of  translation  or  rotation  or  both  combined  is  equal  to  zero. 

The  same  holds  for  a  system  of  rigid  bodies  connected  by  inextensible  strings. 

We  have  then,  from  (i),  for  static  equilibrium  the  condition 

Flpl  -f-  F2p2  -j-  F3ps  -f-  .    .   .  =  ^Fp  =o, (8) 

where  A,  /2,  /3,  etc.,  are  the  displacements  in  the  direction  of  each  force,  of  its  point  of 
application,  and  each  term  is  to  be  taken  positive  or  negative  according  as  the  forces  are  in 
the  direction  of  their  displacements  or  in  the  opposite  direction. 

But  in  static  equilibrium  every  point  is  at  rest  and  there  is  no  real  displacement.  The 
principle  nevertheless  holds  good  for  any  supposed  indefinitely  small  displacement.  Such  a 
supposed  indefinitely  small  displacement  which  does  not  really  take  place  we  call  a  virtual 
displacement.  The  work  of  any  force  for  such  a  displacement  is  then  virtual  work. 

The  principle  expressed  by  (8)  is  called,  therefore,  the  principle  of  virtual  work,  and  may 
be  expressed  as  follows  : 

If  a  rigid  body  is  in  static  equilibrium,  the  algebraic  sum  of  the  virtual  works  for  any 
virtual  displacement  is  zero. 

If  the  forces  do  not  change  in  magnitude  and  direction  with  the  supposed  displacement, 
the  virtual  displacement  need  not  be  taken  indefinitely  small. 

Examples,— (i)  Find  the  condition  for  static  equilibrium  for  a  screw,  neglecting  friction. 
ANS.  This  is  example  (10),  pasje  213.     We  have  solved  it  there  by  resolution 
of  forces.     By  virtual  work  the  solution  is  as  follows  : 

If  the  arm  /  turn  through  a  small  angle  9,  the  work  of  P  is  P.  $.     The  mass 

Q  is  raised  a  distance  ^9.     Hence  by  virtual  work 


-=0t    or          =. 

2TC  27T/. 


(2)  Solve  the  differential  screw  given  in  example  (f-f),  page  214,  by  virtual 
work. 

ANS.  We  have  at  once 


=0.     or     f  = 


(3)  Solve  the  wedge  given  in  example  (12),  pa%e  214,  by  virtual  work. 

ANS.   For  a  small  vertical   displacement  d    we   have   work   of  P  given  by  P&     The  work  of  N  is 

—  Nd  sin  2.     Hence  by  virtual  work 


Pd  —  2Nd"  sin  -  =  o,     or    P  =  2N  sin  — . 


2l8 


STATICS.    GENERAL  PRINCIPLES. 


[CHAP.  II. 


(4)  Sohc  the  wheel  and  axle  given  in  example  (t  j),  page  214,  by  virtual  work. 
ANS.     For  a  small  angle  0  we  have 


PaQ  -  QW  =  o,    or    Q  =     P. 

o 

(5)  In  the  single  movable  pulley  shown  in  the  figure  find  the  relation  between  P  and  O 
for  static  equilibrium,  disregarding  friction  and  rigidity  of  rope. 

ANS.  For  a  displacement    s   of    P    downwards,    each  .rope   passing   around    the 

movable  pulley  is  diminished  —  and  Q  is  raised  —  .     We  have  then 


—  Q—  =  o,     or 


r=8. 


(6)  In  the  system  of  pulleys  shown  in  the  figure  find  the  relation  between  P  and 
r  ^atic  equilibrium,  disregarding  friction  and  rigidity  of  ropes. 
ANS.  Let  tn  be  the  mass  of  each  pulley. 
For  a  displacement  s  of  P  downwards, 

the  first  movable  pulley  is  raised  — . 


second 
third 


"    »th 


We  have  then 


„        ms      ms  ms      Qs 

~  ~2    ""    4  '    '   '   2»  ~  ~Tn 


where  n  is  the  number  of  movable  pulleys. 

(7)  In  the  system  of  pulleys  shown  in  the  figure  find  the  relation  between  P  and  Q  for 
static  equilibrium,  disregarding  friction  and  rigidity  of  ropes. 

ANS.    For   a  vertical    displacement  s  of  P  each    rope  from  the  lower  block  will  be 

shortened  by  s.  and  the  displacement  of  Q  upwards  will  be  -,  where  n  is  the  number  of 
ropes  from  the  lower  block. 

If  m  is  the  mass  of  the  lower  block,  we  have 

Ps-(Q  +  m)  -  =  o,     or     P  =  *LLHL. 

ft  H 

(8)  A  body  of  mass  m  rests  upon  „   smooth  inclined  plane  AB  which  makes  an  angle  a 
with  the   horizontal  and  is  acted  upon   by  a  force  P  which   makes  the  angle  ft  with  the 
plane.     Find  the  conditions  of  static  equilibrium. 

ANS.  Consider  the  body  as  a  particle  at  any  point  O  on 
the  plane.  We  have  acting  upon  the  particle  the  weight  of 
m,  the  force  P  and  the  nominal  reaction  N  of  the  plane,  and 
these  three  forces  must  be  in  equilibrium.  Let  the  angle 
FOB  =  ft  be  positive  when  above  the  plane,  and  negative 
when  below. 

IST  SOLUTION.      Bv    RESOLUTION  OF  FORCES.— We 
have,  putting  the  algebraic  sum  of  components  at  right  angles  to 
zero,  for  gravitation  measures 


N  cos  ft  —  m  cos  (a  +ft)  =  o,     or 


m  cos  (a  + 

cos  ft 


CHAP.  II.]  EXAMPLES.  219 

Putting  the  algebraic  sum  of  components  parallel  to  the  plane  equal  to  zero,  we  have  for  gravitation 
measures 

P  cos  ft  -  m  sin  a  =  o,     or     P  =  *^~-jf (2) 

When  ft  =  90°  —  a,  P  is  vertical  and,  from  (i),  N  =  o.  For  any  greater  value  of  positive  ft  we  have, 
from(i),  .W  negative  and  no  equilibrium  is  possible.  For  negative  ft  we  must  have/3  less  than  90°.  Equations 
(i )  and  (2)  hold  good,  then,  for  all  values  of  ft  between  +  (90=  —  a)  and  —  90°.  Outside  of  these  limits  there 
is  no  equilibrium.' 

The  minimum  value  of  P  is  for  ft  =  o  and  equal  to  P  =  m  sin  a. 

Again,  we  can  put  the  algebraic  sum  of  components  along  the  plane  and  perpendicular  to  the  plane  equal 
to  zero.  We  have  then 

P  cos  ft  —  m  sin  a  =  o, 

N  +  P  sin  ft  —  m  cos  a  =  o. 

From  these  two  equations  we  can  obtain  (i)  and  (2). 

Again,  we  can  put  the  algebraic  sum  of  horizontal  and  vertical  components  equal  to  zero.     We  have  then 

P  sin  (a  +  ft)  +  N  cos  a  —  m  =  o, 
P  cos  (a  +  ft)  —  N  sin  a  =  o, 

and  from  these  equations  we  can  obtain  (i)  and  (2). 

20  SOLUTION.  BY  VIRTUAL  WORK.— In  order  to  find  P,  suppose  a  virtual  displacement  d  along  the 
plane  from  O  towards  B.  This  displacement  is  at  right  angles  to  N,  and  hence  the  virtual  work  of  N  is  zero. 
We  have  then 

Pd  cos  ft  —  md  sin  a  —  o,     or    P  =  m    '   _a. 

cos  ft 

In  order  to  find  N,  suppose  a  horizontal  virtual  displacement  d.  Then  the  virtual  work  of  IV  is  zero, 
and  we  have 


CHAPTER  III. 

STATIC  FRICTION. 

Friction. — Every  natural  surface  offers  a  resistance  to  the  motion  of  a  body  upon  it. 
Part  of  this  resistance  is  due  to  ADHESION  between  the  body  and  surface  and  part  is  due  to 
FRICTION. 

Friction,  then,  is  always  a  retarding  force  or  resistance,  and  acts  always  in  a  direction 
opposite  to  that  in  which  the  body  moves  or  would  move  if  there  were  no  resistance. 

When  one  surface  moves  upon  another,  the  surfaces  in  contact  are  compressed  and  pro- 
jecting points  and  irregularities  are  bent  over,  broken  off,  rubbed  down,  etc. 

The  resistance  due  to  friction,  therefore,  evidently  depends  upon  the  materials  of  which 
the  surfaces  are  composed,  and  also  upon  the  roughness  or  smoothness  of  the  surfaces  in 
contact. 

It  may  also  evidently  vary  for  the  same  surfaces,  according  to  their  condition  or  state 
or  material  constitution. 

Thus  it  may  not  be  the  same  for  surfaces  of  dry  wood  or  iron  as  for  the  same  surfaces 
under  the  same  conditions  when  wet.  It  may  not  be  the  same  for  two  surfaces  of  wood 
with  their  fibres  parallel  as  for  the  same  surfaces  under  the  same  conditions  when  their 
fibres  are  not  parallel. 

Unguents  also  have  a  great  influence.  Such  fluid  or  semi-fluid  unguents  as  oil,  tallow, 
etc.,  fill  up  interstices  and  diminish  the  effect  of  irregularities  of  surfaces  ,  or  a  film  of  unguent 
may  be  interposed  between  the  surfaces  and  thus  the  resistance  of  friction  greatly  diminished. 

Adhesion. — We  must  not  confound  the  resistance  due  to  friction  with  that  due  to 
adhesion.  Adhesion  is  that  resistance  to  motion  which  takes  place  when  two  different  sur- 
faces come  in  contact  at  many  points  without  pressure.  Adhesion  increases  with  the  area 
of  surface  of  contact  and  is  independent  of  the  pressure,  while,  as  we  shall  see  (page  221), 
friction  increases  with  the  pressure  and  is  in  general  independent  of  the  area  of  surface  of 
contact.  When  the  pressure,  then,  is  very  small,  adhesion  may  be  great  compared  with 
friction.  , 

If,  however,  the  pressure  is  great,  adhesion  may  be  neglected  compared  to  the  friction, 
and  the  resistance  to  motion  is  practically  that  due  to  the  friction  only. 

Kinds  of  Friction. — Surfaces  may  slide  or  roll  on  one  another.  We  distinguish 
accordingly  SLIDING  FRICTION  and  ROLLING  FRICTION. 

It  is  also  found  by  experiment  that  the  friction  which  just  prevents  motion  is  greater 
than  that  which  exists  after  actual  motion  takes  place.  The  friction  which  just  prevents 
motion  is  called  friction  of  repose  or  quiescence,  or  STATIC  FRICTION.  The  friction  which 
exists  after  actual  motion  takes  place  is  called  friction  of  motion,  or  KINETIC  FRICTION. 

We  have,  then,  two  kinds  of  static  friction,  viz.,  STATIC  SLIDING  FRICTION  and  STATIC 
ROLLING  FRICTION. 

We  have  also  two  kinds  of  kinetic  friction,  viz.,  KINETIC  SLIDING  FRICTION  and 
KINETIC  ROLLING  FRICTION, 


CHAP.  III.] 


COEFFICIENT  OF  FRICTION. 


In  any  case,  whether  of  sliding  or  rolling,  the  kinetic  friction  is  always  less  than  the 
static  friction. 

We  have  to  do  in  this  portion  of  our  work  with  static  friction  only. 

Coefficient  of  Friction.  —  When  two  surfaces  are  in  contact  and  there  is  friction  and 
normal  pressure  at  every  point  of  contact,  the  sum  of  the  frictions  at  every  point  of  contact 
is  the  total  friction,  and  the  sum  of  the  normal  pressures  at  every  point  of  contact  is  the 
total  normal  pressure. 

The  ratio  of  the  total  friction  to  the  total  normal  pressure  when  motion,  either  sliding  or 
rolling,  is  just  about  to  begin,  is  called  the  COEFFICIENT  OF  STATIC  FRICTION,  either  of  sliding 
or  rolling. 

The  same  ratio  after  motion  has  taken  place  is  called  the  COEFFICIENT  OF  KINETIC  FRIC- 
TION, either  of  sliding  or  rolling. 

We  denote  the  coefficient  of  friction  in  general  by  //.  We  have  then,  in  general,  for  all 
cases  * 

^  =  N'     °r     F  =  >*N' 

where  F  is  the  total  friction  and  N  the  total  normal  pressure  when  motion  either  sliding  or 
rolling  is  just  about  to  begin,  or  else  when  motion  either  sliding  or  rolling  has  taken  place.  In 
the  first  case  /*  is  the  coefficient  of  static  friction  of  sliding  or  rolling.  In  the  second  case  ju 
is  the  coefficient  of  kinetic  friction  of  sliding  or  rolling.  We  have  to  do  in  this  portion  of 
the  work  with  static  friction  only. 

Limiting  Equilibrium.  —  The  student  should  carefully  note  that 


does  not  give  the  actual  resistance  of  friction  in  all  cases  of  equilibrium,  but  only  the  resist- 
ance which  exists  when  the  surfaces  are  on  the  point  of  motion. 

Friction  acts  always  in  a  direction  opposite  to  the  force  which  tends  to  cause  motion, 
and  so  long  as  there  is  equilibrium  it  is  always  equal  in  magnitude  to  this  force.  But  when 
this  force  has  the  magnitude  l^N  motion  \sj2ist  about  to  begin,  and  the  body  is  said  to  be  in 
LIMITING  EQUILIBRIUM.  If  this  force  is  less  than  ^V,  there  will  still  be  equilibrium,  what- 
ever its  magnitude,  and  the  body  is  in  NON-LIMITING  EQUILIBRIUM. 

Coefficient  of  Static  Sliding  Friction—  Experimental  Determination.  —  Let   a  body  of 
weight  W,  acting  at  the  centre  of  mass  C,  rest  in  equilibrium  upon 
a  rough  plane  AB,  the  surfaces  of  contact  being  plane. 

Then  for  equilibrium  the  reaction  R  of  the  plane  is  equal  and 
opposite  to  Wand  in  the  same  vertical  line,  and  the  sum  TV  of  all 
the  normal  pressures  acting  at  every  point  of  contact  must  be  equal 
and  opposite  to  the  normal  component  of  W,  and  the  sum  F  of  all 
the  frictions  at  every  point  of  contact  must  be  equal  and  opposite 
to  the  component  T  of  £F  parallel  to  the  plane. 

We  have,  then,  when  sliding  is  about  to  begin,  for  the  coeffi- 
cient of  sliding  friction 


and  we  see  from  the  figure  that  -^  is  the  tangent  of  the  angle  which  the  reaction  R  makes  with 
the  normal  when  sliding  is  about  to  begin.    Now  the  reaction  at  every  point  of  contact  is  paral- 


222  STATICS.    GENERAL  PRINCIPLES.  [CHAP.  III. 

lei  to  A' or  IV,  and  sliding  begins  at  all  points  of  contact  simultaneously.  Hence  the  angle 
which  R  makes  with  the  normal  when  sliding  is  about  to  begin  is  evidently  the  same  as  the 
angle  which  the  plane  makes  with  the  horizontal.  Therefore 

F 

P  =        =  tan  0. 


We  call  the  angle  0  which  the  plane  makes  with  the  horizontal  when  motion  is  about  to 
begin  the  ANGLE  OF  REPOSE. 

That  is,  the  coefficient  of  static  sliding  friction  is  equal  to  the  tangent  of  the  angle  of 
repose. 

If,  then,  we  place  a  body  upon  a  rough  plane  and  then  gradually  incline  the  plane  until 
sliding  just  begins,  the  inclination  of  the  plane  at  this  instant  gives  the  angle  of  repose  0. 
The  tangent  of  this  angle  gives  the  coefficient  ju  of  static  sliding  friction  for  plane  surfaces. 

We  obtain  the  same  result  by  resolution  of  forces.  Thus  let  0  be  the  inclination  of 
the  plane  when  sliding  begins. 

Then  for  equilibrium  W  cos  0  =  N,  and  W  sin  0  =  F.     Hence 

F 
M  =  TF  =  tan  0. 


We    can    thus    make    use    of    the    inclined   plane    as   an  apparatus  for  determining  /< 
by  experiment. 

Again,  if  we  place  a  body  of  weight  If  on  a  horizontal  plane 
and  measure  the  horizontal  force  /''just  necessary  to  cause  it  to 
begin  to  slide,  we  have 

M  =  W  =  tan  0» 

where  0  is  the  angle  of  the  reaction  R  with  the  normal  when  sliding  begins,  or  the  angle  of 
repose. 

Such  an  apparatus  should  be  so  constructed  that  the  friction  of  the  pulley  and  other 
resistances  due  to  the  string,  etc.,  can  be  disregarded  or  else  allowed  for. 

Cone  of  Friction.— If  we  revolve  the  line  representative  of  R  in  the  figure  page  221 
about  the  vertical,  it  will  describe  a  cone.  This  cone  is  called  the  CONE  OF  FRICTION.  The 
reaction  of  R  is  equal  and  opposite  to  the  resultant  of  F  and  W  acting  at  C. 

We  see,  then,  that  no  force  acting  at  C,  horuever  great  in  magnitude,  can  cause  sliding 
to  begin  if  its  line  representative  lies  within  the  cone  of  friction. 

Laws  of  Static  Sliding  Friction. — The  following  laws  of  static  sliding  friction  have  been 
established  by  experiment  as  holding  true  within  the  limits  indicated : 

i .  Other  things  being  the  same,  within  certain  limits  of  the  normal  pressure,  static  sliding 
friction  is  proportional  to  the  total  normal  pressure  and  independent  of  the  area  of  the  surfaces 
in  contact. 

In  other  words,  within  the  limits  of  normal  pressure  referred  to,  the  coefficient  of  static 
sliding  friction  fit  is  constant  for  the  same  two  surfaces  in  the  same  condition,  whatever  the  area 
of  the  surfaces  of  contact  and  whatever  the  total  normal  nressure.  i 


CHAP.  III.]  LIMITATIONS  OF  THE  LAW.  223 

Thus,  if  the  normal  pressure  N  over  a  given  area  is  increased  or  decreased,  the  friction 
^increases  or  decreases  in  the  same  proportion  and  p  =  —  is  unchanged. 

It  follows  directly  that  if  the  area  increases  or  decreases,  N  remaining  the  same,  the 
number  of  points  of  contact  is  correspondingly  increased  or  decreased,  but  the  normal  pressure 
at  each  point,  and  therefore  the  friction  at  each  point,  is  correspondingly  decreased  or 

p 
increased.     The  sum  of  all  the  frictions  F  remains,  then,  the  same  and  /*  =  --  is  unchanged. 

Limitations  of  the  Law.  —  The  limitations  of  normal  pressure  referred  to  are  as  follows: 

If  the  normal  pressure  per  unit  of  area  approaches  the  crushing  strength  or  becomes  so 
great  as  to  break  up  the  film  of  interposing  unguent,  the  friction  F  increases  more  rapidly 
than  the  normal  pressure,  and  the  law  fails. 

In  properly  designed  structures  the  normal  pressure  per  unit  of  area  is  much  less  than 
this  limit,  and  the  law  applies. 

Again,  if  the  normal  pressure  per  unit  of  area  is  very  small,  adhesion  may  constitute 
the  larger  portion  of  the  resistance.  This  adhesion  increases  with  the  area  of  contact 
(page  220). 

In  all  practical  cases,  however,  the  influence  of  adhesion  may  be  neglected. 

Hence  in  practical  appplications  the  friction  is  the  only  resistance  which  is  considered, 
and  it  is  assumed  that 


gives  the  resistance,  where  /*  is  in  practice  a  constant  for  the  same  two  surfaces  in  the  same 
condition,  whatever  the  area  of  the  surfaces  in  contact  and  whatever  the  total  normal 
pressure  N. 

2.  Other  things  being  the  same,  within  certain  limits  of  the  normal  pressure,  the  static 
sliding  friction  of  greased  surfaces  is  less  than  that  of  ungr  eased  and  depends  less  upon  the 
surfaces  than  upon  the  unguent. 

Here,  again,  if  the  normal  pressure  per  unit  of  area  becomes  so  great  as  to  break  up 
the  film  of  interposing  unguent,  surface  comes  in  contact  with  surface  and  the  friction  may 
depend  more  on  the  surfaces  than  upon  the  unguent. 

In  properly  designed  structures  the  normal  pressure  per  unit  of  area  is  much  less  than 
this,  and  the  law  applies. 

Again,  if  the  normal  pressure  per  unit  of  area  is  very  small,  adhesion  may  constitute 
the  larger  portion  of  the  resistance,  and  this  adhesion  is  increased  by  the  unguent. 

In  all  practical  cases,  however,  the  influence  of  adhesion  may  be  neglected. 

Hence  in  practical  applications  the  friction  is  the  only  resistance  which  is  considered, 
and  it  is  assumed  that 

F=  nN 

gives  the  resistance,  where  /*  is  in  practice  a  constant  for  the  same  two  surfaces  in  the  same 
condition,  whatever  the  area  of  the  surfaces  in  contact  and  whatever  the  total  normal 
pressure  N. 

Upon  these  two  laws  depend  the  value  and  use  of  the  values  for  the  coefficient  of 
static  sliding  friction  given  in  the  next  article. 

Values  of  Coefficient  of  Static  Sliding  Friction.  —  The  following  table  gives  a  few 
values  of  /*  as  determined  by  experiment  for  static  sliding  friction. 


STATICS.    GENERAL  PRINCIPLES. 
COEFFICIENTS   OF   STATIC   SLIDING   FRICTION   /*  =  tan  0. 


[CHAP.  III. 


Substances  in  Contact. 

Condition  of  Surfaces  and  Kind  of  Unguent. 

Dry. 

Wet. 

Olive  Oil. 

Lard. 

Tallow. 

Dry  Soap. 

Polished 
and  Greasy. 

{minimum 

0.30 
0.50 
0.70 
0.15 
O.l8 
0.24 
O.6o 
0.50 
0.63 
0.80 
0.47 

0.65 
0.68 
0.71 

0.14 
0.19 
0.25 

O.II 
0.12 

O.22 
0.36 
0.44 

0.30 
0-35 
0.40 

0.15 
O.IO 

0.28 

0.21 

maximum 
i  minimum 

O.II 

0.12 

O.I') 

O.IO 

O.IO 
O.I2 

1 

maximum 

0.65 
0.87 

Hemp  ropes  or  plaits  on  wood  - 
Leather     belts      over     drums 

minimum 
mean  
maximum 
wood  

Stone    or   brick    on    stone    or 

minimum 
maximum 

0.67 
0.75 
0.65 
0.74 
0.40 
0.7  too.  3 
0.51 
0-33 
0.25  to  i 

I.O 

Masonry  and  brickwork,  damp  mortar.  .  .  . 

D            1             d            1 

More  extensive  tables  will  be  found  in  treatises  on  Friction.  It  will  be  noted  that 
the  coefficient  of  static  sliding  friction  is  practically  always  less  than  unity.  In  only  one 
case  given  in  the  table,  viz.,  for  damp  clay  on  damp  clay,  is  p  =  i,  corresponding  to  an 
angle  of  repose  of  0  =  45°.  Rankine  gives  for  "  shingle  on  gravel  "  a  maximum  /*  =  1.  1  1, 
corresponding  to  an  angle  of  repose  0  =  48°. 

Static  Friction  for  Pivots.  —  In  all  cases  of  the  sliding  of  two  surfaces  we  denote  the 
coefficient  of  static  sliding  friction  by  yu  and  take  the  value  of  p  as  given  by  the  table.  We 
have  then,  in  all  cases  of  sliding  friction,  for  the  friction  when  sliding  is  about  to  begin 

F  =  ^N=Ntan  0, 

where  N  is  the  total  normal  pressure  and  0  is  the  angle  of  repose,  and  //  is  given  by  the  table. 
The  direction  of  the  friction  is  always  opposite  to  the  direction  of  motion  if  motion  were 
to  take  place. 

The  application  to  pivots  is  then  simple. 

i.  SOLID  FLAT  PIVOT.  —  Let  ACB  be  the  base  of  a  solid  flat  pivot,  and  A^the  total  normal 
pressure  upon  the  base. 

We  have  then  for  the  static  friction 

F=pN,       ........     (i) 

where  H  is  given  by  the  table. 

If  we  divide  the  base  into  a  very  large  number  of  very  small  equal 
triangles,  such  as  ACD,  the  friction  on  each  can  be  considered  as  the 
resultant  of  equal  parallel  forces  distributed  over  the  surface.  The 
point  of  application  for  each  triangle  is  then  at  the  centre  of  mass  for 
that  triangle.  The  point  of  application  of  the  entire  friction  is  then  at  a 

The  moment  of  the  entire  friction  with  reference  to  the 


distance  Cs  =  -  r  from  the  centre. 
axis  is  then 


CHAP.  III.] 


STATIC  FRICTION  FOR  PIYOTS. 


225 


Since  for  any  point  s  of  the  base  there  is  a  corresponding  point  s'  for  which  the  friction 
is  equal  and  opposite,  the  moment  of  the  friction  is  the  moment  of  a  couple  and  is  therefore 
the  same  for  every  point  in  the  plane  of  the  base  (page  185). 

2.  HOLLOW  FLAT  PIVOT.  —  If  the  rubbing  surface  is  a  flat  nngADEB,  we  have  as  before 


F=pNt       .     .-•._;.     .     ...     (i) 

where  N  is  the  total  normal  pressure  on  the  base,  and  ju  is  the  coefficient 
of  static  sliding  friction  as  given  by  the  table  page  224. 

Let  the  outer  radius  be  rl  and  the  inner  radius  rz.  Then  any 
small  portion  of  the  base  is  a  circular  ring  for  which  the  length  of  chord 
and  arc  AD  may  be  taken  equal.  The  centre  of  mass  (page  29)  for 
each  small  portion  is  then  at  a  distance  Cs  from  the  axis  given  by 

3 


Hence  the  moment  of  the  friction  with  reference  to  the  axis  is 


(2) 


Since  for  any  point  s  there  is  a  corresponding  point  s'  for  which  the  friction  is  equal  and 
opposite,  the  moment  of  the  friction  is  the  moment  of  a  couple  and  is  therefore  the  same  for 
any  point  in  the  plane  of  the  base  (page  185). 

3.  CONICAL  PIVOT. — In  the  case  of  a  conical  pivot  let  R  be  the  pressure  along  the  axis, 
and  let  the  half  angle  of  convergence  ADC  be  a. 

If  we  divide  the  conical  surface  into  a  large  number  «  of  very  small 
triangles  with  their  vertices  at  the  point  D,  each  will  sustain  the  vertical 


r> 

load  — ,  and  the  normal  pressure  on  each  will  be 


R 


If  we  denote 


the  radius  ClAl  =  ClBl  of  the  pivot  at  the  point  of  entrance  by  rlt  the 
resultant  normal  pressure  upon  each  small  elementary  triangle  acts  at  a 

distance  —  rl  from  the  axis. 


We  have  then  for  the  total  friction 


R 

sin  a' 


where  p  is  the  coefficient  of  static  sliding  friction  as  given  by  the  table  page  224,  and  the 
moment  of  the  friction  with  reference  to  the  axis  is 

,_       2      Rr, 
M  =  —fjt  - — -, 
3    sin  a 

or,  since    .    1    =  the  side  DA,  of  the  cone  of  contact  =  a,  we  have 
sin  a. 


M=2- 


(2) 


,,0 


STATICS.    GENERAL  PRINCIPLES. 


[CHAP.  111. 


This  is  also  the  moment  of  a  couple  and  hence  the  same  for  any  point  in  the  plane  per- 
pendicular to  the  axis  at  a  distance  above  the  point  D  equal  to  two  thirds  the  height  of  the 
cone  of  contact. 

4.   PIVOT  A  TRUNCATED  CONE.  —  Let  R  be  the  pressure  along  the  axis,  and  let  the  half 
angle  of  convergence  ADC  be  a. 

Let  Ra  be  the  pressure  sustained  by  the  flat  base,  and  Rv  the  pressure 
sustained  by  the  conical  surface. 
Then 


s^& 


Also,  if  rl  is  the  radius  ClAl  at  the  point  of  entrance,  and  r2  the  radius 
of  the  base, 

r2 
R3:R::  nr*  :  nr?,     or      R2  =  -\R, 

7*1 


and  hence 


-  r* 
r2 


We  have  then  as  in  Case  I,  page  224,  for  the  flat  pivot,  the  friction  F2  on  the  base 


and  its  moment  about  the  axis 


For  the  friction  on  the  conical  surface  we  have,  as  in  Case  3,  page  225,  for  the  conical 


pivot 


_       R^     __        r?  -  r22        R 
sin  a  r?         sin  a 


and  for  its  lever-arm,  as  in  Case  2,  page  225,  for  hollow  pivot, 

2  r?  -  ra8 

3  ri2  —  rt 
Its  moment,  then,  about  the  axis  is 


3  r,2         sin  a 

The  total  friction  for  the  truncated  pivot  is  then 

and  its  total  moment  about  the  axis  is 

M  =  M,  +  M.  =  -  fi-t 
3    rt- 


sin  a 


(0 


(2) 


where  n  is  the  coefficient  of  static  sliding  friction  as  given  by  the  table  page  224. 


CHAP.  III.] 


STATIC  FRICTION  FOR  PIVOTS. 


227 


[Pivot  with  Spherical  End.] — Let  R  bz  the  pressure  along  the  axis,  denote  the  radius  AO  of  the  spherical 
surface  by  r,  and  the  radius  AC  by  rt ,  and  let  the  angle  AOC  be  a, 

•p 

Then  the  load  per  unit  of  area  of  horizontal  projection  is  ^75-     Take  any 

element  of  the  surface  at  a,  distant  ab  =  x  from  the  axis,  and  let  Ob  =  y.  The 
horizontal  projection  of  this  element  is  2itxdx,  and  the  load  sustained  by  it  is 
R  2Rxdx 


then  iitxdx   x 


y        A/  r"*  _ 
The  cosine  of  the  angle  aOb  is  cos  aOb  =  ^  =  3- 


The  normal  pressure  on  the  element  at  a  is 


then 


and  the  static  friction  is 


zRxdx 


•zuRr 


xdx 


Integrating  between  the  limits  of  x  =  o  and  x  =  r\ ,  we  have  for  the  total  friction 


F  = 


or,  since   j/ra  —  r*  =  r  cos  a  and  r\  =  sin  a, 


sin'  a  •  i  +  cos  a. 

where  ju  is  the  coefficient  of  static  sliding  friction  as  given  by  the  table,  page  224. 
For  hemispherical  end  a=  90°  and  F  =  2juA\     For  flat  end  ct=  o  and  F  =  uR. 
The  moment  about  the  axis  of  the  friction  on  an  element  is 

2juRr        x*dx 


Integrating  between  the  limits  x  =  o  and  x  =  r*,  we  have  for  the  total  moment  of  the  friction  about  the 


M 


WRrr  r*     .       Tr,       r,     ,—  -  -H 
-  r-     -sin"1—  --  4/r"-ri*    , 
r?   \_2  r        2    v 


or,  inserting  the  values  of   J/V"  —  r\*  =  r  cos  a  and  r\  =•  r  sin  a  and  reducing, 

fa.  \ 

M  —  nRr\  — - —  —  cot  a  1 

\sm2  a  I 


(2) 


For  hemispherical  end  a  =  — ,  sin  a  =  i,  cot  a  =  o,  and  this  becomes  M  =  -. 


Static  Friction  of  Axles. — In  all  cases  of  the  sliding  of  two  surfaces,  we  denote  the 
•coefficient  of  static  sliding  friction  by  yu  and  take  the  value  of  jj.  as  given  by  the  table,  page 
224.  We  have  then,  in  all  cases  of  sliding  friction,  for  the  friction  when  sliding  is  about 
to  begin 

F=  uN  =  Af  tan  0, 

where  N  is  the  total  normal  pressure  and  0  is  the  angle  of  repose,  and  n  is  given  by  the 
table,  page  224. 


228 


STATICS.    GENERAL  PRINCIPLES. 


[CHAP.  111. 


The  direction  of  the  friction  is  always  opposite  to  the  direction  of  motion  if  motion 
were  about  to  take  place. 

The  application  to  axles  is  then  simple. 

i.  AXLE  IN  PARTIALLY  WORN  BEARING. — Let  the  bearing  be 
partially  worn,  then  the  axle  at  the  moment  when  sliding  begins  touches 
the  bearing  at  a  point  A,  and  the  resultant  pressure  R  at  this  point  makes 
the  angle  of  repose  0  with  the  normal.  We  have  then  for  the  normal 
pressure  N  =  R  cos  0,  and  for  the  friction 


F  =  N  tan  0  =  R  sin  0, 


where  0  is  the  angle  of  repose  as  given  by  the  table,  page  224. 

Let  r  be  the  radius  AC  of  the  axle.      Then  the  moment  of  the  friction  with  reference 
to  the  axis  is 


=  Rr  sin  0. 


(2) 

If  the  axle  is  well  greased,   the  angle  of  repose   0  is  very  small,   and  we  may  take 
=  tan  0  =  sin  0.      In  the  practical  case  of  a  well-greased  axle,  then,  we  have 


where  n  is  given  by  the  table,  page  224. 

If  the  wheel  AB  revolves,  as  shown,  about  a  fixed  axle  AC,  the  friction  is  the  same  as 

before,  but  the  lever- arm  of  the  friction  is  not  the  radius  of  the  axle,  but 

the  inner  radius  of  the  wheel. 

2.  AXLE — TRIANGULAR  BEARING. — If  the  bearing  is  triangular,  the 
axle  is  supported  at  two  points  A  and  B. 
The  resultant  pressure  R  can  be  resolved 
into  two  components  R^  and  R2 ,  and  when 
sliding  begins  each  of  these  makes  the  angle 
of  repose  0  with  the  normals  at  A  and  B. 
The  normal  pressure  at  A  is  then  Nl  =  Rt  cos  0,  and  the  fric- 
tion at  A  is 

^"i  =  NI  tan  0  =  Rv  sin  0. 
The  friction  at  B  is  in  like  manner  F2  =  R2  sin  0.     The  total  friction  is  then 


Let  the  angle  ACS  =  20.    Then  the  angle  AOR  =  ft  -  0,  and  the  angle  BOR  =  0  +  0. 
We  have  then 

/?, :  R :  :  sin  (ft  +  0) :  sin  20,     or     Rt  =  — 


/?2:/?::  sin  (0-0):  sin  20,      or     R,  =--. 

sin  2ft 


Hence  the  total  friction  is 


F=  [sin  (ft  +  0)  +  sin  (ft  -  0)]^^- 


CHAP.  111.] 


STATIC  FRICTION  FOR  AXLES. 


229 


But  sin  (/?  +  0)  +  sin  (ft  —  0)  =  2  sin  ft  cos  <p,  and  sin  2  #  —  2  sin  0  cos  /?.     Hence  we 


have 


_ 


R  sin  0  cos  0 


/?  sin  2  0 
2  cos  ?  ' 


(I) 


where  0  is  the  angle  of  repose  as  given  by  the  table,  page  224. 

The  moment  of  friction  with  reference  to  the  axis,  if  r  is  the  radius  of  the  axle,  is 

Rr  sin  20 


M=Fr  = 


2   COS 


If  the  axle  is  well  greased,  the  angle  of  repose  0  is  very  small,  and  we  may  take 
sin  20  =  2  sin  0,  also  /*  =  tan  0=  sin  0.  In  the  practical  case  of  a  well-greased  axle,  then, 
we  have 

R  Rr 


cos  ft1 


cos  ft ' 


where  fA.  is  given  by  the  table,  page  224.      If  the  angle  ft   is  small,  cos  ft  may  be  taken  as 
unity,  and  F  and  M  are  then  the  same  as  in  the  preceding  case, 

F  =  *R          M 


[3.  AXLE— NEW  BEARING.] — When  the  bearing  is  new  and  unworn,  the  axle  touches  it  at  all  points. 

Let  R  be  the  resultant  vertical  pressure  acting  at  the  centre  O  of  the  axle. 
Denote  the  radius  AO  of  the  axle  by  r,  the  distance  AC  by  r\,  and  let  the 
angle  AOCbe  ft. 

Then  the  load  per  unit  of  horizontal  projection  is — .     Take  any  element 

of  the  surface  of  the  axle  at  a,  distant  ab  =  x  from  R,  and  let  Ob  —  y.     The 
horizontal  projection  of  this  element  is  dx,  and  the  load  sustained   by  it  is 

- ---.     At  a'  we  have  a  similar  element. 

The  friction  on  these  two  elements  is,  from  the  preceding  article, 

sin  20.  Rdx 
2ri  cos  aOb ' 

But  cos  aOb  =  y-  —  ,  hence  the  friction  for  the  two  elements  is 

r  r 

Rr  sin  20  fix 


Integrating  between  the  limits  x  =  r\  and  x  =  o,  we  have  for  the  entire  friction 


„  _  Rr  sin  20    .     -t  r\ 

r  :=-  sin       — . 


Inserting  the  value  of  r\  =  r  sin  fi, 


F  = 


R  sin  20     J3 
2         '  sin  ft 


where  0  is  the  angle  of  repose  as  given  by  the  table  page  224. 
The  moment  of  the  friction  with  reference  to  the  axis  is  then 


M  = 


Rr  sin  20 


sin  ft 


(2) 


•(2) 


230  ST/1T1CS.    GENERAL  PRINCIPLES.  [CHAP.  III. 

If  the  axle  is  well  greased,  the  angle  of  repose  <f>  is  very  small,  and  we  may  take 

sin  20  =  2  sin  <p,     also     n  =  tan  <p  =  sin  <p. 
In  the  practical  case  of  a  well-greased  axle,  then,  we  have 

F  -HR  .  _ ft    ,      M  -  MXr  -f-* , 

am  ft  sin  ft 

where  n  is  given  by  the  table  page  224. 

If  the  angle  ft  is  small,  we  may  take  ft  =  sin  ft,  and  then  /"and  jl/  are  the  same  as  in  the  two  preceding 
cases, 

F  =  /M',        M  —  vRr. 

4.  FRICTION  WHEELS. — By   the  use  of  friction  wheels  instead  of  bearing  blocks,  the 
friction  of  an  axle  can  be  greatly  diminished. 

Thus  let  the  axle  AC  rest  upon  the  circumferences  of  the  friction 
wheels  ACX  and  BC.2,  touching  them  at  the  points  A  and  B.  The 
vertical  pressure  R  on  the  axle  C  causes  the  pressures  Nlt  N2  at  A  and 

Let  the  angle  A  CB  =  ft.      Then 


If  the  axles  of  the  friction  wheels  are  well  greased,  then,  as  we  have  seen,  the  least 
friction  may  be  written 


where  n  is  given  by  the  table  page  224. 

If  the  radius  of  the  axles  of  the  friction  wheels  is  r,  the  moment  of  the  friction  is 


Fr  = 


nRr 
cos  ft' 


The  moment  of  the  friction  at  the  points  A  and  B  must  be  the  same.      If  we  call  this 
lt  we  have,  if  the  radius  of  the  friction  wheels  is  a, 


,,  -       r  -      r     i* 

F.a  =  Fr,     or     F,  =  -  F  =  -  . 

a  a    cos  ft 


By  making  ft  small,  we  can  take  cos  ft  =  I,  and  have 


By  taking  a  large  with  respect  to  r,  we  may  thus  make  the  friction  Fl 
very  small.  If  the  axle  C  rests  on  bearings,  its  least  friction  is  j*R,  as  we 
have  seen. 

If  we  have  a  single  friction  wheel  C^A,  then  ft  =  o,  anci  we  have 
accurately 


CHAP.  III.] 


STATIC  FRICTION  FOR   CORDS  AND   CHAINS. 


231 


Static  Friction  of  Cords  and  Chains. — Let  a  perfectly   flexible  cord  stretched  by  a 
weight  (2  be  laid  over  the  edge  C  of  a  rigid  body  ABO,  Fig.   I. 

Let   the    force   at   the   other   end    of    the  FIG.  i.  FIG.  2 

cord  be  P,  and  the  angle  of  deviation  DCP  — 
AOB  =  a. 


Draw    CT  making    the    angle    TCP  —  -, 

and  CN  perpendicular  to  CT.  Then,  when 
motion  is  about  to  begin,  the  resultant  R  of  P 
and  Q  makes  the  angle  of  repose  0  with  CN. 

If  the  weight  Q  is  about  to  sink,  the 
friction  F  acts  opposed  to  the  motion,  and  we 
have 


We  have  then,  from  Fig.  2, 

F:  2(2  sin  -^  :  :  sin  0:  sin  [^90  —  (0  —  jjj, 


or 


2(2  sin  —  sin  0 


2(2  sin  —  sin  0 


COS      0 


a  a 

cos  0  cos \-  sin  0  sin  — 


90-f 


Dividing  numerator  and  denominator  by  cos  0,  we  have,  since  tan  0  =  ft  =  coefficient 
of  static  sliding  friction,  for  the  friction  Fl  when  the  weight  Q  is  about  to  sink 


ct 

2//<2  sin  —               2^ 

lOt       - 
ly  en  2 

a                    a 

cos  1-  n  sin  —        I  -) 

-  V  tan  - 

When  the  weight  Q  is  fust  about  to  rise  we  have 

P  =  Q  +  F,      or     Q=  P-  F, 


and  hence 


tan 


771    


a 
—  /i  tan  — 


In  the  first  case,  then,  when  the  weight  Q  is  about  to  sink, 

(  «\ 

Q(i  -  yu  tan  -J 


a 

1  -j-  ,w  tan 


(0 


(2) 


(3) 


«3a  STATICS.    GENERAL  PRINCIPLES.  [CHAP.  III. 

and  in  the  second  case,  when  the  weight  Q  is  about  to  rise, 

Q(I  +  11  tan 


(4) 


—  n  tan  - 


If  the  cord  passes  over  several  edges,  the  force  Pl  can  be  calculated  by  repeated  appli 
cation  of  these  formulas. 

Thus  let  the  number  of  edges  be  n  and  the  deviation  at  each 
edge  be  the  same  and  equal  to  a.  When  the  weight  Q  \sjust  about 
to  sink,  the  tension  of  the  first  portion  of  the  cord  is,  from  (3), 

r"'  / 

Q\i  —  n  tan  - 
i  -j-  /*  tan  |- 


p  _  __ 


-  »  tan  ~ 


a\z 
-J 

That  of  the  last  is 


tan  -  i  tan  - 


l  -  »  tan  j 

(5) 


+  w  tan  - 

If  the  'weight   (2  's  y«J^  about  to  rise,   we  have  simply  to  interchange  P  and   <2  and 
we  have 


/>,  =  -  —  ..........     (6) 


In  the  first  case,  when  the  weight  is  about  to  sink,  we  have  for  the  friction 

/        (,  -  „  tan  ")"\ 

*,-fl-W«-         -4il (;) 


If  the  weight  is  about  to  rise, 


r-r.-Q--      —.->       ......(«) 

\('-f>  tan  -j          / 


CHAP.  III.] 


STATIC  FRICTION  FOR   CORDS   AND   CHAINS. 


233 


Formulas  (5),  (6),  (7)  and  (8)  are  also  applicable  to  the  case  of  a  chain  composed  of 
links  which  is    passed   round   a  cylindrical   surface,    where   n  is    the 
number  of  links  in  contact.      If  the  length  of  each  link  is  AB  =  /, 
and  the  distance  CA  of  the  axis  A  of  a  link  from  the  centre  C  is  r, 
we  have  for  the  angle  of  deviation  DBL  =  ACB  —  a 

A 


. 

sin  —  =  — , 
2       2r 


[If  a  flexible  cord  lies  in  contact  with  a  rough  surface,  let  ACB  =  a  be  the  arc        rs_, 
of  contact. 

If  7'  is  the  tension  at  any  point  of  contact  D  for  the  indefinitely  small  portion  of  the  cord  Dd,  the  fric- 
tion at  this  point  is  dT.  Let  the  indefinitely  small  angle  DCa  be  da. 
Then,  from  equation  (i),  page  231, 


dT- 


da 
2u  T  tan  — 

i  +ju  tan  — 


But  since  da  is  indefinitely  small,  we  may  take  the  arc  equal  to  the 
tangent  and  disregard  u  tan  —  with  reference  to  i.     We  have  then 


dT 

—  =  nda. . 

Integrating  between  the  limits  a  =  o  and  a,  we  have,  since  for  a  =  o,  T  =  Q,  and  for  a  =  a,  T  =  P, 

p 
logn  P  =  //a  +  logn  Q,     or     logn  —  =  jua. 

We  have  then,  when  motion  in  the  direction  of  P  just  begins, 

P  =      ^a, 


where  e  =  2.3026  =  base  of  Naperian  system  of  logarithms. 

When  motion  in  the  direction  of  <2'just  begins,  we  have,  by  interchanging  P  and  Q, 


Also,  inversely, 


a=    2-3Q26(log  /'-log  2) 


(9) 

(10) 
(n) 


where  common  logarithms  are  taken. 

If  the  arc  a  of  the  cord  is  given  in  degrees  instead  of  radians,  we  must  substitute  «=  —^r-0it.       If    the 

surface  is  cylindrical  and  the  number  of  coils  n  of  the  rope  is  given,  we  have  a  =  2icn. 

We  see  from  (9)  and  (10)  that  the  friction  of  a  cord,  F  =  P —  Q  or  F  •=•  Q  —  P,  upon  a  surface  does  not 
depend  at  all  upon  the  radius  of  curvature,  but  only  upon  the  arc  of  contact  a,  or  upon  the  number  of  coils, 
2Ttn,  if  the  surface  is  cylindrical. 

If  we  take  u  =  -,  we  have  for  a  cylindrical  surface  : 

for  -  coils,/7  —  1.692; 
4 
i 

2 
"2        "        P  =  65.94^; 

"4      "      P  =  4343-56(2. 
The  friction  can  thus  be  increased  to  any  amount  by  increasing  the  number  of  coils.] 


'34 


ST4T1CS.    GENERAL  PRINCIPLES. 


[CHAP.  III. 


Rigidity  of  Ropes.  —  When  a  rope  is  perfectly  flexible  it  offers  no  resistance  to  bending. 
iVhen  a  rope  is  not  perfectly  flexible  it  offers  a  resistance  by  reason  of  its  rigidity  when 
wound  on  to  a  drum,  pulley  or  axle,  though  none  is  offered  when  il  is 
wound  off.     Thus  let  a  rope  whose  tension  is  T  be  on  the  point  of 
being  wound  on  to  a  pulley. 

Let  a  —  AC  =  BC  be  the  radius  of  the  pulley,  and  t  the  thick- 
ness of  the  rope.     Then  the  lever-arm  of  the  axis  of  the  rope  on  the 

is  Cb  =  a  -j  --  . 

The  distance  Ac  from  the  pulley  to  the  rope  on  the  on  side  will 
T+T'    depend  on  the  kind  of  rope  and  will  be  less  as  is  greater.      Thus  for 
hemp  ropes  we  can  put 


where  cl  is  a  constant  to  be  determined  by  experiment  for  the  kind  of  rope;  and  iorwire  ropes 


AC  — 


that  is,  Ac  increases  with  the  lever-arm  a  -\  --  and  decreases  as  T  increases. 

It  is  also  evident  that  those  fibres  farthest  out  on  the  on  side  are  stretched  more  than 
those  nearer  the  pulley.  The  resultant  tension  T  will  therefore  act  further  from  the  pulley 
than  the  central  axis  of  the  rope.  We  denote  the  distance  of  T  from  the  central  axis  by  CY 

Let  the  tension  along  the  central  axis  on  the  ^side  be  T  -\-  T  .  Then  we  have  for 
equilibrium,  for  hemp  ropes, 


or 


and  for  wire  ropes, 


or 


cnT 


We  have  then 


Ct=(T+T')Cd,    or     Cc  =    i  -f  ^}Cb. 


The  rope  can  be  considered,  then,  as  without  rigidity  if  we  increase  the  lever 

T 


.     .     (3) 
arm  of  the 


tension  on  the  on  side  by  the  amount 


T' 


CHAP.   IK  J  RIGIDITY  OF  ROPES. 

Hemp  Ropes.  —  For  tarred  hemp  ropes  experiment  gives 


235 


IOO  +O.222T 

1  —  -  --  pounds, 


where  T  is  to  be  taken  in  pounds,  and  a  and  /  in  inches. 
For  new  hemp  ropes,  untarred, 

4 +0.0645  7  T 
T  =  = —  —  pounds, 


where  T  is  to  be  taken  in  pounds,  and  a  and  t  in  inches. 
Wire  Ropes,  —  For  wire  ropes  we  have 


=  ,.  08  +  pounds, 


where  T  is  to  be  taken  in  pounds,  and  a  and  t  in  inches. 

Static  Rolling  Friction.  —  Let  ACB  be  a  roller  resting  on  a  plane  surface. 


By  reason 


of  the  pressure  N  of  the  roller  on  the  plane,  the  roller  is  compressed. 
Let  a  force  F  be  applied  at  the  centre  C  parallel  to  the  plane.  When 
the  resultant  R  of  /''and  N  just  passes  through  the  edge  D  of  the  base, 
rolling  begins  and  the  force  F  is  equal  and  opposite  to  the  friction. 

Let  the  distance  AD  —  d.      Then,  when   rolling  is  about  to  begin, 
the  angle  A  CD  is  the  angle  of  repose   0.      Let  r  be  the  radius.     Since 

the  compression  is  small  compared  to  the  radius,  we  have  tan  0  =  —  =  jt 
=  coefficient  of  static  rolling  friction.      Hence  for  equilibrium  Fr  =  Nd,  or 


The  distance  d  depends  on  the  materials  in  contact. 

The  theory  of  rolling  friction  is  not  yet  well  established  and  but  few  experiments  upon 
it  have  been  made. 

In  all  practical  cases  of  rolling  we  usually  have  to  do  with  axle  friction,  which  has 
already  been  discussed  (page  228). 

Examples  —  (i)  A  body  of  mass  m  rests  upon  a  rough   inclined  plane  which  makes  an  angle  a.  with  the 
horizontal  and  is  acted  ubon  by  a  force  P  which  makes  the  angle  ft  with  the 
plane.     Find  the  conditions  for  equilibrium.     (For  .smooth  plane  see  ex.  (8), 
page  2 1 8.) 

ANS.  Consider  the  body  as  a  particle  at  any  point  O.  Let  the  angle 
POD  —  ft  be  positive  when  above  the  plane,  and  negative  when  below. 

i.  BODY  ON  THE  POINT  OF  MOTION  UP  THE  PLANE. — In  this  case  we 
have  the  component  of  P  parallel  to  the  plane  acting  up  the  plane,  and  the 
friction  uN acting  down.  We  have,  then,  P,  N,  m  and  uNin  equilibrium. 
If  we  put  the  algebraic  sum  of  forces  along  the  plane  and  perpendicular  to 
the  plane  equal  to  zero,  we  have  in  gravitation  measure 


236  57//77C5.    GENERAL  PRINCIPLES.  [CHAP.  111. 


P  cos/5  -,„  sin  a-uN  =  o,   or    j,  *  + 


cos  p 
Psin  ft  +  N  —m  cos  a  =  o,     or    N  =  m  cos  a  —  P  sin  # 

where  u  is  the  coefficient  of  friction. 
From  these  equations  we  have 


_  sin  a  +  u  cos  a        _  sin  (a  +  0  ) 
~  cos  ft  +  //  sin  fl'm  ~  cos  (/*  -  <f>)m' 


where  0  5s  the  angle  of  repose  whose  tangent  is  equal  to//.  If//  =  o,  there  is  no  friction  and  we  have 
equation  (2),  example  (8),  page  219.  If  ft  =  90°  -  a.  P  =  m  and  N  =  o.  For  any  greater  value  of  positive 
ft,  N  is  negative  and  no  equilibrium  is  possible.  For  negative  ft  we  must  have  ft  less  than  90  —  0.  If 
negative  ft  is  greater  than  this,  we  have  N  negative  and  no  equilibrium  is  possible.  Equation  (i)  holds  good, 
then,  for  all  values  of  ft  between  -|-  (90  —  a)  and  —  (90  —  0).  The  force  P  is  a  minimum  when  cos  (ft  —  0) 
is  a  maximum  or  when  ft  =  <f>.  This  minimum  value  of  P  is  then 

P  =  m  sin  (a  -t-  <p) 

2.  BODY  ON  THE  POINT  OF  MOTION  DOWN  THE  PLANE—  a  GREATER  THAN  0.—  In  this  case  we  have 
the  friction  fiN  acting  up  the  plane  and  the  component  of  P  up  the  plane.     Hence 


or     P 


. 
cos  p 

P  sin  ft  +  N  —  m  cos  a  =  o,    or     N  —  m  cos  a  —  P  sin  ft. 
From  these  e  quations  we  have 


_    sin  a  —  u  cos  a        _  sin  (a  —  0) 

~  ~ 


cos  ft  —  u  sin  ft          ~  cos 


Here,  again,  we  see  that  equation  (2)  holds  for  all  values  of  ft  between  +  (90  —  a)  and  —  (90  —  0).     Outside 

of  these  limits  no  equilibrium  is  possible.     The  force  P  is  a  minimum  when  ft  =  —  0.  This  minimum  value 
of  /'  is  then 

P  =  m  sin  (a  —  0). 

3.  Bot)Y  ON  THE  POINT  OF  MOTION  DOWN  THE  PLANE  —  a  LESS  THAN  0.  —In  this  case  we  have  the 
friction  uN  acting  up  the  plane,  and  the  component  of  P  down  the  plane.      Hence 

nN  —  m  sin  a 
—  P  cos  ft  —  m  sin  a  +  nN  =  o,     or     P  =  — 

cos  ft 

P  sin  ft  +  N  —  m  cos  a  =  o,  N  =  m  cos  a  —  P  sin  ft. 

From  these  equations  we  have 

«cnsa-sina  in- 


cos  ft  +  u  sin  ft 

We  cannot  have  4-  ft  greater  than  90°  or  the  component  of  P  will  not  act  down  the  plane.  If  —  ft  - 
90  —  0,  P  =  m  and  N  =  o.  For  any  greater  value  of  negative  ft,  N  is  negative  and  no  equilibrium  is 
possible,  liquation  (3)  holds  for  all  values  of  ft  between  +  90  and  —  (90  —  0). 

The  force  P  is  a  minimum  when  ft  =  0.     This  minimum  value  of  P  is  then 

P  =  m  sin  (0  —  a). 

(2)  Find  the  conditions  of  equilibrium  for  a  rough  screw.     (For  smooth  screw  see  example  (i),  page  217). 
ANS.  If  N  is  the  normal  pressure  on  the  thread,  we  have,  by  resolving  O  normally  and  horizontally, 

N  =  -Q  —    The  friction  is  then  uN  =  —Q-. 
cos  a  cos  a 

If  P  has  a  virtual  displacement  of  0  radians,  Q  is  raised  a  distance  —  ,  the  friction  moves  through 
—  —  and  we  have,  by  virtual  work, 


CHAP.  III.] 


STATIC  FRICTION— EXAMPLES. 


237 


Hence,  since  —  =  tan  a,       /*  =  tan 

" 


If  we  neglect  friction  we  have  /<  =  o,  and  P  =  ^>,  which  is  the  same  result  as  in  ex.  (i),  page  217. 

27T/ 

(3)  Find  the  conditions  for  equilibrium  for  the  differential  screw  given  in  example  (2),  page  217,  taking 
ftiction  into  account. 

ANS.    />  =  g 


(4)  Find  the  conditions  of  equilibrium  for  the  wedge  given  in  example  (j),  page  217,  taking  friction  into 

account. 


.  P  = 


in  f  ±  *  co.5)  -  ^£  .in  (f  ±  ,). 


where  the  (+)  sign  is  taken  for  wedge  on  point  of  entering,  and  the  (—  )  sign  on  point  of  sliding  out. 

ct 
If  P  is  between  these  values,  the  wedge  is  neither  on  the  point  of  entering  nor  sliding  out.     If  ~  =  <t>, 

there  is  no  force  required  to  keep  the  wedge  from  sliding  out.     The  angle  of  the  wedge  should  not,  then, 
exceed  20, 

(5)  A  rod  rests  with  its  ends  against  a  rough  vertical  and  horizontal  plane.  The  mass  of  the  rod  is  m, 
and  its  weight  acts  at  its  middle  point.  Find  the  conditions  of  equilibrium. 

ANS.  Let  9  be  the  angle  with  the  horizontal,  and  JVi  ,  N,  the  normal  pressures  on  the  horizontal  and 
vertical  planes  respectively.  Then 


tan  9  =  cot.  20, 


i  =  m  cos2  0,         N*  =  m  sin  0  cos  0. 


(6)  In  a  wheel  and  axle  the  radius  of  the  wheel  is  a,  and  of  the  axle  b.  Find  the  conditions  for  equilibrium, 
taking  into  account  friction  and  the  rigidity  of  the  rope,  when  a  mass  P  hung  from  the  wheel  just  balances  a 
mass  Q  hung  from  the  axle.  (Without  friction  and  rigidity  see  example  (13),  page  214.) 

ANS.  We  have  seen  (page  228)  that  for  well-greased  axle  and  small  surface  of 
contact  we  can  take  in  all  cases  of  axle  friction  the  friction  F  =  nR  =  u(P  -j-  Q), 
where  n  is  the  coefficient  of  static  sliding  friction. 

Let  the  radius  of  the  journal  be  r,  and  let  t  be  the  thickness  of  the  rope. 

Then  when  P  is  just  about  to  fall,  we  have  (page  234)  for  the  lever-arm  of  Q, 


—  \ib  H J,  and  hence  for  equilibrium 


p  = 


where  (page  235) 


for  hemp  ropes  T'  =  c-l±^- 


for  wire  ropes  T'  =  c\  H *—j 

2 


the  values  of  c\  and  c?  being  given  on  page  235. 


238  STATICS.    GENERAL  PRINCIPLES.  [CHAP.  Ill 

When  Q  is  just  about  to  fall,  we  have  (page  234)  for  the  lever-arm  of  P,  ^i  +  -^J  \a  +  -  J,  and  hence 


•+1 


where  (page  235) 


for  hemp  ropes  T'  =  —   —  —  ; 


for  wire  ropes  T'  =  c\  -\  —          ; 


the  values  of  ct  and  ct  being  given  on  page  235. 

For  values  of  P  less  than  the  first  and  greater  than  the  second  we  have  non-limiting  equilibrium,  and 
the  wheel  and  axle  is  not  upon  the  point  of  rotating  in  either  direction. 

>+'- 

If  we  neglect  friction  and  rigidity,  we  have  P  =  --  -Q,    or,     neglecting    the    thickness  of  the   rope 

«  +  i 

P  =  -Q,  as  in  ex.  (13),  page  214. 

If  b  =  a,  we  have  the  case  of  the  single  pulley. 

For  partially  worn  bearing  (page  228)  we  can  put  more  accurately 

sin  <p  in  place  of  u, 

where  <p  is  the  angle  of  repose. 

For  triangular  bearing  (page  228)  we  can  put 


where  ft  is  the  half  angle  of  the  bearing. 
For  new  bearing  (page  229)  we  can  put 

8  sin  20  . 

-        ~  in  place  of  n, 

2  sin  ft 

where  ft  is  the  half  angle  of  contact. 

(7)  In  the  single  movable  pulley  find  the  relation  between  the  force  P  and  the  mass  Q  for  equilibrium, 
taking  into  account  friction  and  the  rigidity  of  the  rope.     (Without  friction  and  rigidity  see  ex.  (5),  page  218.) 
ANS.  Let  r  be  the  radius  of  the  axle  of  each  pulley,  a  the  radius  of  each  pulley,  /  the  thickness  or  rope 
H  the  coefficient  of  static  sliding  friction,  and  c\,  c%  as  given  on  page  235. 
For  convenience  of  notation  let 

u  =  a  H  ---  \-ur-\-Ct,    tv  =  a  -f-  -  —  Mr. 
P 

Then  from  the  preceding  example,  making  b  =«  a,  we  have,  when  P  in  just  about 
to  fall,  for  hemp  ropes 


where  T\  is  the  tension  in  the  first  rope  as  shown  in  the  figure. 
We  have  in  the  same  way 


CHAP.  III.] 
We  have  also 

Eliminating  T^  and  Tt,  we  have 


STATIC  FRICTION— EXAMPLES. 


=  Q. 


239 


p_ 


(w 


w(w  -\-  u) 

In  the  same  way  we  find  when  P  is  on  the  point  of  rising 
P=(^^- 


—  cl) 


(u>  -j-  u)(u  —  2nr) 

For  values  of  P  less  than  the  first  and  greater  than  the  second  we  have  non-limiting  equilibrium  and  P 
is  not  on  the  point  of  falling  or  rising. 


For  wire  ropes  we   have  only  to  substitute  Ci\ a  +  —  J  in  place  of  c\. 


For  partially  worn   bearing  or  new  bearing  we  can  replace  /*  by  the  values  given  in  the   preceding 
example. 

If  we  neglect  friction  and  rigidity,  we  have  P  =  —  as  in  ex.  (5),  page  218. 

(8)  In  the  system  of  pulleys  shown  find  the  relation  between  the  force  P  and  the  mass  Q  for  equilibrium, 
taking   into   account  friction    and  rigidity   of  the  rope.     (Without  friction  and 
rigidity  see  ex.  (6),  page  218.) 

ANS.  Let  m  be  the  mass  of  each  movable  pulley,  and  «the  number  of  movable 
pulleys.  Let  r  be  the  radius  of  the  axle  of  each  pulley,  a  the  radius  of  each 
pulley,  u  the  coefficient  of  static  sliding  friction,  /  the  thickness  of  the  rope,  and 
Ci  and  c*  as  given  on  page  235. 

For  convenience  of  notation  let 

«  =  a  H \-  vr  +  Ci\      w  =  a  -\ jur; 

v  =  u  +  w  =  2a  4-  /  +  c-i. 

Then,  from  the  preceding  example,  we  have,  when  P  is  just  about  to  fall,  for 
hemp  ropes 

T\  =  : 1-  ~  • 


rj  =  "(7*  +  m\      £1 

S 

=  u(  T,  +  m)      c2 

V  V 

and  so  on.     Inserting  the  values  of  T\  and  T<>,  we  have  in  general 


T  =          4- 
vn   "* 


But  from  the  preceding  example  we  have 


Hence,  since  u  —  v  =  —  w, 


P  = 


uTn   .  d 


240 


STATICS.     GENERAL  PRINCIPLES. 


[CHAP.  III. 


For  wire  ropes  we  have  only  to  substitute  c\  \a  -j-  -1  in  place  of  c\. 

For  partially  worn  bearing  or  new  bearing  we  replace  u  by  the  values  given  in  ex.  (6). 


reduces  to  P 


Q  +  (2"  —  \)m 


/A 
\a  -f  -j,  u 


.  whiclj  is  the  same  result  as  given  in  ex.  (6),  page  218. 


(9)  In  the  system  of  pulleys  shown  find  the  relation  between  the  force  P  and  the  mass  Q  for  equilibrium, 
taking  into  account  friction  and  the  rigidity  of  the  ropes.     (Without  friction  and  rigidity  see  ex.  (7),  page  218.) 
ANS.  Let  m  be  the  mass  of  the  lower  block.  ;ind  «  the  number  of  ropes  coming  from  the  lower  block. 
Let  r  be  the  radius  of  the  axle  of  each  pulley,  n  the  coefficient  of  static  sliding  friction, 
/  the  thickness  of  the  rope,  and  d  and  ft  as  given  on  page  235. 
Let  a  be  the  mean  radius  of  the  pulleys. 
For  convenience  of  notation  let 

u  =  a  +  -   +  ur  +  ft,        w  =  a  H ur. 

Then  we  have  for  hemp  ropes,  when  P  is  about  to  descend, 


For  wire  ropes  we  have  only  to  substitute  c\(a-\  —  ]  in  place  of  c\. 

For  partially  worn  bearing  or  new  bearing  we  replace  u  by  the    values  given  in 
ex.  (6). 

If  we  neglect  friction  and  rigidity,  we  have  u=w  and  <:,  =  o.     The  value  of  P 

reduces  then  to  P  =  -  ;   but  if  we  divide  numerator  and  denominator  by  u  —  w  and 


then  make  u  =  w,  we  have 


P= 


which  is  the  same  result  as  given  in  ex.  (7),  page  218. 

(10)  In  the  system  of  pulleys  shown  find  the  relation  between  the  force  P  and  the 
mass  Q  for  equilibrium,  taking   into   account  friction   and  the  rigidity  of  the  ropes. 
ANS.   Let  m  be  the  mass  of  each  pulley  and  n  the  number  of  pulleys.     Letr  be  the 
radius  of  the  axle  of  each   pulley,  n  the   coefficient  of  static 
sliding  friction,  /  the  thickness  of  the  rope  and  c\ ,  c*  as  given 
on  page  235. 

Let  a  be  the  radius  of  each  pulley,  and  for  convenience  of  notation  let 

u  =  a  H \-ur  +  ft,        w  =  a  ^ ur. 

Then  we  have,  when  P  is  about  to  descend,  for  hemp  ropes 


For  wire  ropes  we  have  only  to  substitute  fJa  +  -  J  in  place  of  c\. 

For  partially  worn  bearing  or  new  bearing  we  replace  n  by  the  values  given  in  ex.  (6). 
If  we  neglect  friction  and  rigidity,  we  have  u  =  w  and  c\  =  o,  and 


Q  +  nm  —  (2*  —  \)m 


CHAP.  III.] 


STATIC  FRICTION-EXAMPLES. 


241 


(i  i)  In  the  differential  piilley  shown  in  the  figure  an  endless  chain  passes  over  a  fixed  pulley  A,  then  under 
a  movable  pitlley  to  which  the  mass  Q  is  attached,  and  then  over  another  fixed  pulley  B,  a  little  smaller  but 
co-axial  with  A.  The  two  pulleys  A  and  B  are  in  one  piece  and  obliged  to  turn  together  through  the  same  angle. 
The  two  ends  of  the  chain  are  joined  so  as  to  form  a  loop.  The  force  P  is  applied  to  the'  right-hand  portion  of 
the  loop.  To  prevent  the  chain  from  slipping,  there  are  cavities  in  the  circumferences  of  the  upper  pulleys 
into  which  the  links  of  the  chain  fit.  Find  the  relation  of  P  to  Qfor  equilibrium,  taking  into  account  friction. 

ANS.  Let  a  be  the  radius  of  the  pulley  A,  and  b  the  radius  of  the  pulley  B,  m  the  mass  of 
each  pulley  above  and  below,  r  the  radius  of  each  axle,  and  fi  the  coefficient  of  friction. 
Since  the  pulley  is  worked  by  a  chain,  we  can  disregard  rigidity  and  have  only  friction  to 
take  into  account.  Let  T  be  the  tension  of  the  chain.  Then  for  equilibrium 

Q  +  ™ 


2T=Q  +  m,     or    T  = 

Let  -Fbe  the  friction.     Then  taking  moments  about  C,  we  have  for  equilibrium 

-  Pa  +   Ta  —  Tb  +  Fr  =  o. 
The  pressure  on  each  axle  is  Q  +  2m  +  P.     Therefore  the  friction  is 

F  =  2/*(Q  +  2m  +  P). 
Substituting  this  value  of  /'"and  the  value  of  T  in  the  preceding  equation,  we  have 


p  _ 


2 


+  2ur(Q  +  2m) 


—  2/j.r 


For  partially  worn  bearing,  or  for  new  bearing,  we  can  replace  n  by  the  values  given  in  ex.  (6).     If  we 
neglect  friction  and  the  mass  of  the  pulleys,  we  have  P  =  ^      ~ — -. 


KINETICS  OF  A  PARTICLE. 


CHAPTER  I. 

DEFLECTING    FORCE. 

Kinetics. — That  portion  of  Dynamics  which  treats  of  forces  with  reference  to  the 
change  of  motion  of  bodies  caused  by  them  is  called  KINETICS. 

Kinetics  of  a  Particle. — We  have  seen  (page  190)  that  when  a  rigid  body  is  acted  upon 
by  any  number  of  forces,  the  motion  of  the  centre  of  mass  is  the  same  as  if  it  were  a  par- 
ticle of  mass  equal  to  that  of  the  body,  all  the  forces  acting  upon  the  body  being  transferred 
without  change  in  direction  or  intensity  to  this  particle. 

When,  therefore,  we  consider  only  the  motion  of  translation  of  a  body  without  reference 
to  its  rotation,  we  can  always  consider  the  body  as  a  particle  of  equal  mass  concentrated  at 
the  centre  of  mass. 

It  will  therefore  be  convenient  to  first  consider  the  KINETICS  OF  A  PARTICLE. 

Impressed  and  Effective  Force  on  a  Particle. — Let  F  be  the  resultant  of  any  number 
of  forces  Flt  F2,  F3,  etc.,  acting  upon  a  particle.  These  forces  we  have  called  impressed 
forces  (page  169),  and  the  resultant  F  is  the  resultant  impressed  force. 

Let  m  be  the  mass  of  the  particle,  and /"its  acceleration.  Then /"will  be  in  the  direc- 
tion of  F,  and  we  have  from  equation  (2),  page  170, 

F  =  mf. (i) 

We  call  the  quantity  w/the  effective  force  of  the  particle.      We  see  from  equation  (i),  then, 
that  the  resultant  impressed  force  is  equal  to  the  effective  force. 

D'Alembert's  Principle  Applied  to  a  Particle. — We  can  write  equation  (i) 

F-mf=o (2) 

That  is,  if  we  reverse  the  direction  of  the  effective  force  it  will  hold  the  resultaTit  impressed 
force  in  equilibrium. 

Hence  the  impressed  forces  acting  on  a  particle  and  the  reversed  effective  force  of  the 
particle  constitute  a  system  of  concurring  forces  in  equilibrium. 

This  is  D'Alembert's  principle  as  applied  'to  a  particle,  it  reduces  any  kinetic  prob- 
lem to  one  of  equilibrium  between  actual  ("  impressed"}  forces  and  a  fictitious  ("  reversed 
effective  ")  force. 

Examples.— (I)  Let  a  mass  m  be  moved  on  a  smooth  horizontal  plane  oy  i  rope  which  passes  over  the  edge  of 
the  plane  on  a  pulley  and  has  a  mass  P  hung  at  its  end.  Disregarding  all  friction  ami  mass  of  pulley  and 
rope  and  rigidity  of  rope,  find  the  acceleration. 

ANS.  This  example  has  already  been  solved  (see  example  (12).  pape  177).  We  solve  it  here  by  D'Alem- 
bert's principle. 

242 


CHAP.  I.] 


DEFLECTING  FORCE. 


243 


The  impressed  force  on  m  is  the  tension  of  the  rope  P(g  —f).    The  effective  force  on  m  is  m/.    Revers- 
ing this,  we  have,  by  D'Alembert's  principle,  for  equilibrium 


P(g~  /)  -m/=o,     or    /  = 


P+  m* 


(2)  Two  masses  P  and  Q,  P  being  the  greater,  are  hung  by  means  of  a 
rope  over  a  pulley.  Disregarding  friction  and  the  mass  of  the  pulley  and  rope, 
find  the  acceleration. 

ANS.  This  example  has  already  been  solved  (see  example  (13),  page  178).    P(9r-/j 
We  solve  it  here  by  D'Alembert's  principle. 

The  impressed  forces  on  Q  are  Qg  downwards  and  the  upward  tension 
of  the  string  P{g — /).  The  effective  force  of  Q  is  Qf  upward.  Reversing 
we  have,  by  D'Alembert's  principle,  for*  equilibrium 


P  +  Q  ' 


Deflecting  Force. — We  have  seen  (page  77)  that  when  a  particle  moves  in  a  curve 
whose  radius  of  curvature  is  p  with  a  velocity  v  at  any  instant,  there  must  be  a  central  accel- 
eration fp  always  directed  toivards  the  centre  of  curvature  and  given  by 


This  central  acceleration  causes  no  change  of  speed,  but  only  change  of  direction  of  the 
velocity. 

If  m  is  the  mass  of  the  particle,  we  must  then  have  a  central  force  always  directed 
towards  the  centre  of  curvature  given  by 

mv* 


This,  then,  is  the  force  which  causes  the  particle  to  move  in  a  curve.  If  this  force  did 
not  act,  the  particle  would  move  in  a  straight  line.  We  therefore  call  this  force  the 
DEFLECTING  FORCE,  and  denote  it  by  Ty 

Since  the  centre  of  mass  of  a.  body  moves  as  if  the  whole  mass  m  were  concentrated  at 
its  centre  of  mass,  we  have,  when  the  centre  of  mass  of  a  body  moves  in  a  curve  with 
velocity  v,  the  deflecting  force  for  the  entire  body 


(i) 


where  p  is  the  distance  from  the  centre  of  curvature  to  the  centre  of  mass,   and  GO  is  the 
angular  velocity,  so  that  pco  =  v. 

If  the  path  is  a  circle,  p  is  constant  and  equal  to  the  radius  r  and  we  have 


F  ==  --  = 


=  mvco. 


If    there   is   no  tangential   acceleration  /„    there   will    be   no  change  of  speed  and    v 
will  be  constant  in  magnitude,  changing  only  in  direction. 

In  such  case,  if  T  is  the  time  of  revolution  in  the  circle,  we  have 


KINETICS  Of  A  P ARTICLE 


[CHAP.  1. 


27T  271T 

=  -~>      or     v  =  -~-t 


and 


(3) 


All  equations  give  the  force  in  poundals.     For  gravitation  units  divide  by  g  (page  171). 

Simple  Conical  Pendulum.  —  The  simple  conical  pendulum  consists  of  a  particle  of  mass 
m  attached  to  a  fixed  point  P  by  a  massless  inextensible  string  of 
length  /,  and  moving  with  uniform  speed  v  in  a  circular  path 
about  a  vertical  axis  through  the  fixed  point. 

In  this  case  the  particle  is  acted  upon  by  two  forces,  its 
weight  mg  vertically  downwards  and  the  stress  5  of  the  string 
directed  towards  the  fixed  point  P.  If  the  particle  moves  with 
uniform  speed  v  in  the  circle  whose  radius  is  r  —  I  sin  6,  where  0 
is  the  inclination  of  the  string  to  the  vertical,  the  vertical  com- 
ponent 5  cos  ff  of  the  stress  S  must  balance  the  weight  mg,  and 

imfl 

the  horizontal  component  S  sin  6  of  the  stress  5  must   be  equal  to  the  deflecting  force  — 

necessary  to  make  the  particle  move  in  a  circle  with  uniform  speed.      We  have  then 


—  mg  -j-  5  cos  0  =  o,     or     5  cos  6  =  mg, 
and  from  equation  (2),  page  243, 

mvz         mv* 

=  mr      ~ 


(i) 


(2) 


where  co  is  the  angular  speed. 

We  can  find  5  from  either  (i)  or  (2).     Squaring  (i)  and  (2)  and  adding,  we  have  also 


5  = 


Also  dividing  (2)  by  (i),  we  have 


=  m 


(3) 


For  5  in  gravitation  units  we  must  divide  (3)  by  g  as  usual. 

Let  //  be  the  distance  of  the  fixed  point  P  above  the  plane  of  motion,  or  the  height  of 
the  pendulum.     Then,  since  h  tan  0  =  r,  we  have,   from  (4), 


(5) 


If,  then,  GO  is  the  angular  speed  of  the  particle  about  the  centre  0,  roo  =  v,  and,  from  (5), 
V  £.      If  /  is  the  time  of  a  revolution, 


This  is  the  same  as  the  time  of  oscillation  of  a  simple  pendulum  of  length  h  (page  138). 


245 


CHAP.  I.]  DEFLECTING  FORCE. 

COR.  i. — If  0  is  indefinitely  small,  //  and  /are  equal  and 

<£•' 
and  we  have  the  case  of  the  simple  pendulum. 

COR.  2. — Since  GO  =  |/±  ,  we  see  that  the  greater  the  angular  ve- 
locity the  less  h,  and,  as  /  is  constant,  the  greater  r.     This  fact  is 
taken  advantage  of  in  the  steam-engine  governor.      As  the  piston  speed 
increases,  the  spindle  PO  revolves  more  rapidly,  the  balls  separate  and  the  slide  at  B  rises 
and  by  means  of  levers  acts  upon  the  valves  of  the  engine  to  diminish  the  supply  of  steam. 

Centrifugal  Force. — Let  us  now  solve  the  preceding  problem   of    the    simple   conical 
pendulum  by  applying  D'Alembert's  principle  (page  242). 

The    impressed    forces    acting    upon    the    particle    are    the 
weight  mg  and  the  stress  5  of  the  string.     The  effective  force  is 

mvz 
the  deflecting  force acting  towards  O.    This  is  an  actual  force. 


If  now  we  reverse  the  direction  of  this  effective   force,  we  have 
This  is  not  an  actual  but  a  fictitious 


mv* 
—  acting  away  from  u. 


force.      There  is  in  reality  no  force  acting  in  this  direction  upon 
m.    The  actual  force  upon  m  is  towards  O.     But  by  D'Alembert's 


principle  the   impressed   and  reversed  effective  forces  form  a  system  in  equilibrium, 
have  then,  by  reversing  the  effective  force, 


We 


mi? 
S  sin  0 — •  =  o. 

We  thus  evidently  obtain  the  same  results  as  in  the  preceding  article,  where  we  took  — 
correctly  as  a  force  acting  towards  the  centre    O. 

This  reversed  effective  force  • acting  away  from  0  is  often  called  CENTRIFUGAL  FORCE, 

and  as  thus  used  the  term  is  allowable. 

But  it  should  be  borne  in  mind  that  it  is  a  fictitious  and  not  an  actual  force.  There  is 
really  no  force  acting  on  m  away  from  the  centre,  and  there  really  is  a  force  acting  on  m 
towards  the  centre.  If  we  really  had  a  centrifugal  force  in  equilibrium  with  S  and  mg,  the 
particle  m  would  move  in  a  straight  line  and  not  in  a  circle.  When,  then,  a  particle  moves  in 

a  curve,  since  its  direction  of  motion  changes,  there  must  be  an  unbalanced  force acting 

towards  the  centre  of  curvature,  that  is,  a  deflecting  force.  If  we  consider  this  force 
reversed,  we  can  apply  D'Alembert's  principle,  and  the  term  "  centrifugal  force  "  only  means, 
then,  this  reversed  force,  which  is  a  purely  fictitious  force  not  really  existing. 

A  particle  moving  in  a  curve  is  often  incorrectly  represented,  however,  as  possessing  an 
inherent  "centrifugal  force"  by  virtue  of  which  it  "tends  to  fly  away  fiom  the  centre." 
Indeed  it  is  sometimes  represented  that  it  is  acted  upon  simultaneously  by  such  a  force,  and 


246 


KINETICS  OF  A  PARTICLE. 


[CHAP.  I. 


also   by   an    equal   and    opposite    deflecting  force  towards  the   centre.*     Both    views   are 
erroneous  and  misleading. 

The  student  would  do  well,  therefore,  to  discard  the  term  "centrifugal  force,"  since  it 
answers  no  real  purpose.  It  is  sufficient  in  all  cases  to  take  all  forces,  both  impressed  and 
effective,  as  they  really  are,  and  then  to  apply  D'Alembert's  principle. 

Deflecting  Force  at  the  Earth's  Surface. — Suppose  the  earth  to  be  a  homogeneous 
sphere  of  radius  r  and  centre  C.  Let  WME  represent  the  equator,  NS  the  axis,  and  let  P 

be  any  point  of  the  surface  on  the  meridian  PMS,  so 
that  the  latitude  is  A  =  PCM. 

Let  a  particle  of  mass  m  rest  on  the  surface  at  P. 
Let  G  be  the  actual  acceleration  of  gravity  acting 
towards  the  centre  C  along  the  radius  PC. 

Then  mG  is  the  actual  force  of  attraction  towards 
C  acting  upon  the  particle.  Let  g  be  the  observed  ac- 
celeration of  gravity  at  P.  Then  the  observed  weight 
of  the  particle  at  P  is  mg.  The  pressure  of  the  earth 
on  the  particle  is  then  mg  acting  away  from  C. 

The  difference  between  w/6"and  mg,  or  mG  —  mg, 
is  that  part  of  the  earth's  attraction  necessary  to  keep 
the  particle  on  the  earth  without  pressure.  This 
force  mG  —  nig  acting  towards  C  can  be  resolved  into 
a  component  Ft  tangent  to  the  meridian  at  P,  and  a 
component  F0  in  the  direction  PO.  This  latter  com- 
ponent is  the  deflecting  force  which  makes  the  particle  move  in  the  latitude  circle  whose 
radius  is  PO  =  r  cos  A  This,  then,  is  the  effective  force,  while  m(G  —  g)  towards  C  and  Ft 
are  impressed  forces. 

If  we  reverse  the  effective  force,  we  have,  by  D'Alembert's  principle, 

—  m(G  —  g)  -f-  FQ  cos  A  =  o, 

—  F0  sin  A  —  Ft  =  o. 

But  (page  244)  since  the  angular  velocity  GO  =  -=-,,   where  Tis  the  time  of  rotation, 


Hence 


F0  =  tnrar  cos  A  =  — —  cos  A. 


mg  =  mG  —  mra?  cos2  A  =  mG —3—  cos2  A (i) 


f,—, 


2  n*ntr 
sin  2A  =  —  —  y-2  —  sin  2  A. 


(2) 


"  When  I  was  about  nine  years  old,  I  was  taken  to  hear  a  course  of  lectures  by  an  itinerant  lecturer  in  a 
country  town,  to  get  as  much  as  I  could  of  the  second  half  of  a  good,  sound  philosophical  omniscience.  .  .  . 

"'You  have  heard  what  I  have  said  of  the  wonderful  centripetal  force,  by  which  Divine  Wisdom  has 
retained  the  planets  in  their  orbits  round  the  sun.  But,  ladies  and  gentlemen,  it  must  also  be  clear  to  you  that 
if  there  were  no  other  force  in  action,  this  centripetal  force  would  draw  our  earth  and  the  other  planets  into  the 
sun,  and  universal  ruin  would  ensue.  To  prevent  such  a  catastrophe,  the  same  Wisdom  has  implanted  a 
centrifOgal  force  of  the  same  amount,  and  directly  opposite.'  .  .  . 

"  I  had  never  heard  of  Alfonso  X.  of  Castile,  but  I  ventured  to  think  that  if  Divine  Wisdom  had  just  let  the 
planets  alone  it  would  come  to  the  same  thing  with  equal  and  opposite  troubles  saved." — DE  MORGAN,  Budget 
of  Paradoxes. 


CHAP.  1.]  CENTRIFUGAL  FORCE.  247 

If  the  earth  were  a  homogeneous  sphere,  the  effect  of  this  tangential  component  Ft  upon 
liquid  particles  on  the  surface  would  be  to  force  them  towards  the  equator  and  thus  increase 
the  equatorial  and  diminish  the  polar  diameter.  The  fact  that  the  earth  is  not  a  sphere 
thus  indicates  that  the  now  solid  portions  may  once  have  existed  in  a  plastic  condition.  The 
equatorial  diameter  is  found  to  exceed  the  polar  by  about  26  miles.  The  ratio  of  this  dif- 
ference to  the  equatorial  diameter,  called  the  ELLIPTICITY  of  the  earth,  is  about  ^-^. 

The  earth  is  considered,  then,  as  an  ellipsoid  of  revolution  with  this  ellipticity,  so  that 
the   direction    of   the    observed   force  of    gravity,  or  of  the 
plumb-line  ZP,    is  always  normal  to  the  surface  and  hence 
does  not  pass  through  C  except  at  the  equator  and  poles. 

The  force  of  attraction  mG  acting  towards  C  is  then 
resolved  into  two  components,  one  normal  to  the  surface 
along  PC  and  one  F0  along  PO.  This  latter  is  the  deflect- 
ing force  necessary  to  keep  the  particle  on  the  latitude  circle 
whose  radius  is  PO  —  r  cos  A.  The  former  is  balanced  by 

the  pressure  of   the  earth  upon  the  particle.      There  is   then  no  tangential  force  Ft,  and  no 
tendency  of  the  particle  at  P  to  move  towards  the  equator. 

The  effective  force  is  then  F^  as  before  acting  towards  O,  and  the  impressed  forces  are 
mG  acting  towards  C  and  nig  acting  towards  Z.  Let  0  be  the  angle  CPC' '.  If  we  reverse 
the  effective  force,  we  have,  by  D'Alembert's  principle, 

mg  -|-  F0  cos  (A  -(-  0)  —  mG  cos  0  =  o, 
or,  substituting  for  F0  its  value, 

mg  =  mG  cos  0  —  mra?  cos  A  cos  (A  -{-  0)  =  mG  cos  0 ^—  cos  A  cos  (A  -f-  0). 

Since  in  the  case  of  the  earth  the  deviation  from  a  sphere  is  small,  the  angle  0  is  very 
small  and  this  equation  reduces  practically  to  (i).  We  can  then  treat  the  earth  as  a  sphere 
of  mean  radius  r  and  neglect  the  tangential  component  Ft. 

COR.  i. — If  we  take  the  mean  radius  r  =  3960  miles  and  T  =  86164  seconds  for  a 
sidereal  day,  we  have 

4?r2r 
rGi32  — ___-  —  0.111255  ft.-per-sec.  per  sec (3) 

We  have  then,  from  (i),  for  the  total  acceleration  of  gravity  G  at  any  point  P  in  latitude  A 

G  —^-+0.11 1255  cos2  A (41 

At  the  poles  A  =  90°  and  G  —  g,  or  the  observed  acceleration  of  gravity  g  at  the  pole  , 
is  equal  to  the  total  acceleration  of  gravity  G  of  the  earth. 

At  the  equator  A  =  o  and  here  the  observed  value  of  ^at  sea-level  is  found  to  be  about 

g  =  32.09022  ft.-per-sec.  per  sec. 
Hence  from  (4)  we  obtain,  assuming  the  earth  as  a  sphere, 

G  —  32.20148  ft.-per-sec.  per  sec. 
The  resultant  central  force  mG  —  mg  acting  towards  C  is  then,  from  (4), 

,     mG  ( i  —  —  }  =  o.  1 1 1 2  5  5 m  cos2  A  =  mr<*?  cos2  A  =  — ~2—  cos2  A. 
\         (j '  J- 


248  KINETICS  OF  A  PARTICLk.  [CHAP.  I. 

I 
At  the  equator  \  —  o,  and  I  -  ~  =  —  g-.     Hence  at  the  equator 


(5) 


That  is,   the  deflecting  force  at  the  equator  is  about  ^  of  the  total  force  of  gravity. 

COR.  2.  —  To  find  the  time  of  rotation  T0  of  the  earth  in  order  that  a  particle  at  any 
point  P  may  have  no  observed  weight,  i.e.,  exert  no  pressure  on  the  surface,  we  have  from 
(i),  by  putting  mg  =  o  and  T=  T0, 


cos2  A,     or     T9  =  A  - 


9  =  A 
V 


But,  from  (5),  we  have  for  the  actual  time  of  rotation  T 

I  4n2r  289  X  4rrzr 

_£=,     or    G=      -—. 


Substituting  this  value  of  G,  wex  obtain 


At  the  equator  we  have   A  =  o  and    T0  =  —  T.      In  order,  then,  that  a  body  at  the 

equator  may  have  no  weight,  the  earth  should  rotate  in  about  one  seventeenth  of  a  day.     If 
the  earth  were  to  rotate  faster  than  this,  bodies  at  the  equator  would  not  stay  on  the  surface. 

Examples.—  f  I)  A  string  just  breaks  with  a  weight  of  20  pounds.  It  is  fastened  to  a  fixed  point  at  one  end 
and  at  the  other  to  a  mass  of  5  Ibs.,  which  revolves  round  t/te  point  in  a  horizontal  plane.  Find  the  greatest 
number  of  complete  revolutions  in  a  minute  before  the  string  breaks,  if  the  radius  to  the  centre  to  mass  is  j 
/"/.  <g=J*>) 

ANS.  To  break  the  string  requires  a  force  of  2og  poundals.     Let  n  be  the  number  of  revolutions  per 

~.  2itrn       nn     .  mv*  — 

minute.      Then  v  =  —  —  =  —    ft.  per  sec.     The  stress  in  the  string   is  -  =  20^-,  or  v 


Hence  n  =  —-  =  4§  complete  revolutions  and  a  fraction  over. 

(2)  Suppose  the  mass  revolves  in  a  vertical  plane  f 

ANS.  Then  at  the  lowest  point  the  stress  is  ^-  +  mg  poundals  and  —  -•  +  mg  =  2qf  ,  or  v  =  ^  \  tf=  -g- 

Hence  n  =  —  ^   *&  =  41  complete  revolutions  and  a  fraction  over. 

(3)  What  portion  of  their  weight  do  bodies  lose  at  the  equator,  taking  the  radius  of  the  earth  4000  miles 
and  g  =  32  ft.  -per-  sec.  per  sec. 

ANS.  About  ^ 

(4)  Find  the  length  of  day  in  order  that  a  body  in  latitude  60"  may  possess  no  weight. 
ANS.  One  thirty-fourth  of  the  present  day. 

(5)  A  skater  moves  in  a  curve  of  100  feet  radius  with  a  speed  of  40  feel  per  tec.     Find  his  inclination  to 
the  ice.     (g  =  3*-) 

ANS.  60  degrees. 


CHAP.  I.] 


DEFLECTING   FORCE— EXAMPLES. 


249 


mg 


(6)  Find  the  necessary  elevation  of  the  outer  rail  on  a  railroad-track  on  a  curve  of  radius  r,  so  that  an 
engine  weighing  m  Ibs.  moving  with  a  speed  v  can  pass  witliout  lateral 
pressure  on  the  rails  by  the  wheel-flanges.     Also  find  the  pressures  on 
the  rails. 

ANS.  Let  C  be  the  centre  of  mass  of  the  engine  and.  h  =  CO  the 
distance  of  the  centre  of  mass  above  the  rails.     We  have  acting  at  Cthe 
weight  mg,  downwards.     Since  there  is  no  lateial  pressure  on  the  rails,    pw 
the  rail  pressures  are  A\  and  R,  at  right  angks  to  AB,  as  shown  in  the      '\ 
figure.     The  impressed  forces  are  then  mg,Ri  and  R* ,  and  the  effective         \ —  — f —  Q 

force  is  F  =  '^-  horizontal  and  acting  towards  the  centre  of  curvature. 

By  D'Alembert's  principle,  if  we  reverse  F  we  have  a  system  of  forces 
.R,       in  equilibrium.    Let  a  be  the  angle  of  elevation  BAD,  and  w  the  width  of 
track  AB. 

We  have  then,  putting  the  algebraic  sum  of  vertical  components  equal 
to  zero, 

Ri  cos  a  +  R*  cos  a  —  mg  =  o,     or     (A'i  +  AY)  cos  a  =  mg.    .     .     (i) 
Putting  the  algebraic  sum  of  the  horizontal  components  equal  to  zero, 

—  A'i  sin  a  —  R,  sin  a  +  "—  =  o,     or     (A'i  +  AY)  sin  a  =  '^~.       .     (2) 
r  r 

Taking  moments  about  A  and  putting  the  algebraic  sum  of  the  moments  equal  to  zero, 


m/v 


R^isj  _  mg\  —  cos  a  —  h  sin  a 
From  (i)  and  (2)  we  have 


)fnv*  / ,  TV    .       \ 

—  ( h  cos  a  +  —  sin  a\ 


and  substituting  this  in  (3),  we  obtain 
and  hence,  from  (i), 


v* 
tan  a  =  —  ; 


m 


2  cos  a 
mg 


2  cos  a. 
Equations  (5)  and  (6)  give  Ri  and  /?,  in  poundals.      For  gravitation  measure  we  divide  by^  and  obtaii 


Ri  =  R*  = 


in 


From  (4)  we  obtain 


•/' 


V* 


cos  a  — 


(4) 

(5) 

(6) 

in 

(7) 


7/* 


We  have  then  for  the  elevation  DB  =  w  sin  a. 

DB=  — 


(8) 


and  for  Ri  and  /?-» ,  in  pounds, 


A", 


(9) 


Equations  (8)  and  (9)  are  accurate.  They  admit  of  practical  simplification.  Thus  if  we  take  the  maxi- 
mum speed  a  mile  per  minute  or  88  feet  per  second,  and  a  ten-degree  curve  for  which  r=  573-69  ft.,  we  have 
for  Ar  =  32,  —  =  L,  or  ~a  =  ~-,  and  this  value  is  far  greater  than  it  will  ever  actually  be.  We  see,  then, 


6, 


v 
that  —  ^  can  be  disregarded  in  (8)  and  (9),  and  we  have  practically 


25°  KINETICS  Oh'  A  PARTICLE. 

and  for  the  elevation  of  the  outer  rail  the  practical  formula 


[CHAP.  1. 


or,  taking^  =  32  and  iv  —  4  fj.  8$  in., 


DB  =  7-?-  inches,  nearly, 


(10) 


where  r  is  to  be  taken  in  ft.  and  v  in  ft.  per  sec. 

If  r  is  taken  in  ft.  and  v  in  miles  per  hour,  we  have 


DB  =      ~  inches,  nearly (n) 

4r 

Equations  (10)  and  (n)  are  then  practical  formulas  giving  the  elevation  in  inches  for  r  in  feet,  and  v  in 
ft.  per  sec.  or  in  miles  per  hour. 

(7)  Find  the  elevation  of  outer  rail  for  radius  of  300  yards  and  speed  of  45  miles  per  hour. 
ANS.  8.43  inches. 

(8)  Find  the  speed  v  of  an  engine  on  a  curved  /eve/  track  of  radius  r  and  gauge  w  when  it  is  just  on  the 
Point  of  overturning,  the  centre  of  mass  being  h  above  the  rails. 

ANS.   v=\/^. 

(9)  A  vessel  containing  water  revolves  with  uniform  angular  velocity  (a  about  a  vertical  axis  through  the 
centre  of  mass.     Find  the  curve  of  the  surface  of  the  water. 

ANS.   Let  a  particle  of  mass  ///  be  at  any  point  P  of  the  surface   whose  co-ordinates  are  ON  =y  and 
NP  =  x. 

The  impressed  forces  are  the  weight  mg  of  the  particle  and  the  pres- 
sure A* of  the  surface  upon  the  particle.  The  surface  at  P  must  be  at 
right  angles  to  A'.  Hence  the  tangent  PT  is  at  right  angles  to  R. 

The  effective  force  is  mxv?  acting  towards  N.  If  we  reverse  this 
force,  we  have,  by  D'Alembert's  principle, 

mxa?  —  R  sin  NPT  =  o, 
R  cos  NPT  -  mg  =  o. 
Hence  we  have 


g 
Produce  the  direction  of  R  to  S,  so  that  SP  is  the  normal  and  5Aris  the  subnormal  at  P.     Then 

SN X  tan  NPT  =  x,     or     SN  =  -^. 

The  property  of  a  parabola  is  that  the  subnormal  is  constant  and  equal  to  the  parameter. 
Therefore  \\\t  curve  is  a  parabola  and  its  equation  is 


(10)  A  vessel  containing  water  is  moved  horizontally  with  an  acceleration  f.     Find  the  inclination  of  the 
water-surface. 

ANS.  Let  a  particle  of  mass  m  be  at  any  point  P  of  the  surface. 

The  impressed  forces  are  the  weight  mg  of  the  particle  and  the  pres- 
sure R  of  the  surface  upon  the  particle.  The  surface  must  be  at  right 
angles  to  R.  Let  a  be  the  angle  of  the  surface  with  the  horizontal. 

The  effective  force  is  mf  in  the  direction  of/.  If  we  reverse  this  force, 
we  have,  by  D'Alembert's  principle, 


Hence 


R  cos  a  —  tng  =  o, 
R  sin  «  —  mf  =  o. 

tan  ft  =  — . 


CHAP.  I.] 


PARTICLE  MO'/ING   ON   THE  EARTH'S  SURFACE. 


Particle  Moving  on  the  Earth's  Surface. — Let  the  particle  P,  instead  of  being  at  rest  on 
the  surface   of  the  earth,  have  a  velocity  v  relative  to  the 
earth   at   any    instant    in    any    direction    tangent    to    the 
earth's  surface. 

Take  the  point  P  as  origin,  the  axis  of  X  towards 
the  east,  the  axis  of  F  towards  the  north,  the  axis  of  Z 
along  the  radius  through  P. 

Let  vx  and  vy  be  the  components  of  v  along  the  axes 
of  X  and  Y,  so  that  vx  is  positive  towards  the  east  and 
negative  towards  the  west,  and  vy  is  positive  towards  the 
north  and  negative  towards  the  south. 

Let  P^P2  —  vyt  be  the  distance  south  along  the 
meridian  described  by  P  in  north  latitude,  in  an  indefinitely 
small  time  t.  If  there  were  no  rotation,  P-^P.2  would 
coincide  with  the  meridian  through  Pr  But  owing  to 
the  rotation  of  the  earth  this  meridian  moves  to  P{M, 
while  Pl  moves  to  P2',  so  that  if  P^'O'  is  parallel  to  the 

axis  NO,  the  angle  MO'P^  =  otf,  where  03  is  the  angular  velocity  of 
rotation.  The  angle  O'P{P^  =  A  =  the  latitude  of  Pr  We  have 
then  O'P2'  —  Vyt  sin  A  and 

MP2'  =  vyt  sin  A  .  cot. 

But  if  fx  is  the  acceleration  due  to  rotation   of  P  with  reference  to 
the  meredian  M,  we  have  (page  92) 

MPZ'  =  l-fxt*  =  vycot*  sin  A. 

Hence    we    have    for    the  acceleration  of  P  with  reference  to  the 
meridian  M,  due  to  rotation  and  the  velocity  vyj 


fx  = 


sin  A, 


(I) 


Equation  (i)  is  general  if  we  take  vy  positive  towards  the  north  and  negative  towards 
the  south,  and  A  positive  or  negative  according  as  the  point  is  north  or  south  of  the  equator. 
Thus  in  the  figure  vy  is  south  or  negative,  and  hence  for  north  latitude  fx  is  negative  or 
towards  the  west. 

Again,  let  P^PZ  =  vxt  be  the  distance  east  described  by  P  in  north  latitude  in  an 
indefinitely  small  time  t.  The  meridian  moves  to  M  while  Pl  N 

moves  to  Pz,  so  that  the  angle  MP  P  —  out.      We  have  then 


and  its  projection  on  the  meridian  is  MPZ  sin  A  =  covxt2  sin  A 
towards  the  south.  If  fy  is  the  acceleration  of  /'with  reference 
to  the  meridian,  due  to  rotation  and  the  velocity  vx,  we  have 


sin  A,      or     fy  =  —  2< 


A. 


(2) 


252 


KINETICS  OF  A  PARTICLE. 


[CHAP.  I. 


Equation  (2)  is  general  if  we  take  vx  positive  towards  the  east,  negative  towards  the 
west,  latitude  north  positive,  south  negative,  and  fy  positive  towards  the  north,  negative 
towards  the  south. 

Again,  we  have  in  the  preceding  figure  for  the  projection  of  MI\  along  the  radius  r, 
J//*,  cos  A  =  oov^  cos  A  upwards.  If  ft  is  the  radial  acceleration  due  to  rotation  and  the 
velocity  vx  of  P  with  reference  to  the  meridian,  we  have 


-/,/*  =  <uvj*  cos  A,      or    /,  =  2<avx  cos  A. 


(3) 


But  we  also  have  the  acceleration  of  P  with  reference  to  the  meridian  when  there  is  no 

*      v*    I    V  2 

—          —due  to  the  velocity  v,  and  also   the  downward  acceleration  g  due  to 

gravity.      Hence  the  total  radial  acceleration   of  P  with  reference  to  the  meridian,  due  to 
rotation  and  the  velocities  vx  and  vy  ,  is 


rotation,  —  = 


-j-  2  GOVX  cos  A. 


(4) 


Equation  (4)  is  general  if  we  take  vx  positive  towards 
the  east,  negative  towards  the  west,  and  i>y  positive 
towards  the  north,  negative  towards  the  south,  A  positive 
for  north,  negative  for  south  latitude,  and  f,  positive 
upwards,  negative  downwards. 

Deviation  of  a  Falling  Body  by  Reason  of  the 
Rotation  of  the  Earth.  —  Let  a  particle  be  projected 
upwards  along  the  radius  of  the  earth  with  a  relative 
velocity  vz,  and  let  P^P2  =  vst  be  the  distance  described 
in  an  indefinitely  small  time  t.  If  there  were  no  rotation, 
/>1/">2  would  coincide  with  the  radius  through  Pr  But 
owing  to  the  rotation  of  the  earth  the  meridian  moves  to 
P^M  while  />,  moves  to  />,',  so  that  if  PjO"  is  parallel 
to  the  axis  NO,  the  angle  MO"P2'  =  cot,  where  GO  is  the 
angular  velocity  of  rotation.  The  angle  O"P^P.^  =  90  —  A,  where  A  is  the  latitude  of  Pr 
tt  .  cos  A  and 

—  vtt  cos  A   .  oat: 


\Ve  have  then  O"P2'  = 


But  if/,  is.  the  acceleration  due  to  rotation  of  P  with  reference  to  the  meridian,  we  have 


MP2'  =  -//=-  z'z/ 


cos  A. 


Hence  we  have  for  the  acceleration  in  longitude  of  P  with  reference  to  the  meridian, 
due  to  rotation  and  the  velocity  rt, 

fx—  —  2<ovf  cos  A  ...........      (i) 

Equation  (i)  is  general  if  we  take  i\  positive  upwards  and  negative  downwards,  north 
latitude  positive,  south  latitude  negative,  and/,  positive  towards  the  cast,  negative  towards 
the  west. 

For  a  falling  body,  then,  vt  is  negative  and  we  have  /,  essentially  positive  or  towards 
the  east.  Hence  a  falling  body  falls  to  the  cast  of  the  point  vertically  beneath  it  at  the  start. 


CHAP.  I.] 


PARTICLE  MOVING   ON   THE  EARTH'S  SURFACE. 


Let  t  be  the  time  of  fall.     Then  if  the  particle  starts  from  rest,  we  have  for  the 
fall  (page  92) 


253 
height  of 


=  -/*•,      or      /=A/^. 


The  resultant  acceleration  at  any  point  of  the  path  is  then 


But  GO  — 


For    ordinary   falls,    then,    we 


60  X  60  X  24 

may  neglect  42^0^  cos2  A,  and  we  have  practically  the  accelera- 
tion at  any  point  of  the  path  equal  to  g.  The  velocity  at  any 
p'oint  of  the  path  is  then  practically  v  =  gt,  and,  from  (i), 

/*=  2G»gt  cos  A .     (3) 

The  mean  acceleration  is  then 


2jx 

ana  the  final  velocity  in  longitude  is  then  —fxt  =  cagtz  cos  A.  The  velocity  in  longitude 
varies,  then,  as  the  square  of  the  time,  or  as  the  ordinate  to  a  parabola.  The  mean  velocity 
in  longitude  is  then  —Gogf*  cos  A,  and  hence  the  distance  in  longitude  is 

i 
x  =  —  oogt*  cos  A. 


Inserting  the  value  of  /  from  (2),  we  have 


x  =  —goo  cos  A 


//2/;\3      2  /2h 

\/{  =  —  h<o  cos AA  / — -, 

V  v*/      3  V  £• 


(4) 


Equation  (4)  gives  the  deviation  in  longitude  of  a  falling  body  by  reason  of  the  earth's 
rotation.  It  is  always  towards  the  east  or  positive.  For  a  body  projected  vertically  upwards 
it  is  towards  the  west. 

Examples. — (i)  An  engine  weighing  27  tons  runs  at  the  rate  of  4.5  miles  per  hour  on  a  straight  track  in 
latitude  30°  north.  Find  the  pressure  on  the  rails  (a)  when  it  runs  north ;  (b)  south ;  (c)  east ;  (d)  west. 
(g  =  32.) 

ANS.  (a)  290  poundals  or  about  9  pounds  on  the  east  rail ;  (b)  the  same  on  the  west  rail ;  (c)  the  same  on 
the  south  rail ;  (d)  the  same  on  the  north  rail. 

(2)  In  the  preceding  example  let  the  latitude  be  30°  south. 

ANS.  (a)  290  poundals  or  about  9  pounds  on  the  west  rail ;  (B}  the  same  on  the  east  rail ;  (c)  the  same  on 
the  north  rail;  (d)  the  same  on  the  south  rail. 

(3)  In  example  {/)  find  the  -vertical pressure  on  the  rails  when  the  engine  runs  (a)  north;  (b)  south  ;  (c)  east; 
(d)  west,     (r  =  3960  miles.) 

ANS.  (a)  Less  by  12.6  poundals  or  about  0.4  pounds;  (b)  the  same;  (c)  less  by  515  poundals  or  about  16 
pounds ;  (d)  increased  by  490.2  poundals  or  about  15.3  pounds. 

(4)  Find  the  velocity  of  a  body  in  order  that  it  may  have  no  weight  when  it  moves,  in  latitude  60°,  (a)  north; 
{b}  south  ;  (c)  east ;  (d)  west,     (r  =  3960  miles,  g  =  32.) 

ANS.   (a)  and  (b)  about  5  miles  per  sec.  ;  (c)  about  4,85  miles  per  sec.;  (d)  about  5.14  miles  per  sec. 

(5)  A  particle  in  latitude  jo°  north  has  a  Telocity  of  60  feet  per  sec.  and  m<n>es  on  a  perfectly  smooth 
horizontal  plane.     Disregarding    resistance   of  the   air,  find  the   acceleration  and  the  distance  described  in 


«S4  KINETICS  OF  A  PARTICLE  [CHAP.  I. 

latitude  and  longitude  in  4  seconds  (a)  when  the  velocity  is  north  ;  (b)  south  ;  (r)  east ;  (if)  west,     (r  =  3960 

mites.) 

ANS.  (<*)/*  =  0.00436  ft.-per-sec.  per  sec.  east,  distance  240  ft.  north  and  0.035  h-  ca*1- 
(6)  /„  =  0.00436  ft.-per-sec.  per  sec.  west,  distance  240  ft.  south  and  0.015  ft.  west. 

(c)  fy  =  0.05236  ft.-per-sec.  per  sec.  south,  distance  240  ft.  east  and  0.42  ft.  south. 
(d)f,  =  0.05236  ft.-per-sec.  per  sec.  north,  distance  240  ft.  west  and  0.42  ft.  north. 

(6)  A  cannon-ball  is  fired  in  latitude  jo9  north  with  a  -velocity  of  1440  feet  per  sec.     Neglecting  resistance 
of  the  air.  find  the  acceleration  and  distance  described  in  latitude  and  longitude  in  4  seconds  (a)  when  the 
velocity  is  north  ;  (b)  south  ;  (c)  eas^t ;  (a)  west. 

ANS.  (a)fx  =  0.10472  ft.-per-sec.  per  sec.  east,  distance  5760  ft.  north  and  0.84  ft.  east. 
(b)  fx  =  0.10472  ft.-per-sec.  per  sec.  west,  distance  5760  ft.  south  and  0.84  ft.  west. 
(f)  fy  =  0.15272  ft.-per-sec.  per  sec.  south,  distance  5760  ft.  east  and  1.22  ft.  south. 

(d)  fy  —  0.15272  ft.-per-sec.  per  sec.  north,  distance  5760  ft.  west  and  1.22  ft.  north. 

(7)  A  particle  in  latitude  60°  north  fulls  from  rest  a  distance  of  1296  ft.  to  the  ground.    Find  the  deviation 
in  latitude,  disregarding  resistance  of  the  air. 

ANS.  0.2827  ft.  towards  the  east. 

(8)  In  latitude  jo°  north,  find  the  angular  velocity  of  rotation  of  the  plane  of  a  pendulum. 

ANS.  0.0000363  radians  per  sec.  in  a  direction  clockwise  to  one  facing  the  north.     The  plane  rotates 
through  180*  in  24  hours. 

(9)  A  locomotive  weighing  32  tons  runs  at  the  rate  of  45  miles  per  hour  in  latitude  jo°  north  in  a  direction 
S.  30"  E.  on  a  curve  of  one  mile  radius  in  a  counter-clockwise  direction  to  one  looking  north.    Find  the  pressure 
on  the  outer  rail. 

ANS.   1868  pounds.     If  we  disregard  rotation  of  the  earth,  the  pressure  would  be  1848  pounds. 

(10)  In  the  preceding  example  suppose  the  direction  is  clockwise  to  one  looking  north. 
ANS.  1825.6  pounds. 


CHAPTER   II. 

TANGENTIAL  FORCE.     MOMENTUM.     IMPULSE. 

Tangential  Force — We  have  seen  (page  77)  that  when  a  particle  P  moves  in  a  curve 
whose  radius  of  curvature  is  p,  with  a  velocity  v  and  an  acceleration  / at  any  instant,  this 
acceleration  /can  be  resolved  into  a  central  acceleration  p 

v^  dv  dv 

fp=  —  and  a  tangential  acceleration  ft  =  — r  ,  where  -=-    is  the 
p  at  at 

rate  of  change  of  speed. 

If  m  is  the  mass  of  the  particle,  then  the  force  F  acting 
on  it  is  F  —  mf  in  the  direction  of/",  and  this  force  can  be  re- 
solved into  a  central  or  deflecting  force  Fp  —  mfp  =  — —  ,  which  causes  change  of  direction  of 

dv 

motion  and  a  tangential    force   Ft  =  mft  =  m  ^- ,  which  causes  change  of  speed. 

The  deflecting  force  Fp  we  have  discussed  in  the  preceding  chapter. 

Let  us  now  consider  the  tangential  force  Ft. 

Momentum. — Let  the  mass  of  a  particle  be  in,  and  its  velocity  v.  Then  we  call  the 
quantity  mv  the  MOMENTUM  of  the  particle. 

Hence  the  momentum  of  a  particle  is  the  product  of  its  mass  and  velocity. 

Momentum,  then,  has  magnitude  and  direction,  and  we  can  represent  it  by  a  straight 
line,  just  like  velocity,  and  all  the  principles  which  hold  for  velocity  hold  also  for 
momentum. 

We  can  therefore  combine  and  resolve  momentums,  and  we  have  the  triangle  and  poly- 
gon of  momentum  just  the  same  as  for  velocity  or  force. 

Hence  the  momentum  of  a  particle  in  any  direction  is  the  product  of  its  mass  and  velocity 
in  that  direction, 

We  can  also  have  the  moment  of  a  momentum,  just  the  same  as  for  a  velocity,  and  the 
same  principles  apply. 

The  unit  of  momentum  is,  then,  evidently  the  momentum  of  one  unit  of  mass  moving 
with  one  unit  of  velocity.  We  may  call  a  unit  of  velocity,  or  one  unit  of  length  per  unit  of 
time,  a  "  velo." 

The  English  unit  of  momentum  is  then  \he  pound-veto  (Ib.-velo),  or  the  momentum  of  a 
mass  of  one  Ib.  moving  with  a  velocity  of  one  foot  per  second. 

If,  then,  a  particle  of  mass  40  Ibs.  has  a  velocity  in  any  given  direction  at  any  instant  of  8  ft.  per  sec.,  the 
momentum  in  the  direction  of  the  velocity  is  320  Ib.-velos. 

In  the  C.  G.  S.  system,  we  would  have,  then,  the  gram-velo,  or  kilogram-velo,  that 
is,  the  momentum  of  a  mass  of  one  gram  or  one  kilogram  moving  with  a  velocity  of  one 
centimeter  per  second.* 

*  A  committee  of  the   British  Association  have  proposed  for  this   the  name   "dole."     This   has,    however, 
never  come  into  use. 

255 


KINETICS  OF  A  PARTICLE.     '  [CHAP.  II. 

Significance  of  Momentum. — Let  a  particle  of  mass  m  move  in  any  path  from  the  posi- 
v,  tion  />  to  P  in  the  time  /.     Let  the  velocity  at  /\  be  vv  and  at 


^  be 


We  have  (page  255)  the  tangential  force  Ft  given  by 
dv 


Now  if  Ft  is  constant  in  magnitude,  the  tangential  acceleration  ff=~    Js    also    con- 

dt 

stant  in  magnitude.     In  such  case  the  instantaneous  rate  of  change  of  speed  —  is  equal  to 

the  mean  rate  of  change  of  speed  -y-^1  for  any  interval  of  time  /  (page  74).      We  have 
then  for  Ft  constant  in  magnitude 


mv  — mv. 


0) 


Now  mv  —  mvl  is  the  change  of  momentum  in  the  path  in  the  time  /,  (v  —  v^)  is  the 
chajige  of  speed  in  the  time  /,  and *  is  the  time-rate  of  change  of  momentum  in 

the  path. 

Hence,  whatever  the  path  and  however  the  actual  tangential  force  may  vary  in  magnitude 
during  the  time,  the  time-rate  of  change  of  momentum  in  the  path  gives  a  tangential  force  of 
constant  magnitude  which,  acting  for  that  time,  would  cause  the  change  of  speed  in  the  path. 

Again,  let  us  resolve  v^  and  v  into  components  in  any  given  direction  P^a  and  at  right 
angles  to  this  direction.      Let  z\  make  the  angle  a^  and  v 
the  angle  a  with  the  direction  P^a.     Then  the  component          ~ 
velocities  in  this  direction    are  i\  cos  al ,   v  cos  a,  and  the 
mean   time-rate   of   change   of   speed   in   this  direction   is 

v  cos  a  —  v.  cos  a. 

— - — •         — .      If  the  force  F  in  this  direction  were 

constant,  the  instantaneous  rate  of  change  of  speed  would 
be  the  same  as  the  mean  for  any  interval  of  time,  and  we 
should  have 


_ 

H  — 


mv  cos  «  —  mv  cos  or 


(2) 


Hence  the  time-rate  of  change  of  momentum  in  any  direction  gives  the  uniform  force 
Fu  in  that  direction  which,  acting  for  that  time,  would  cause  the  change  of  velocity  in  that 
direction. 

No  matter,  then,  how  the  actual  force  in  the  given  direction  may  vary  during  the  time  /, 
we  can  find  from  (2)  in  any  case  that  equivalent  uniform  force  FH  which,  acting  for  that  time 
in  that  direction,  would  produce  the  change  of  velocity  in  that  direction. 

From  equation  (2)  we  see  that  as  the  time  /  decreases  the  force  FH  increases  for  the 
same  change  of  momentum.  If  /  is  zero,  FH  becomes  infinitely  large.  That  is,  change  of 
momentum  requires  time,  and  the  less  the  time  the  greater  the  force. 


CHAP.  II.]  RELATION  BETWEEN  IMPULSE  AND  MOMENTUM.  257 

If  the  time  is  one  second  and  the  initial  velocity  vl  cos  al  in  any  direction  s  zero,  we 
have,  from  (2), 

mv  cos  a 

Jr..  =  —        -. 

I  sec. 

If  the  time  is  one  second  and  the  final  velocity  v  cos  a  in  any  direction  is  zero,  we 
have,  from  (2), 

mvl  cos  a 

*=       I  sec.     ' 

That  is,  the  momentum  in  any  direction  is  NUMERICALLY  equal  to  that  uniform  force  which, 
acting  in  the  direction  of  the  momentum,  would  give  the  particle  starting  f  rout  rest  its  velocity 
in  that  direction  in  one  second ;  or  which,  acting  opposite  to  the  direction  of  momentum,  would 
bring  the  particle  to  rest  in  one  second. 

If  we  take  mass  in  Ibs.  and  velocity  in  feet  per  sec.,  the  force  in  equations  (i)  and  (2) 
is  given  in  poundals.  For  gravitation  measure  divide  by  g  (page  172). 

If,  then,  a  particle  of  mass  40  Ibs.  has  a  velocity  in  any  given  direction  at  any  instant  of  8  ft.  per  sec., 

320 
the  momentum   in   that  direction  is  320  Ib.-velos  (page  255),  and  a  uniform  force  of   320   poundals   or  — 

pounds  acting  in  the  direction  of  the  velocity  would  give  the  particle  starting  from  rest  its  given  velocity  in 
one  second.  Acting  opposite  to  the  velocity,  it  would  bring  the  particle  to  rest  in  one  second. 

Impulse. — The  product  Ft  of  a  uniform  force  F  by  its  time  of  action  /  is  called  the 
IMPULSE  of  the  force. 

Hence  impulse  is  the  product  of  a  uniform  force  by  its  time  of  action,  and  it  acts  in  the 
direction  of  the  force. 

Impulse,  then,  has  magnitude  and  direction,  and  we  can  represent  it  by  a  straight  line 
just  like  force,  and  all  the  principles  which  hold  good  for  force  hold  also  for  impulse.  We 
can  therefore  combine  and  resolve  impulses  and  have  the  triangle  and  polygon  of  impulse 
just  the  same  as  for  force. 

We  can  also  have  the  moment  of  an  impulse  just  the  same  as  for  force,  and  the  same 
principles  apply. 

The  unit  of  impulse  is  then,  by  definition,  the  impulse  of  one  unit  of  force  acting  for 
one  unit  of  time. 

The  English  unit  is,  then,  the  poundal-sec.  or  the  pound-sec.  The  C.G.S.  unit  is  the 
dyne-sec. 

If,  then,  a  uniform  force  of  320  poundals  acts  for  3  sec.,  the  impulse  is  960  poundal-sec.  or  —  pound-sec. 

Relation  between  Impulse  and  Momentum. — Let  a  particle  of  mass  m  move  in  any 
path  through  the  positions  Pl ,  P2,  jP3,  etc.  Let  vlt 
v2,  vs,  etc.,  be  the  corresponding  velocities,  tlt  tz, 
(3,  etc.,  the  corresponding  times  in  passing  from 
point  to  point,  and  Flt  Fz,  F^,  etc.,  the  correspond- 
ing forces  in  any  given  direction,  as  Pva. 

Let  the  points  Plt  P2,  P3  be  consecutive,  so  that 
the  times  tlt  /2,  /3  are  indefinitely  small.  Then,  how- 
ever the  force  in  the  direction  P^a  may  vary,  we  can 
consider  Fl  as  constant  for  the  time  tr  Its  impulse 
is  then  Fjr  The  impulse  of  F2  is  Fztz,  and  so  on. 


S58  KINETICS  OF  A  t>ART\CL£. 

But  from  equation  (2),  page  256,  we  have 

/r^  =  mvt  cos  aa  —  im\  cos  av 
Fatt  =  *«t-8  cos  tfs  —  w2  cos  a2, 

^5/3  =  mi'i  COS  «4  —  W2/s  COS  (Xy 


[CHAP.  11. 


If  we  sum  these  equations  and  let  the  final  velocity  be  i>,  making  the  angle  a  with  the 
direction  Plat  we  obtain 

2(Fi'\  4-  ^'2  4-  ^s's  4"  etc-)  =  mv  cos  a  ~  mvi  cos  ai  .....     (!) 
Let  /  be  the  ei.tire  time,  so  that 


Let  Fm  be  the  equivalent  uniform  force  in  the  direction  /\0  ,  whose  impulse  FHt  for  the 
entire  time  /  is  the  same  as  the  sum  of  all  the  actual  impulses  during  that  time,  so  that 


We  have  then,  from  (i), 


FJ  =  iX/Vi  +  FA  4-  ^3  +  etc.). 


etc.).  =  tf**'  cos  a-  — 


(2) 


This  is  equation  (2;  of  page  256. 

Hence  the  change  of  momentum  in  any  direction  for  any  time  t  gives  the  sum  of  the 
impulses  in  that  direction  during  that  time,  no  matter  liow  the  force  in  that  direction  may  vary. 
Examples,  -(i  )  A  ball-player  catches  a  ball  moving  with  a  velocity  of  50  ft.  per  sec.      The  mass  of  the 
ball  is  jj  oz.     If  the  time  of  coming  to  rest  is  fa  sec.,  find  the  equivalent  uniform  pressure. 

ANS.  We  have  m  =  —  Ib.      The  equivalent  uniform  pressure  is  then  Fn  =  —r-  =  —  —  =  859  - 

poundals,  or,  taking^-  =  32  ft.-per-sec.  per  sec.,  26.85  pounds. 

This  is  a  uniform  force  which  would  stop  the  ball  in  the  given  time.  The  actual  force  acting  at  any 
instant  we  cannot  tell  without  a  full  knowledge  of  the  law  of  variation  of  the  pressure  with  the  time.  If  we 
assume  the  pressure  at  first  contact  to  be  zero  and  to  increase  directly  with  the  time,  then  the  final  pressure 
would  be  twice  as  great  as  the  equivalent  uniform  pressure. 

(2)  An  So-ton  gun  fires  a  shot  of  5b  Ibs.  with  a  horizontal  muzzle  velocity  v  of  1800  fl.  per  sec.  Find  the 
•velocity  of  recoil  V. 

ANS.  Mass  of  gun  M  =  80  x  2240  =  179200  Ibs.  If  the  velocity  of  the  shot  is  imparted  in  the  time  /, 
the  sum  of  the  impulses  on  the  shot  during  that  time  is  mv  =  56  x  1800  =  ioo8oo-lb.  velos.  Since  action 
and  reaction  are  equal  at  every  instant,  the  sum  of  the  impulses  on  the  gun  during  the  time  is  the  same. 


Hence 


MV=mv,     or     V  = 


=  -     ft  .per  sec. 
179200       1  6 


(3)  A  horizontal  stream  of  water  "those  cross-section  is  a  and  velocity  v\  meets  a  surface  moving  in  the 
same  direction  with  a  velocity  v.  Find  the  pressure  exerted  on  the 
surface. 

ANS.  Let  the  water  pass  off  the  surface  in  a  direction  making  the 
angle  a  with  the  direction  of  motion.  The  volume  of  water  in  any 
time  /  is  av\t.  If  y  is  the  density  or  mass  of  a  unit  of  volume  of 
water,  the  mass  in  the  time  /  is  yavj  =  m. 

The  velocity  relative  to  the  surface  just  before  impact  is  vi  —  v. 
After  impact  the  velocity  relative  to  the  surface,  in  the  direction 
of  motion  is  (z>,  —  v)  cos  a. 

We  have  then  the  impulse  Ft  =  mv\  —  m  (v,  —  v)  cos  a,  or 
,._  f(r',  -  i>)  —  md't  —v)  cos  a  . 


CHAP  II.] 


MOMENTUM— EXAMPLES. 


Inserting  the  value  of  m  =  yav\t, 

F=  yav*(vi  —  v)(i  —  cos  a). 

If  the  surface  is  plane  and  at  right  angles  to  the  stream,  a  =  90°,   cos  a  =  o  and 

F  =  yavi(vi  —  v). 

If  the  surface  is  curved  so  that  the  water  is  reversed 
in  direction,  a  =  180°,  cos  a  =  —  i  and  the  pressure  is 
twice  as  great,  or 

F  =  2yav^(vi  -  v). 

These  values  of  F  are  given  in  poundals.     For  gravitation  measure  divide  by^. 

(4)  A  rifle-bullet  of  one  ounce  mass  is  shot  into  a  block  of  wood  of  53  Ibs.  and  gives  the  block  a  velocity  of 
2  ft.  per  sec.     Find  the  velocity  of  the  bullet. 

ANS.   We  have  mv  for  the  bullet  equal  to  (JM-\-m)v  for  the  combined  mass  of  block  and  bullet,  or 

-TV  =  ( 53 +  -7)2,    or    v=  1698  ft.  per  sec. 


16 


16 


(5)  A  mass  moving  with  a  velocity  of  'j  ft.  per  sec.  is  brvught  to  rest  by  a  uniform  opposing  force  of  one 
pound  in  2  sec.     Assuming  g  =  32  ft.-per-sec.  per  sec.,  find  the  mass. 

ANS.  A  force  of  one  pound  is  i  Ib.  Kg,  or  g  poundals.     We  have  then   F=  ^,   or  g  =  — -,   or 

m  =  —  =  2ii  Ibs. 
3 

(6)  A  uniform  force  of  i  o  pounds  acts  for  2  sec.  upon  a  mass  of  10  Ibs.  and  then  ceases.      With  what 
velocity  will  the  mass  continue  to  move  in  the  direction  of  the  force  ? 

ANS.  A  force  of  10  pounds  is  log  poundals.     We  have  then  F=  — ,  or  \og  =  — ,  or  v  =  2g  ft.  per  sec. 

(7)  A  particle  of  10  Ibs.  mass  has  an  initial  velocity  of  20  ft.  per  sec.  north  and  is  acted  upon  by  two  uni- 
form forces,  one  of  3  pounds  in  a  direction  northeast  and  the  other  of  the  same  magnitude  in  a  direction  north- 
west.    Find  its  velocity  after  one  minute. 

_ANS.  The  resultant  force  is  3^2  pounds,  or  31/2  .g  poundals,  north.     We  have  then  Ft  =  m(v  —  vt),  or 
34/2.^x60  —  \o(v  —  20).     Hence  v  =  18 1/2  .  ^  +  20,  or  for,^  =  32.  v  =  834.46  ft.  per  sec.  north. 

(8)  A  particle  of  mass  m  is  moving  east  with  a  velocity  v.     Find  the  uniform  force  necessary  to  make  it 
move  north  with  an  equal  velocity  in  t  seconds. 

ANS.   In  the  time  /  the  velocity  east  is  zero.     The  impulse  towards  the  west  is  then  mv.  •   In  the  same 
time  we  must  also  have_an  impulse  mv  towards  the  north.     The  resultant  impulse  is  then  Ft  =  mv .  |/2 

northwest,  or  F  =  ^- —  northwest. 


CHAPTER  III. 

WORK.     POWER. 

Work. — Let  a  uniform  force  Fact  upon  a  particle  which  moves  in  any  path  from  Pl  to 
Let  the  projection  Pji  of  the  path  along  the  line  of  the  force  be  denoted  by/.     Then 

the  product  Fp  is  called  WORK. 

If  the  projection/  of  the  path  is  in  the  direction  of 
the  force,  work  is  done  by  the  force  and  Fp  is  positive. 
If  the  projection  /  is  opposite  to  the  direction  of  the 
force,  work  is  done  against  the  fofce  and  Fp  is  negative. 
Let  the  displacement  PVP2  =  d  make  the  angle  ft 
with  F.  Then  we  have 

/  =  d  cos  0. 
The  work  of  F  is  then 


if  (.he  projection  Pji  =  p  is  in  the  direction  of  the  force  F,  and 

W=  -  Fp=  -  Fdcose 

if  the  projection  /*,«  =/  is  opposite  to  the  direction  of  F. 
We  have  then,  in  general, 

W=  ±F.dcosB=± 


But  d  cos  6  is  the  displacement  along  the  line  of  the  force,  and  F  cos  0"  is  the  force 
along  the  displacement. 

We  can  then  define  WORK  generally  as  follows: 

Work  is  the  product  of  a  uniform  force  by  the  component  displacement  along  the  line  of  the 
force  (W  —  ±  F.  dcos  #);  or,  the  product  of  the  displacement  by  the  component  force  along 
the  line  of  the  displacement.  (  W  =  ±  d  .  F  cos  0.) 

COR.  i.  —  It  is  evident  that  if  the  displacement  is  at  right  angles  to  the  force,  the  work 
is  zero. 

COR.  2.  —  We  see  that  work  is  independent  of  time.  A  given  uniform  force  and  displace- 
ment give  the  same  work  no  matter  what  the  time  occupied  by  the  displacement. 

COR.  3.  —  The  work  done  in  raising  a  body  is  equal  to  the  weight  of  the  body  acting  at 
the  centre  of  mass,  multiplied  by  the  vertical  displacement  of  the  centre  of  mass.  This 
work  is  done  against  the  weight  and  is  therefore  negative. 

Also,  the  work  of  lowering  a  body  is  equal  to  its  weight  multiplied  by  the  vertical  dis- 
placement of  the  centre  of  mass.  This  work  is  done  in  the  direction  of  the  weight  and  is 
positive. 

260 


CHAP.  Ill  ]  UKIT  op  u/ORK.     EXAMPLES.  261 

If  31  is  the  mass  of  the  body,  then  mg  is  its  weight.  If  d  is  the  vertical  displacement, 
then 

W—±  mgd, 

and  this  is  the  same  for  the  same  weight  and  vertical  displacement  whatever  the  time  or 
whatever  tlic  path. 

Unit  of  Work.  —  The  unit  of  work  is  then  evidently  the  work  of  one  unit  of  force  with 
one  unit  of  displacement. 

The  English  unit  is  then  the  foot-poundal,  or  a  uniform  force  of  one  poundal  acting 
through  one  foot;  or,  in  gravitation  units,  the  foot-pound,  or  the  weight  of  one  Ib.  acting 
through  one  foot.  '  This  latter  is  of  course  variable,  since  the  weight  of  one  Ib.  varies  at 
different  localities. 

If,  then,  we  take-in  feet  and  m  in  Ibs., 


gives  work  in  foot-poundals.      For  foot-pounds  we  divide  by^-  and  have 

.  W  =  ±  md  foot-pounds. 

The  C.  G.  S.  unit  of  work  is  the  uniform  force  of  one  dyne  acting  through  one  centi- 
meter.     It  is  called  an  ERG. 

A  multiple  of  this  equal  to    loooooooergs,  or   io7  ergs,  is  used  in  electrical  measure- 
.ments  and  called  a  JOULE,  after  Dr.  James  Prescott  Joule. 

Examples.  —  (i)  Find  the  work  expended  in  raising  16000  Ibs.  through  a  distance  of  20  feet. 

ANS.  320000^  ft.-poundals,  or,  if  g  =  32,  10000  ft.  -pounds,  or  the  work  of  raising  10000  Ibs.  one  foot  at 
a  locality  where  _£•  is  32  ft.-per-sec.  per  sec.,  whatever  the  time  or  the  path. 

(2)  A  body  of  80  Ids.  mass  moves  along  a  rough  horizontal  plane  with  a  speed  of  30  ft.  per  sec.  If  the 
retarding  force  of  friction  is  constant  and  equal  to  20  pounds,  find  the  work  done  against  friction  in  the  first 
second  and  in  coming  to  rest. 

ANS.  For  motion  in  a  straight  line  with  uniform  acceleration,  we  have  (page  92)  for  the  distance 
described  in  anv  time  / 


arid  for  the  distance  described  in  coming  to  rest 
The  force  Ft  in  the  path  is 


.  Ft  =  mft. 


In  the  present  case  this  force  is  a  retarding  force  of  20  pounds,  and  therefore  Ft  =  —  20^  poundals. 
Since  m  =  80  Ibs.,  we  have 


The  work,  then,  in  any  time  /  is 


-  2cg  =  Soft,     or    ft  =  — *> 


Fts  =  —  20?-  (7/1  /  +   -ft?)  ft'.-poundals,     or     —  2o(z/i/  —  -ft?)  ft. -pounds,     • 

and  the  work  in  coming  to  rest  is 

•v*  v  a 

Fts  =  —  l/Qg--£  ft.-poundals,    or     — 20  — ^  ft.  -pounds. 
2Jt  2ft 

Taking  g  —  32  ft.-per-sec.  per  sec.,  Vi  =  50  ft.  per  sec.  and  ft  as  found,  we   have  for  the  work  in  one 
second 

Fts  =  —  920  //.  -pounds, 
and  for  the  work  in  coming  to  rest; 

F$  =  •*-  3125 //. -pounds. 


a6z  KINETICS  OF  A  PARTICLE.  [CHAP.  III. 

(3)  A  mass  of  *  tons  (2*40  Ibs.)  is  pulled  up  too  feet  of  an  incline  which  rises  t  foot  in  25  ft.  of  length. 
Taking  the  resistance  of  friction  at  'jo  pounds  for  each  ton  of  weight,  find  the  work  done. 

ANS.  Let  a  be  the  angle  of  inclination  of  the  plane,  and  in  the  mass.  The  weight  is  then  nig,  and  the 
component  of  the  weight  down  the  plane  is  mg  sin  a  poundals. 

The  friction  is  300^-  poundals  also  down  the  plane.    The  total  force  down  the  plane  is  then 

(tng  sin  a  +  yyog)  poundals,     or    (;«  sin  a  +  300)  pounds. 

The  work  is  then 

FtS  =  —  (tn  sin  a  +  300)100 ft. -founds. 

In  the  present  case  m  =  4480  Ibs.,  sin  a  —  — ,  and  hence 

Fa  =  —  47920  ft. -pounds. 

(4)  Find  the  work  done  by  a  crane  in  lifting  the  materials  for  a  stone  wall  100  feet  long,  10  feet  high  and 
3  feet  thick,  the  average  density  being  fjo  Ibs.  per  cubic  foot. 

ANS.  i  500  ooo  ft. -pounds. 

Rate  of  Work — Power. — Work,  as  we  have  seen,  is  independent  of  time.  If  we  raise 
one  Ib.  one  foot,  we  do  the  same  work  whether  the  time  of  raising  is  one  second  or  one 
minute.  But  the  rate  at  which  the  work  is  done  is  not  the  same. 

If  then  we  take  time  into  consideration,  the  time-rate  of  work  is  called  POWER.  We 
exert  more  power  when  we  raise  one  Ib.  one  foot  in  one  second  than  when  we  raise  one 
Ib.  one  foot  in  one  minute,  although  we  perform  the  same  work  in  both  cases. 

The  unit  of  power  is  then  one  unit  of  work  per  unit  of  time.  The  English  absolute 
unit  of  power  is  then  one  ft.-poundal  per  sec.,  and  the  C.  G.  S.  absolute  unit  is  ona  erg  per 
sec. 

In  gravitation  units  we  have,  in  English  measures  one  foot-pound  per  sec.,  and  in 
French  measures  one  meter-kilogram  per  sec. 

These  units  are,  however,  inconveniently  small  in  most  cases. 

The  gravitation  unit  employed  in  English  engineering  calculations  is  therefore  taken  at 
550  foot-pounds  per  sec.  This  is  called  a  HORSE  POWER  and  is  denoted  by  H.  P.  In  French 
engineering  calculations  the  gravitation  unit  employed  is  75  meter-kilograms  per  sec.  This 
is  Called  FORCE  DE  CHEVAL. 

In  electrical  measurements  the  unit  adopted  is  io7  ergs  per  sec.  This  is  called  a  WATT, 
after  James  Watt.  The  watt  is  therefore  one  joule  per  sec.  (page  261). 

Examples.— (i)  The  area  of  the  piston  of  a  steam-engine  is  A  sq.  inches,  the  length  of  stroke  L  feet,  the 
steiini  pressure  in  pounds  per  sq.  inch  is  P,  the  number  of  strokes  per  minute  N.  Find  the  work  per  minuti 
and  the  horse-power. 

ANS.  Work  per  min.  =  P.  L.  A.  N.     H.  P  =  R  L"  A>  N . 

(i\  Find  the  work  done  against  grtivitv,  neglecting  friction,  in  pulling  a  car  of  a. 5  tons  (2240  Ms.)  loaded 
with  30  passengers  weighing  *54  Ibs.  each,  up  an  incline  the  ends  of  which  differ  in  level  by  120  feet ;  also  the 
horse-power  if  the  time  is  half  an  hour. 

ANS.  i  226400  ft. -pounds;  1.24  horse-power. 

($)  In  the  transmission  of  pou<er  bv  a  belt,  the  wheel  carrying  the  belt  is  14  feet  in  diameter  and  makes  S'J 
revolutions  per  minute,  the  tension  of  the  belt  being  too  pounds.  Find  the  horse- power  transmitted. 

ANS    4  horse-power. 

(4)  Check  this  statement :    Fiftv-ftue  pounds  mean  effective  pressure  at  boo  ft.  per  sec.  piston  sfleed  gives 
one  H.  P.  for  each  square  foot  of  piston  area. 

(5)  A  train  weighing  7S  tons  is  running  uf>  an  incline  of  r  in  $00  with  a  uniform  speed  of  40  miles  an 
hour.     Assuming  friction  to  be  equivalent  to  6  pounds  per  ton,  find  the  rate  at  which  the  engine  is  working. 

ANS.  70.4  H.  P. 


CHAPTER   IV. 

KINETIC  FRICTION. 

Kinetic  Friction.  —  The  friction  which  justs  prevents  motion,  that  is,  the  friction  which 
exists  when  motion  is  just  about  to  begin,  we  have  called  STATIC  FRICTION,  and  we  have 
fully  discussed  it  already  (page  220). 

The  friction  which  exists  after  motion  has  taken  place  is  called  KINETIC  FRICTION. 

Coefficient  of  Kinetic  Friction.  —  We  have  the  same  laws  for  kinetic  as  for  static 
friction  (page  222).  The  ratio  of  the  total  friction  to  the  total  normal  pressure  when 
motion  is  just  about  to  begin  we  have  called  the  coefficient  of  static  friction.  The  same 
ratio  after  motion  has  taken  place  is  the  coefficient  of  kinetic  friction. 

We  denote  the  coefficient  of  friction  in  general  by  /*.      We  have  then,  in  all  cases, 

P  =      >     or     F=pN, 


where  F  is  the  total  friction  and  N  the  total  normal  pressure. 

Angle  of  Kinetic  Friction. — If  N  is  the  normal  pressure  for  a  body  moving  on  a  rough 
surface,  then  ^N  is  the  friction.     The  resultant  reaction  of  the  sur- 
face is  then  R,  making  an  angle  0  with  the  normal  given  by 


We  call  the  angle  0  which  the  reaction  R  makes  with  the  normal  the  ANGLE  OF  FRIC- 
TION. For  static  friction  it  is  the  ANGLE  OF  REPOSE  (page  222).  Hence  the  coefficient  p  of 
kinetic  friction  is  equal  to  the  tangent  of  the  angle  of  friction  0. 

Kinetic  Friction  of  Pivots,  Axles,  Ropes,  etc.—  The  application  of  the  equation 

F—  /sN  =  Ntan  0 

to  pivots,  axles,  ropes,  etc.,  is  then  precisely  the  same  as  for  static  friction  (page  224).  We 
have  only  to  let  JA  stand  for  the  coefficient  of  kinetic  instead  of  static  friction. 

With  this  change  we  have  in  each  case  the  same  value  for  the  friction  and  moment  of 
the  friction  as  already  given  for  static  friction, 

Experimental  Determination  of  Coefficient  of  Kinetic  Sliding  Friction.—  We  may 
determine  the  coefficient  of  kinetic  sliding  friction  by  means  of  various  contrivances,  some 

of  which  we  shall  now  describe. 

263 


264  KINETICS  OF  A  PARTICLE.  [CHAP.  IV. 

I.  BY  SLED  AND  WEIGHT.  —  Let  a  sled  rest  upon  a  horizontal  plane  and  be  dragged  along 
_  ^     j          _._,  ky  means  of  a  string  passing  over  a  pulley,  to 

jimg<       ^'         m          ^     >  C  ^\  t^ie  enc*  °*  wmcl1  a  weight  is  hung.      In  order 

to  obtain  coefficients  for  different  substances, 
the  runners  and  plane  can  be  of  the  materials 
desired. 

In  such  an  apparatus  the  mass  of  string  and  pulley,  and  friction  of  string  and  pulley,  as 
well  as  rigidity  of  the  string,  should  all  be  insignificant,  or  else  they  must  be  allowed  for. 

Let  us  suppose  them  insignificant  and  let  m  be  the  mass  of  the  sled,  and  P  the  suspended 
mass.  The  normal  pressure  is  then  m^,  the  friction  ywm^,  and  the  weight  of  the  suspended 
mass  is  Pg.  Let  the  acceleration  of  the  masses  P  and  m  be  f.  The  impressed  forces  acting 
on  m  are  then  the  friction  //m^,  and  the  horizontal  tension  of  the  string  P(g  —  f}.  The 
effective  force  on  m  is  m/.  By  D'Alembert's  principle  (page  242),  if  we  reverse  the  effec- 
tive force  we  have 

-  m/  =  o, 


m 


But   for  uniformly  accelerated  motion  we   have   the  distance  described   in  any  time 
falling  from  rest  (page  92) 

I  2S 


Substituting  this  in  (i)  we  have 


m  mgf* 


If  then  we   observe  the  fall  s  of  P  starting  from   rest    in   any  time  /,  we  have  jw,  from 
equation  (2),  for  kinetic  friction. 

If  motion  is  just  about  to  begin,  f  =•  o,  and  we  have,  from  (i),  for  static  friction 


We  see,  then,  that  the  coefficient  for  kinetic  friction  is  less  than  the  coefficient  for  static 
friction. 

2.  BY  SLED  ON  INCLINED  PLANE — If  we  place  the  sled  on  an  vN 

inclined  plane  and  the  sled  slides  down  with  an  acceleration  f,  we 
have  the  impressed  forces  mg  sin  a  clown  the  plane,  the  friction  ^N 
—  /.ttog  cos  «  up  the  plane,  and  the  effective  force  taf  down  the 
plane.  If  we  reverse  the  effective  force,  we  have,  by  D'Alembert's  T^_ 

principle  (page  242), 

/*m£-  cos  a  —  mg  sin  a  -f-  mf  —  o, 
or 


CHAP.  IV.]  BY  SLED  ON  INCLINED  PLANE. 

The  distance  s  described  in  the  time  t  is,  as  before, 


265 


or    /=_. 


Substituting  in  (i),  we  have 


=  tan  or  —  — * 


2S 


gr  cos  o 


If  then   we  observe  the  distance  s  described  by  the  sled  starting  from  rest  in  any  time 
,  we  have  //,  from  equation  (2),  for  kinetic  friction. 

If  motion  is  just  about  to  begin,  f  —  o  and  we  have,  from  (i),  for  static  friction 


=  tan  a  =  -7  =  tan  0, 


where  //  is  the  height  and  b  the  base  of  the  plane,  and  <p  the  angle  of  repose  (page  222). 

Again,  we  see  that  the  coefficient  for  kinetic  friction  is  less  than  for  static  friction. 

3.  BY  FRICTION  BRAKE. — The  friction  brake  consists  of  a  lever  AB  of  equal  arms 
AO  —  BO  =  /,  with  a  bearing-block  at  the  centre.  It  is  placed  on  an  axle  horizontally  so 
that  the  entire  weight  of  the  lever  with  attached  scale-pans,  bearing-block,  etc.,  acts  at  the 
centre  O  of  the  axle. 

If  the  axle  turns  as  indicated  in  the  figure,  the 
lever  tends  to  turn  in  the  same  direction.  If  we  put  a 
mass  Q  in  one  pan  and  a  mass  P  in  the  other  and  make 
P  just  large  enough  to  keep  the  lever  horizontal  while  the 
axle  is  turning,  we  have  the  weights  Pg  and  Qg  and  the 
friction  F  in  equilibrium. 

If  m  is  the  mass  of  the  apparatus,  the  pressure  on 
the  axle  is  (P  -f  Q  -f  m^  =  R. 

For  a  new  bearing  ^page  229)  the  friction  is 


F= 


f> 


sin 


sin  0' 


where  /?  is  the  bearing  angle  aOb. 
Let  r  be  the  radius  of  the  axle. 


Then  we  have  for  equilibrium 


Hence  we  have  for  the  coefficient  of  kinetic  friction 

(P-  0V sin  /? 


A*  = 


(P+  Q 


ft 


266  KINETICS  OF  A  PARTIBLE.  [CHAP.  IV. 

Friction-brake  Test.  —  The  friction  brake  can  be  used  for  measuring  the  work  done 
by  an  engine  when  working  uniformly.  Thus  suppose  the  axle  is  driven  by  an  engine,  and 
that  by  means  of  a  crank  on  the  axle  some  machine,  as,  for  instance,  a  pump,  is  worked. 

We  first  count  the  number  of  revolutions  n  per  minute  while  the  pump  is  in  action. 
If  then  we  disconnect  the  pump,  we  shall  find  that  the  axle  revolves  much  more  rapidly 
than  before,  since  the  only  work  now  done  by  the  engine  is  against  the  friction  of  the  axle- 
bearing.  We  now  apply  the  brake  and  load  it  at  the  ends  with  P  and  Q  until  it  rests  hori- 
zontally, and  the  axle  is  slowed  up  to  its  former  speed  of  n  revolutions  per  minute.  In  this 
condition  the  work  now  done  against  the  brake  friction  is  equal  to  the  work  before  consumed 
by  the  pump,  provided  the  engine  works  uniformly. 

But  the  friction  F  is  given  by 


or,  in  gravitation  measure,  by 


We  have  then  for  the  work  done  in  one  revolution  2nrF,  and  in  n  revolutions  per 
minute  the  work  per  minute  is  2nrnF.  Taking  Fin  gravitation  measure  or  in  pounds,  and 
r  in  feet,  this  is  foot-pounds  per  minute.  If  we  divide  by  33000,  we  obtain  (page  262) 
horse-power.  Hence 

=  nrnF   =  nn(P  -  Q}1 
16500  16500      ' 

where  F,  P  and  Q  are  in  pounds,  /  and  r  in  feet,  and   n  is  the  number  of  revolutions   per 
minute  made  while  the  pump  was  connected. 

Work  of  Axle-  friction.  —  The  friction  upon  an  axle  in  any  case  when  /i  is  known  is 
given  on  page  228.  Thus  for  a  neiv  bearing  we  have  (page  229) 


where  R  is  the  resultant  pressure  on  the  axle,  and  ft  is  the  bearing  angle.      If  we  substitute 
this  in  the  place  of  Fin  the  preceding  article,  we  have  the  work  per  minute 


and  for  the  horse-power 


ft 

2  nrnF  =  2  ituRrn  - — -a  , 
sin  p 


HP-..  rri'Kritft 

16500  sin  ft  ' 


where  R  is  taken  in  pounds,  r  in  feet,  n  in  revolutions  per  minute.      If  the  bearing  angle  is 
small,  we  have  ft  =  sin  ft  nearly. 


CHAP.  IV.] 


COEFFICIENTS  OF  KINETIC  SLIDING  FRICTION. 


267 


Coefficients  of  Kinetic  Sliding  Friction. — The  following  tables  give  a  few  values  of  the 
value  of  fj.  as  determined  by  experiment  for  kinetic  sliding  friction  and  axle-friction. 

COEFFICIENTS    OF   KINETIC    SLIDING   FRICTION,  yU  =  tan  0. 


Substances  in  Contact. 

Condition  of  Surfaces  and  Kind  of  Unguent. 

Dry. 

Wet. 

Olive 
Oil. 

Lard. 

Tallow. 

Dry 
Soap. 

Polished 
and 
Greasy. 

O.o6 
0.07 
0.07 
0.07 
0.09 

O.II 

0.07 
0.07 
0.08 

O.o6 
O.O7 
0.08 
0.07 
O.OQ 
O.II 
O.O6 
O.O8 
0.10 

0.14 
0.15 
0.16 

O.O8 
O.I2 
0.15 
O.II 
0.13 
0.17 
O    IO 

0.14 

0.16 

Wood   on  )  rj 

0.36 
o  48 

0.25 

wood       1  i]ea"  

o  18 

0.06 
O.O7 
0.08 
0.05 
0.06 
0.08 

Metal  on    ]  ;,                 
metal        )  ^ea?  

0.24 

0.31 

O.2O 

O.26 

Wood  on   (  Minimum  
J  Mean 

0.42 
0.62 
0.45 

0.24 
0-33 

metal       )  ;i      .  '  ' 
(.  Maximum  

Hemp  ropes  (  On  wood  

0.19 

O.2O 

Leather  belts  (  Raw      

0.54 
0.30 

0.36 

0.16 

on  wood  or  •<  Pounded  
metal              (  Greasy    

0.25 

0.34 

0.31 

0.14 

0.14 

COEFFICIENTS    OF   AXLE-FRICTION. 


Dry 
or  Slightly 
Greasy. 

Oil,  Tallow,  or  Lard. 

Damp 
and  Greasy. 

Ordinary 
Lubrication. 

Thorough 
Lubrication. 

O   OQ7 

0.049 
0.054 
0.054 
0.054 
0.054 

Wrought  iron  on  bell-metal  

0.251 

0.075 
0.075 
0-075 
0.075 

0.189 

Cast  iron  on  cast  iron  

0-137 
o.  161 

"        "      "    bell-metal                 .... 

0.194 

Comparing  the  values  in  the  tables  just  given  with  those  in  the  table  given  on 
page  224,  we  see  that  the  coefficient  of  kinetic  is  always  less  than  tJie  coefficient  of  static 
sliding  friction. 

We  see  also  that  for  axle-friction,  in  general  we  have  for  the  coefficient  of  kinetic 
friction : 

for  ordinary  lubrication  JJL  •=  0.070  to  0.080; 

for  tJiorougJi  lubrication  /*  =  0.054.  .. 

Efficiency — Mechanical  Advantage. — In  a  machine  we  have  in  general  a  "moving 
force"  F  acting  at  some  point  called  the  "point  of  application,"  and  at  another  point, 
called  the  "  working  point,"  we  have  a  "  useful  resistance"  F'  overcome  through  a  certain 
distance.  If  there  is  no  friction,  the  rate  of  work  of  the  moving  force  must  always  equal 
the  rate  of  work  of  the  resistance.  Owing  to  friction  it  must  always  be  greater. 

The  ratio  of  the  rate  of  work  of  the  useful  resistance  to  the  rate  of  work  of  the  moving 
force  is  called  the  EFFICIENCY  of  the  machine. 

It  must  always  be  a  fraction  less  than  unity,  and  it  approaches  unity  the  more  perfect 


268  KINETICS  OF  A  PARTICLE,  [CHAP.  IV. 

the  machine  and  the  less  the  friction.      If  we  denote  it  by  e,  and  let  v  be  the  velocity  of 
the  moving  force  F,  and  v'  the  velocity  of  the  resistance  F',  we  have 


If  there  is  no  friction,  e=  I   and  Fv'  =  F'v.     The  ratio  e  ^  is  called  the  MECHANICAL 

ADVANTAGE  of  the  machine. 

If  F'  is  greater  than  F,  v'  must  be  less  than  v  in  nearly  the  same  proportion,   or,  if 

/'"        v 
friction  is  disregarded,  in  exactly  the  same  proportion,  that  is,  —  =  ^  ,. 

Hence  the  familiar  maxim  that  "  What  is  gained  in  force  is  lost  in  speed." 

Examples.  —  (i)  A  body  of  So  pounds  mass  is  projected  along  a  rough  horizontal  plane  with  a  speed  of  50  ft. 
per  sec.  It  slides  155^8  ft.  in  coming  to  rest.  Find  the  coefficient  of  kinetic  sliding  friction,  the  t  etarding 
force  of  friction,  and  the  work  done  against  friction  in  coming  to  rest. 

ANS.  u  =  ^—,  or,  if  g  =  32.2  ft.-per-sec.  per  sec.,  u  —  0.25.     Retarding  force  of  friction  is  20  Ibs.;  work 

done,  3105.6  ft.-lbs. 

(2)  A  tody  of  So  pounds  mass  is  dragged  along  a  rough  horizontal  plane  by  means  of  a  mass  of  186  pounds 
attached  to  a  string  passing  over  a  pulley  (page  264).     //  is  observed  to  slide  to  feet  in  the  first  second  starting 
from  rest.     Disregarding  rigidity  of  string  and  mass  and  friction  of  string  and  pulley,  find  the  coefficient  of 
kinetic  sliding  friction,     (g  =  J2.) 

ANS.  u  —  0.25. 

(3)  A  body  placed  upon  a  rough  inclined  plane  whose  height  is  i  foot  and  base  16  inches  is  observed  to  slide 
6.4  inches  in  I  he  first  second  starting  from  rest.     Find  the  coefficient  of  friction,     (g  =  32.} 

ANS.  u  =  —  kinetic;  u  =  0.75  static. 
3° 

(4)  A  friction  brake  of  'm  =  /j  Ibs.  mass,  and  length  of  4  feet,  is  balanced  on  a  rotating  shaft  of  radius 
r  =  6  inches,  by  masses  of  Q  =  fo  Ibs.  and  P  —  jo  Ibs.  to  oz.     Find  the  coefficient  of  kinetic  friction  and  the 
friction.     Also,  if  the  shaft  makes  60  revolutions  per  minute,  find  the  rate  of  work  of  the  friction. 

ANS.  u  =  0.07,  F  =  2.5  Ibs.     Rate  of  work  of  friction   -  7.854  ft.-lbs.  per  sec.,  or  0.01428  horse-power. 

(5)  A  screw  of  radius  r  =  /  inch  is  acted  upon  by  a  force  of  P  —  -  Ib.  with  a  constant  lever-arm  of  I  —  i  ft- 

and  overcomes  a  resistance  of  Q  =  j  Ibs.  If  the  angle  of  the  thread  is  a  =45°,  find  the  coefficient  of  kinetic 
sliding  f  fiction  if  the  number  of  revolutions  per  minute  is  60.  Also  find  the  efficiency,  and  the  acceleiation  of 
P.  Disregard  the  mass  of  the  screw,  and  take  g  =  32  J  ft.-per-sec.  per  sec. 

ANS.  Let  P  be  the  force  applied  at  the  end  of  the  arm  /,  and  let  the  radius  of  the  screw  be  r,  the  pitch 
/,  and  the  resistance  Q. 

If  N  is  the  sum  of  the  normal  pressures  and  a  the  inclination  of  the  thread  to  the  horizontal,  we  have 

J  _  .       N  =  -  '    -  ,  and  the  friction  F=  uN  —  ~—^,  where  u  is  the  coefficient  of  friction. 
cos  a  cos  a 

Let/  be  the  acceleration  of  P.  Then  the  moving  force  is  P(g  —  /)  poiiiidiils. 
If  s  is  the  distance  passed  through  by  P  in  any  time  /,  then  the  work  of  the  moving 
force  is 

P(g  —  f)s  ft.-poundals. 

*/ 
The  resistance  Q  is  overcome  through  the  distance  -—.s.      The  work  of  over 

coming  the  resistance  is  then 

_  ^  .  J^r  ft.-poundals. 

The  friction  is  overcome  through  the  distance  •  •  .  —  —  -  .    The  work  of  overcoming  the  friction  is  then 

uQ  rs 


COS  a    / cos  a 


CHAP.  IV.]  KINETIC  FRICTION-EXAMPLES.  269 

The  minus  sign  is  used  because  work  is  done  against  friction  and  the  resistance. 

The  work  of  P(g  — /)  must  be  equal  and  opposite  to  the  work  done  against  friction  and  the  resistance. 
Hence  the  algebraic  sum  must  be  zero,  or 


From  this  we  have,  since  -^-—  =  tan  a  and  f  =  — , 


^ (I) 

and  from  (i),  for  the  coefficient  of  kinetic  friction, 

Pt          ,  /  2SPI         \ 

>«  =  77- cos   a  —  sin  acos  a  i  +  —  (2) 

Or  \       gfQr  tan  a) 

For  the  efficiency  we  have 


2TCl  I 

7. .    .    .     (3) 


2jcl        /cos*  a  sin  a  cos  a 

If/  =  o,  we  have  equilibrium,  and  from  (i)  we  have  in  this  case 


or  the  same  as  already  found,  ex.  (2),  page  237. 

In  this  case  (2)  becomes  the  coefficient  of  static  friction, 

PI 
H  =  -Q-  cos  a  —  sin  a  cos  a. 

We  see  from  (3)  that  the  efficiency  is  a  maximum  when  sin  a  cos  a  is  a  maximum,  or  when  sin  a  =  cos  a 
or  a  =  45°. 

If  »  is  the  number  of  revolutions  per  minute,  the  distance  s  described  in  one  minute  is  2itln.  We  have 
then 

•25  _          ytln          _  itln 
gf  ~~  60  x  60  x  g  ~ 

Inserting  in  these  equations  the  values  /=  i  ft.,  r  =  —  ft.,  P  =-lb.,  Q  =  5  Ibs.,  a.  =  45°,  n  =  60, 
g-  =  32^  ft.-per-sec.  per  sec.,  we  have 

/i  =  0.096,     e  =  0.84,   /  =  0.007,    g  =  0.225  ft.-per-sec.  per  sec. 

(6)  A  train  runs  on  a  horizontal  track  with  the  speed  v\  ,  and  by  the  application  of  brakes  to  the  driving- 
wheels  of  the  locomotive  the  speed  is  reduced  to  the  speed  v.  Find  the  distance  and  time  of  running  during  the 
reduction  of  speed,  disregarding  all  resistances  other  than  those  due  to  the  action  of  the  brakes. 

ANS.  Let  m  be  the  mass  of  the  train  in  pounds,  Vi  the  initial  and  v  the  final  speed  in  feet  per  second, 
s  the  distance  in  feet,  and  /  the  corresponding  time  in  seconds. 

Let  n  be  the  number  of  driving-wheels  braked,  and  R  the  pressure  of  each  braked  wheel  on  the  rails  in 
pounds. 

Let  n>,  be  the  coefficient  of  kinetic  sliding  friction,  and  /^  the  coefficient  of  static  sliding  friction. 

ist.  Let  tJie  brakes  be  set  so  that  the  wheels  do  not  tttrn.  In  this  case  the  retarding  force  due  to  friction 
is  n/^kK  pounds,  or  njnkRg  poundals.  We  have  then  for  the  retardation/ 


mf  =  nUkRg,    or   /  = 
From  page  92,  for  uniformly  retarded  motion  the  distance  described  is 


2/ 


2 7°  KINETICS  OF  A  PARTICLE.  [CHAP.  IV. 

2d.  Ltt  the  brakes  be  set  so  that  the  wheels  are  just  on  the  point  of  slipping.     In  this  Case  the  retarding 
force  due  to  friction  is  nptKg  poundals.    We  have  then 


or 
Hence  the  distance  described  is 


If  the  train  is  brought  to  rest,  we  have  i<  =  o  in  these  equations. 

Now  the  coefficient  u,  for  static  sliding  friction  is  always  less  than  the  coefficient  «*  for  kinetic  sliding 
friction  (page  267). 

We  see.  then,  that  the  train  unit  be  stopped  in  the  least  distance  whtn  the  drakes  are  applied  so  that  the 
wheels  are  just  on  the  point  of  slipping  but  do  not  slip. 


CHAPTER    V. 

KINETIC  AND  POTENTIAL  ENERGY.     LAW  OF  ENERGY.    CONSERVATION   OF  ENERGY. 

EQUILIBRIUM    OF  A   PARTICLE. 

Kinetic   Energy. — Let  a  particle  of  mass  m  move  in  any  path  and  have  the  initial 
velocity  vl  at  Pl. 

In  the  indefinitely  small  timer  let  the  particle  move  to  P2, 
so  that  Pl  and  P2  are  consecutive  points,  and  let  the  velocity  at    . ^L^P*_ 


P2  be 


Then  the  tangential  acceleration  is 


and  the  tangential  force  is 


v  -4-  v. 
The  mean  speed  is  —  -  ,  and  hence  the  distance  described  is 


Let  F  be  the  force  on  the  particle.  This  force  can  be  resolved  into  the  tangential  com- 
ponent Ft  already  found,  and  the  force  Fp  in  the  direction  of  the  radius  of  curvature.  The 
work  of  the  force  F  during  the  passage  from  Pl  to  P2  is  equal  to  the  sum  of  the  works  of  the 
components  (page  215).  But  since  the  path  ds  is  at  right  angles  to  Fp,  the  work  of  Fp  is 
zero,  and  hence  the  work  of  F  is  the  same  as  the  work  of  Ft. 

We  have  then  for  the  work  done  in  passing  from  Pl  to  P2  when  Pl  and  P2  are  consecutive 


In  the  same  way,  in  passing  from  P2  to  Ps  when  P2  and  P3  are  consecutive 


z 
For  the  next  two  consecutive  points 


*~  2 

and  so  on. 

271 


272  KINETICS  OF  A  PARTICLE.  [CHAP.  V. 

If  we  add  together  all  these  works  and  denote  the  final  velocity  at  P  by  v  we  have  for 
the  entire  work  W,  if  sl  and  s  are  the  distances  of  Pl  and  P  measured  along  the  path  from 
any  point  O  of  the  path, 


(i) 


If  the  final  velocity  v  is  greater  than  the  initial  velocity  vlt  work  is  done  on  the  particle 
in  giving  it  increased  velocity  and  is  positive.  If  vl  is  greater  than  v,  work  is  done  against  the 
particle  and  is  negative.  In  the  one  case  we  have  a  tangential  force  Ft  acting  in  the  direc- 
tion of  motion.  In  the  other  case  Ft  is  opposite  to  the  direction  of  wotion. 

We  see  that  equation  (l)  is  independent  of  the  time  and  path. 

Hence,  whatever  the  time  or  path,  and  however  the  tangential  force  may  vary,  the  work 
done  in  giving  a  particle  of  mass  m  an  increase  of  velocity  (v  —  v^)  is  equal  to  one  half  the 
product  of  the  mass  m  and  the  difference  of  the  squares  of  the  final  and  initial  velocities. 

If  the  initial  velocity  is  zero,  then  vl  =  o  and 


is  the  work  (positive)  done  in  giving  a  particle  of  mass  m  the  velocity  v  starting  from  rest, 
no  matter  what  the  time  or  path,  or  however  the  accelerating  force  may  vary. 
Conversely, 

W  =  -  -mv* 

is  also  the  work  (negative)  done  on  the  particle,  or  the  work  which  a  particle  of  mass  M 
moving  with  a  velocity  v  can  do  while  being  brought  to  rest  by  opposing  force,  no  matter 
what  the  time  or  path  or  however  the  retarding  force  varies. 

The  work  which  a  particle  or  body   is  capable  of  doing  is  called  its  ENERGY.      Since 

-  mi?  is,  then,  the  work  which  a  particle  of  mass  m  and  velocity  v  is  capable  of  doing  by  reason 

of  its  velocity,  we  call  it  the  KINETIC  ENERGY  of  the  particle. 

We  denote  kinetic  energy  by  the  letter  K.     If  then  the  initial  velocity  is  *',,  the  initial 
kinetic  energy  is 

*,  =  \rnvf. 

If  the  final  velocity  is  v,  the  final  kinetic  energy  is 

i 
«K  =  -mv*. 

2 

The  work  given  by  equation  (i)  can  then  be  written 

^m/Js  =  3C  —  3C,  r:=  -m(ir*  -  Vfi (2) 

That  is,  the  work  (positive]  done  on  the  particle  is  equal  to  the  gain  of  kinetic  energy, 
and  inversely  the  work  (negative)  done  by  the  particle  is  equal  to  the  loss  of  kinetic  energy. 


CHAP.  V.]  KINETIC  ENERGY  OF  A  ROTATING  BODY.  273 

Kinetic  Energy  of  a  Rotating  Body.*  —  Let  a  body  have  the  angular  velocity  a.   about 
any  axis.     Then  the  velocity  of  any  particle  at  a  distance  r  from  the  axis  is  v  =  ra>.     The 

kinetic  energy  of  the  particle  is  then  —  mr2^.      Since  all  the  particles  have  the  same  angular 
velocity  about  the  axis,  we  have  the  kinetic  energy  of  the  entire  body, 

3f=—  GT^wr*. 

But  2mr*  is  the  moment  of  inertia  /'  of  the  body  relative  to  the  axis  (page  3  1).     We 
have  then  for  the  kinetic  energy  of  a  rotating  body 


(3) 


Since  the  motion  of  a  body  at  any  instant  consists  in  general  (page  145)  of  translation 
and  rotation  about  an  axis  through  the  centre  of  mass,  we  have  for  the  kinetic  energy  of  a 
rigid  body  in  general 


(4) 


In  all  equations  5C  is  given  in  ft.-poundals  if  we  take  m  in  Ibs.  and  v  in  ft.  per  sec.    For 
foot-pounds  divide  by  g. 

Examples.—  (  i)  A  fly-wheel  has  a  mass  of  30  tons  and  radius  of  8  ft.  What  work  can  it  do  in  coming  to 
rest,  considering  it  as  a  disc,  if  it  is  making  20  revolutions  per  minute  f 

ANS.  For  a  disc  we  have  for  a  principal  axis  through  the  centre  of  mass  (page  45)  /=  -  .  r*.  The 
angular  velocity  is  oo  =  2°  *  2g.  The  kinetic  energy  is  then 

3C  =  *£  .  **L  =  30  x-  2240^64  _  4x9.87  =  47l6$44  ft._pounda]S) 

49  4  9 

or,  taking^-  =  32  ft.-per-sec.  per  sec., 

3C  =  147392  ft.-pounds. 

(2)  In  the  preceding  example  suppose  the  fly-wheel  starts  from  rest  and  acquires  its  speed  in  one  minute 
iinder  the  action  of  a  constant  force  applied  at  the  extremity  of  a  crank  18  inches  long.  Find  this  force. 

ANS.  The  mean  angular  speed  is  10  revolutions  per  minute.  The  distance  s  passed  through  by  the 
constant  force  is  then,  since  the  length  /  of  the  crank  is  \\  ft., 

S=  IOX27T/  =  307T  ft. 

The  kinetic  energy  is  as  before  147392  ft.-pounds.     Hence  the  force  is 

I473Q2 


6  pounds. 
10  x  27t/          y>it 

(3)  A  ball-player  catches  a  ball  moving  with  a  velocity  of  jo  ft.  per  sec.     Tkt  mass  of  the  ball  is  S\  oz.     If 
the  space  in  which  the  ball  is  brought  to  rest  is  6  inches,  find  the  average  opposing  pressure  and  the  time  of 

stoppage. 

ANS.  We  have  Fts  =  ^  mv*.  or  Ft  =  "—  =  IIX^^°X2  =  859!  poundals.    Taking  g  =  32  ft.-per-sec. 

per  sec.,  we  have  Ft  =  26.85  pounds.    The  mean  velocity  is  —  =  25  ft.  per  sec.  and  *  =  —  /,  or  t—  —  =  —  sec. 

(See  example  (i),  page  258.) 

(4)  If  an  ounce  bullet  with  a  velocity  of  Soo  ft.  per  sec.  strikes  a  man  in  "  bullet-proof"  armor  and  the 
bullet  is  brought  to  rest  in  a  space  of  six  inches,  find  the  average  pressure  and  time  of  stoppage,    (g  =  32.) 

ANS.  Pressure  =  1250  pounds;  time  =  ^  sec. 

*  This  article  properly  belongs  to  the  discussion  of  the  kinetics  of  a  rigid  body  and  not  of  a  particle.     It  Is, 
however,  such  a  direct  result  of  the  preceding  that  we  give  it  here. 


a  74 


KINETICS  OF  A  PARTICLE.  [CHAP.  V. 


(5)  Find  the  tension  of  a  rope  which  pulls  a  car  of  S  tons  up  a  smooth  incline  of  i  in  5  and  causes  a  uni- 
form acceleration  of  3  ft.-per-sec.  per  sec.     If  the  rope  breaks  when  the  speed  is  48  ft.  per  sec. ,  how  far  will  the 
car  continue  to  move  up  the  incline  f     (f  =  32.} 

ANS.  5264  pounds;  1 80  feet. 

(6)  A  bullet  of  a\  os.  leaves  a  gun  with  a  velocity  of  iSSoft.  per  sec.     The  length  of  barrel  is  a\  ft.     Find 
the  average  force  of  the  powder.     ( g  =  3*. ) 

ANS.  2346  pounds. 

(7)  A  train  of  200  tons,  starting  from  rest,  acquires  a  speed  of 40  miles  an  hour  in  three  minutes.     Find 
the  effective  moving  force,  assuming  it  uniform. 

ANS.  2.03  tons. 

(8)  An  engine  exerts  on  a  car  weighing  20000  Ibs.  a  net  pull  of  2  !bs.  per  ton  of  2000  Ibs.      Find  the 
energy  of  the  car  after  going  2\  miles.     If  shunted  onto  a  level  side  track  where  friction  is  10  pounds  per  ton, 
how  far  will  it  run  f    If  this  side  track  has  a  one  per  cent  grade,  how  far  will  it  run  ? 

ANS.  264000  ft.-lbs.;  one-half  mile;  one  sixth  mile. 

Potential  Energy. — Let  a  particle  be  constrained  to  move  in  any  path  from  />,  to  P. 
Let  F  be  the  resultant  of  all  those  forces  acting  upon  the  particle  which  depend  solely  upon 

the  position  of  the  particle. 

For  an  indefinitely  small  displacement  ds  in  the 
path  F  can  be  considered  uniform,  and  its  work  is 
F .  dp,  where  dp  is  the  component  displacement  along  F. 
We  can  resolve  F  into  a  tangential  component  Ft 
and  a  normal  component  ./y  In  the  displacement  ds  from  Pl  to  Pv  the  work  of  Ff  is  zero. 
Since  the  work  of  the  resultant  is  equal  to  the  algebraic  sum  of  the  works  of  the  components 
(page  2  15),  we  have 

F .  dp  -  Ft  .  ds. 

This  can  also  be  proved  as  follows:  If  0  is  the  angle  of  F  with  the  tangent,  we  have 
dp  =  ils  cos  t)  and  F  .  dp  —  F  .  ds  cos  0.  But  /''cos  6  =  Ft.  Hence 

F  .  dp  =  Ft  .  ds. 

The  force  F  and  hence  Ft  depends  solely  upon  position.  It  is  in  general  variable  and 
changes  in  magnitude  and  direction  as  the  particle  changes  its- position.  For  any  two 
consecutive  points  we  have,  however,  F  uniform. 

If,  then,  j,  is  the  distance  of  7^,  and  s  of  P,  measured  along  the  path  from  any  point  O,> 
we  have  for  the  entire  work  on  the  particle  in  passing  from  Pl  to  P,  due  to  change  of 
position  only, 

^F.  dp  =  2*Ft  .ds (5) 

The  work  which  a  particle  is  thus  capable  of  doing  by  reason  of  change  of  position  only 
is  called  its  POTENTIAL  ENERGY. 

The  potential  energy  of  a  particle  or  body,  then,  is  the  work  it  can  do  by  reason  of 
change  of  position  or  configuration  under  the  action  of  forces  which  depend  solely  upon  such 
change. 

Thus  a  mass  suspended  at  a  distance  above  the  earth  can  do  work  if  released,  by  reason 
of  the  force  of  gravity,  which  is  a  force  depending  solely  upon  position.  This  work  is 
potential  energy.  A  bent  spring  can  do  work  when  released,  by  reason  of  the  elastic  forces 
between  its  particles,  which  depend  solely  upon  configuration.  This  work  is  potential 
energy. 


CHAP.  V.] 


TOTAL  ENERGY. 


275 


We  denote  potential  energy  by  the  letter  2.  The  initial  potential  energy  of  the  particle 
at  /\  in  the  figure,  page  274,  is  then 

\  =  2^  F .  dp  =  y '  Ft  .  ds. 
The  final  potential  energy  at  P  is 

2  =  2s F.  dp  =  2* Ft.  ds. 
The  work  done  on  the  particle  due  to  change  of  position  only  is  then 

\-*  =  2tF.dp-^Ft.dS '    .     .     (6) 

That  is,  the  work  (positive)  done  on  the  particle  by  reason  of  change  uf  position  only  is 
equal  to  the  loss  of  potential  energy,  or,  inversely,  the  work  (negative)  done  by  the  particle  by 
reason  of  change  of  position  only  is  equal  to  the  gain  of  pi  tential  energy. 

Total  Energy. — The  total  energy  of  the  particle  at  any  instant  is  the  sum  of  its  poten- 
tial and  kinetic  energy  at  that  •  instant.  Let  us  denote  the  total  energy  by  &.  Then  the 
initial  energy  is  t 


and  the  final  energy  is 


=  %  -f  -mv2   = 


-f-  3C. 


Law  of  Energy. — When  a  particle  or  body  moves  it  generally  encounters  resistance  due 
to  friction,  resistance  of  the  air,  etc.  Such  forces  do  not  depend  upon  position  solely, 
but  upon  velocity  or  upon  the  character  of  the  surfaces  in  contact,  upon  pressure  between  these 
surfaces,  etc.  They  always  resist  change  of  motion,  and  we  call  them  therefore  resistances. 
Let,  then,  in  be  the  mass  of  a  particle  constrained  to  move  in  any  path  from  Pl  to  P,  and 
let  F  be  the  resultant  of  all  those  forces  acting  on  the  par- 
ticle which  depend  solely  upon  its  position,  N  the  normal 
pressure  of  the  path  upon  the  particle  and  R  the  resultant 
of  all  the  resistances.  We  can  resolve  Finto  Ft  and  Fp  tan-  o 
gent  and  normal  to  the  path.  So,  also,  R  can  be  resolved 

into  Rt  and  Rp,  and  Rt  is  always  opposite  to  the  direction 

of  motion.      These  are  impressed   forces.      If  the  particle 

has  the  acceleration  /,  this  can  be  resolved  into  ft  and  fp,  and  the  components  of  the  effective 

force  ;;//~are  then  mft  and  mfp. 

Now  by*  D'Alembert's  principle  (page  242)  the  impressed  and  reversed  effective  forces 

•.an  be  treated  as  a  system  of  forces  in  equilibrium,      Hence 


Ft  -  Rt  -  mft  =  o. 


Multiplying  by  ds,  we  have 


Ftds  =  mft  . 


276  KINETICS  Of  A  PARTICLE.  [CHAP.  V 

If,  then,  5,  is  the  distanceof  />lf  and  s  olP,  measured  along  the  path  from  any  point  O,  we 
have  for  the  entire  work  done  by  F  irom  Pl  to  P,  whatever  the  path,  and  however  the  forces 
may  vary  in  magnitude  or  direction, 


But  we  have  seen  from  equation  (6)  that 

2'tF.  dp  =  X,^  .  «fc  =  2, 

the  loss  of  potential  energy. 

We  have  also  seen  from  equation  (2),  page  272,  that 


or  the  gain  of  kinetic  energy.      We  have  then 

51—  2=3C  —  3Cj+  2tlRt  .ds  .........     (8) 

That  is,  the  lass  of  potential  energy  is  equal  to  the  gain  of  kinetic  energy  plus  the  work  of 
overcoming  all  resistances. 

Since,  as  we  have  seen,  we  can  write 

Sj  =  «B,  -f  3C,  ,          &  =  2  +  M. 
Equation  (8)  can  be  written 

&l-&  =  2fRt.ds  ...........     (9) 

That  is,  the  loss  of  energy  is  equal  to  the  work  done  in  overcoming  resistances. 
This  is  known  as  the  law  of  energy.  -. 

Conservation  of  Energy.  —  If  there  are  no  resistances,  equation  (9)  becomes 

§j  —  §  =  o.     or     2,  -  $  =  K  -  5C,  ........     (10) 

That  is,  if  there  are  no  resistances,  but  only  forces  which  depend  solely  upon  position,  there 
is  no  loss  of  energy,  or  the  loss  of  potential  is  equal  to  the  gain  of  kinetic  energy,  and  con- 
versely, the  gain  of  potential  is  equal  to  the  loss  of  kinetic  energy. 

This  is  called  the  law  of  conservation  of  energy,  and  hence  forces  which  depend  solely 
upon  position  are  called  conservative  forces,  because  for  them  the  law  of  conservation  holds. 
The  force  of  gravity  upon  a  particle  depends  solely  upon  the  position  of  the  particle  and  is 
therefore  a  conservative  force.  So  is  the  elastic  force  of  a  spring  which  depends  upon  con- 
figuration only.  So,  also,  are  the  forces  of  nature  generally.  For  such  we  have  the  law  of 
conservation  of  energy  as  given  by  equation  (10). 

Forces  which  do  not  depend  upon  position  are  called  non-conservative  forces.  The 
resistance  due  to  friction,  and  in  general  all  resistances  to  motion,  do  not  depend  upon 
position  and  are  therefore  non-conservative.  The  resultant  of  such  forces  always  acts 
opposite  to  the  direction  of  motion.  For  such  we  have  the  law  of  energy  as  given  by  equa- 
tion (8). 


CHAP.  V.]  EQUILIBRIUM  OF  A  PARTICLE.  277 

Equilibrium  of  a  Particle. — If  a  particle  is  at  rest  under  the  action  of  forces,  it  is  said 
to  be  in  static  equilibrium.  If  it  moves  with  uniform  speed  in  a  straight  line,  that  is,  with 
uniform  velocity,  it  is  said  to  be  in  molecular  equilibrium. 

Let  Fx,  Fy,  Fz  be  the  components  in  any  three  rectangular  directions  of  all  the  con- 
servative forces  acting  upon  a  particle ;  let  Rt  be  the  resultant  non-conservative  force,  and 
jRA,  Ry,  Rz  its  components. 

Then  for  any  possible  indefinitely  small  displacement  ds,  real  or  virtual,  we  have  the 
work  of  the  conservative  forces,  or  the  loss  of  potential  energy 

2j  -  2  =  Fxdx  +  Fydy  -f  Fsdz, 
and  by  the  law  of  energy,  equation  (8), 

$!  -  2  =  Fxdx  +  Fydy  +  Fsdz  =   -m(v*  -  ^2)  +  Rtds. 
But  Rtds  —  Rxdx  4-  Rvdy  -f  Rzdz.      Hence 

2,  -  *  -  Rtds  =  (Fx  -  Rx}dx  +  (Fy  -  Ry}dy  +  (F,  -  R^dz  =  ±  m(v>  -  vf). 

For  static  equilibrium  v  =  o,  7^  =  o,  and  for  molecular  equilibrium  v  =  vr  Hence  in 
all  cases  of  equilibrium,  static  or  molecular, 

(Fx  -  Rxyx  +  (Fy  -  Ry}dy  +  (F.  -  R^dz  =  o. 
Since  dx,  dy,  dz  are  not  zero,  we  must  have 

Fx  -  Ry  =  o,         Fy  -  Ry  =  o,         F2  -  R2  —  o. 

That  is,  in  all  cases  of  equilibrium,  static  or  molecular,  the  forces,  conservative  and  non-con- 
servative, acting  upon  the  particle  must  constitute  a  system  of  forces  in  equilibrium. 

Since  in  all  cases  of  equilibrium,   static  or  molecular,  —  m(i?  —  z^2),  or  the  change    of. 

kinetic  energy  during  any  indefinitely  small  displacement  is  zero,  the  kinetic  energy  is  zero 
for  static  equilibrium,  and  either  a  maximum  or  a  minimum  when  the  particle  is  in  molecular 
equilibrium. 

Stable  and  Unstable  Equilibrium  of  a  Particle. — The  equilibrium  of  a  particle  is  said 
to  be  stable  if,  when  at  rest  or  when  supposed  to  be  at  rest,  it  would  return  after  any  possi- 
ble indefinitely  small  displacement  to  its  original  position.  Now  in  thus  returning  work  must 
be  done  by  the  conservative  forces,  and  hence  potential  energy  lost.  In  stable  equilibrium, 
then,  the  potential  energy  is  a  minimum. 

If  the  particle  would  not  return  to  its  original  position,  the  equilibrium  is  said  to  be 
unstable.  In  such  case  work  must  be  done  against  the  conservative  forces  in  order  to  bring 
it  back,  and  potential  energy  would  be  gained.  In  unstable  equilibrium,  then,  the  potential 
energy  is  a  maximum. 

If  potential  energy  is  neither  lost  nor  gained,  the  equilibrium  is  said  to  be  indifferent, 
and  the  particle  in  its  new  position  is  still  in  equilibrium. 

Now  when  potential  energy  is  lost,  kinetic  energy  is  gained,  and  conversely. 

Hence  if  a  particle  is  in  equilibrium,  static  or  molecular,  all  the  acting  forces  must  con- 
stitute a  system  of  forces  in  equilibrium.  If  the  potential  energy  is  a  minimum,  the  equilib- 


278 


KINETICS  OF  A  PARTICLE. 


[CHAP.  V. 


rium  is  stable  and  the  kinetic  energy,  if  any,  a  maximum.  If  the  potential  energy  is  a 
maximum,  the  equilibrium  is  unstable  and  t/ie  kinetic  energy,  if  any,  a  minimum.  If  the 
potential  energy  does  not  change,  the  equilibrium  is  indifferent  and  the  kinetic  energy,  if  any, 
does  not  change. 

Illustration. — Take  the  case  of  a  pendulum.      Let  /  be  the  length  and  m  the  mass  of 
the  bob.     Disregard  the  mass  of  the  string. 

When  the  bob  is  at  the  lowest  point  the  potential  energy  is 
mg  .  //.  For  every  possible  displacement  h  increases  and  the  poten- 
tial energy  increases.  The  potential  energy  is  then  a  minimum. 
This,  then,  is  a  position  of  stable  equilibrium.  If  the  bob  is  at  rest, 
it  is  in  stable  static  equilibrium.  If  the  pendulum  swings,  when  it 

arrives  at  its  lowest  point   the    kinetic  energy  is  a  maximum  and 

the  potential  a  minimum,  and  at  the  instant  it  is  in  stable  kinetic  equilibrium. 

When  the  bob  is  at  the  highest  point  the  potential  energy  is  a  maximum.  This,  then, 
is  a  position  of  unstable  equilibrium.  If  the  bob  is  at  rest,  it  is  in  unstable  static  equilibrium. 
If  in  motion,  when  it  arrives  at  the  highest  point  the  kinetic  energy  is  a  minimum  and  the 
potential  a  maximum,  and  at  this  instant  it  is  in  unstable  kinetic  equilibrium. 

Change  of  Potential  Energy. —In  order  to  apply  equations  (8)  or  (10)  it  will  be  con- 
venient to  find  the  value  of  the  change  of  potential  energy  5,  —  2  in  special  cases. 

(a)  FORCE  UNIFORM. — If  a  particle  is  acted  upon  by  a  uniform  conservative  force  F,  the 
work  done  during  a  change  of  position   from   Pl  to  P  is,  by   equation  (6), 
page  275, 

^  -  2  =  F.p=  F.  <tcos  ft (i) 


where  d  is  the  displacement  and  H  the  angle  of  F  with  the  displacement.      If 
p  and  Fare  in  the  same  direction,  Fp  is  positive   and  we  have  loss  of  poten- 
tial energy.      If  p  and  F  are  in  different  directions,   \ve  have  gain  of  potential  energy. 
(b)  CENTRAL  FORCE  — Let  O  be  the  centre  of  force. 

p  Let  />,,  P2,  P3  .  .  .  Pbc  any  path  from  I\  to  P,  and  let  the  posi- 

tions   />,,    7>2,    />,,    etc.,  be    consecutive.      Draw    OPl ,  OP2,  OP3,  etc., 
d/\  \        and   with  O  as  a  centre   describe  arcs  of- circles   through  P2,  P3,  etc., 

intersecting  OP}  at  b,  c,  d,  etc. 

,p  The  force  Fz  at  P2  acting  towards  O  may  be  considered  uniform 

/  *    for  the  indefinitely  small  displacement  P2  to  P3. 
P  The  work,  then,  from  P2  to  Ps  is 

F2  X  P2n  =  F2  X  be. 

Every  element  of  the  path  may  be  treated  in  the  same  way. 
Thus  the  work  from  P{  to  P2  is 

^,  X  P,b; 
from  PS  to  P, 

F,  X  cd, 

and  so  on.     The  entire  work  from  />,  to  P  is  then 

F,  X  Pvb  +  F2  X  be  +  F^  X  cd-\  etc.  =  2F.  dr, 
where  dr  is  the  elementary  distance  along  the  radius  vector  OP{  between  any  two  arcs. 


CHAP.  V.]  CHANGE  OF  POTENTIAL  ENERGY.  279 

The  work,  then,  necessary  to  move  the  particle  from  Pl  to  P  by  any  path,  under  the 
influence  of  a  central  force  always  directed  towards  O,  is  equal  to  that  necessary  to  move  it 
from  Pl  to  d  in  the  straight  line  OPr 

This  work  is  independent  of  the  path  and  depends  only  upon  the  initial  and  final  posi- 
tions and  the  magnitude  of  the  force.  If  this  force  depends  only  upon  position,  this  work 
is  the  loss  of  potential  energy  and  we  have  in  general 


...     (2) 

If  the  central  force  is  opposite  to  dr,  we  have  gain  of  potential  energy,  or 

<%  -  \=^F  .dr. 

(c]  CENTRAL  FORCE  CONSTANT.  —  If  the  magnitude  of  the  central  force  Fis  constant, 
we  have,  from  (2), 

V-.  *  =  F2dr  =  F(r,  -  r),      .      .      .  '  .....     (3) 

where  rl  is  the  initial  and  r  the  final  radius  vector  from  O. 

If  rl  is  less  than  r,  we  have  gain  of  potential  energy,  or 

/ 

%  -  \  =  F(r  -  r,}. 

(d]  CENTRAL  FORCE  PROPORTIONAL  TO  DISTANCE  FROM  CENTRE.  —  If  the  magnitude 
of  the  central  force  varies  directly  as  the  distance  from  the  centre  O,  let  FQ  be  its  known 
magnitude  at  a  given  distance  r0.     Then  the  force  at  any  other  distance  r  is  given  by 


F:F0::r:r0,     or     F  =  -  Fc 


o- 


We    have  then  at  Pl  the    force  Fl  =  —F0,  at  P2  the  force  F2  =  —  F0,  at  P3  the  force 

—  P0,  and  so  on. 

;'o 

The  average  force  between  Pl  and  P2  is  then  —  ---  2  =  —  .  -—  -  -  ,  and  the  work  from 

2  fg  2 

Pl  to  P2  is  then,  from  (2), 


%          2  ^      * 

In  the  same  way  we  have  for  the  Work  from  P2  to  P3 


and  so  on.      The  entire  work  from  Pl  to  P  is  then  the  loss  of  potential  energy  given  by 


where  rl  is  the  initial  and  r.z  the  final  radius  vector  from  O. 


2 So  KINETICS   OF  A  PARTICLE. 

If  rt  is  less  than  r,  we  have  gain  of  potential  energy,  or 


[CHAP.  V. 


2ro 

f>)  CENTRAL  FORCE  INVERSELY  PROPORTIONAL  TO  THE  SQUARE  OF  THE  DISTANCE 
FROM  CERTRE. — If  the  magnitude  of  the  central  force  varies  inversely  as  the  square  of  the 
distance  from  the  centre,  let  F0  be  its  known  magnitude  at  a  given  distance  r0.  Then  the 
force  F  at  any  point  is  given  by 

r*  r* 

We  have  then  at  Pl  the  force  Fl  =  ^FQt  and  at  Pt  the  force  F2  =  ~FQ,  and  so  on. 

For  the  indefinitely  small   displacement  from  Pl  to  P2,  r,  and  r2  are  equal,   and  we  have 
r,2  =  r,r2  and  r22  =  rjv     Hence  the  force  at  Pl  can  be  written  Fl=  -2 — ?,  and  the  force  at 

n        77    __     0       0. 
2  i    •'j    —    ~ 

The  work,  then,  from  (3),  from  Fl  to  Ft  is 


From  P2  to  P3  we  have  in  the  same  way  the  work 


and  so  on.     The  entire  work  from  Pl  to  P  is  then  the  loss  of  potential  energy  and  is  given  by 


If  r,  is  less  than  r,  we  have  gain  of  potential  energy,  or 


Exam  pies.  -(  i) 


ervation  of  energy  to  a  falling  body. 
ANS.  Let   /«  he   the    mass  of  the  body,  v\    its  initial   and  v  its  final 
velocity.     Then  the  gain  of  kinetic  energy  is 


We  have  from  equation  (5),  for  central  force  varying  inversely  as  the  square 
of  ihe  distance,  the  loss  of  potential  energy. 

4,  -2  =  ^,'(--'-11 

\  r       r^' 

where  r,  is  the  initial  distance  and  r  the  final  distance  from  the  centre  of 
force,  and  F,  is  the  known  force  at  a  yiven  distance  r0.     In  the  present  case 


CHAP.  V.] 


ENERG  Y— EXAMPLES. 


281 


the  centre  of  force  is  the  centre  of  the  earth,  r0  is  the  radius  of  the  earth,  and  Fa  at  the  earth  s  surface  :s 
mg.     We  have  then 


But  ri  —  r  is  the  distance  described,  (j,  —  s\,  if  Si  is  the  initial  and  s  the  final  distance  from  the  centre  of 
the  earth. 

We  have,  by  the  principle  of  conservation  of  energy,  gain  of  kinetic  equals  loss  of  potential  energy,  or 

3C  -  3Ci  =  2,  - 


This  is  the  same  as  equation  (5),  page  113. 

If  we  suppose  the  force  to  be  uniform  and  equal  to  mg,  as  it  practically  is  at  the  surface  of  the  earth,  we 
have,  from  equation  (i),  page  278, 

Si  —  S  =  tng(si  -  s), 
and  hence 

m      .       ,  __     nx  ,  V 


—  v?  + 


-  s), 


which  is  equation  (7),  page  92. 

(2)  Apply  the  law  of  conservation  of  energy  to  a  projectile. 

ANS.   Let  the  force  be  vertical  and  equal  to  mg.     The  initial 

energy  g»  at  O  is  ^-.      The  energy  g  at  J°is  —  v1  kinetic  and  mgy 
potential.     If  there  is  no  loss  of  energy,  &i  =  £,  or 

»^=^  +         ,     or     v*  =  vS-*gy. 


This  is  equation  (11),  page  104. 

(3)  Apply  the  law  of  conservation  of  energy  to  the  simple  pendulum. 

ANS.  Let   the    velocity   at   P\    be   zero!,       Then    the    initial    energy  at  />,   is  £1  = 
mglj  _  /  cos  0,)  potential,  relative  to  A.     At  any  point  P  where  the  velocity  is  v  we  have 

the  energy  £,   ™^~  kinetic  and  mg(l  —  I  cos  0)  potential.       If  there  is  no  loss  of  energy 


=  @,  or 


Hence 


mg(l  -  I  cos  0.)  =  ~  +  mg(l—  /cos  9). 


s  9  —  cos  61). 


But  /(cos 9  -  cos9,)  is  the  fall  from  Pi  to  P.     Hence  the  velocity  is  the  same  as  for  a  particle  falling 
through  the  vertical  distance  from  Pi  to  P. 

(4)  Let  a  spring  whose  unstrained  length  is  AB  be  fixed  at  the  end  B  and  compressed  from  A  to  C,  where 
it  presses  against  a  body  of  mass  m.     Disregarding  the  mass  of  the  spring,  find  Jhe  motion  when  released. 

ANS.  Let  the  force  at  any  distance  x  from  A  be  F,  and  at  the  given 
distance  s  =  ACirom  A  be  Fi.     Then  we  have 


F  :  x  : :  F\  :  s,    or     F  =  — •*"• 
The  initial  energy  is  &!  =  FiS,  all  potential.     The  energy  g  at  .F  is 


Fx  potential  and  -  kinetic.     If  there  is  no  loss  of  energy, 


F, 


=  Fx  + 


mv* 


282 


KINETICS  OF  A  PARTICLE. 


[CHAP.  V. 


Substituting  the  value  of  Ft  we  have 


We  see  then,  from  page  126,  that  the  motion  is  simple  harmonic. 

(5)  A  vessel  containing  water  has  a  small  orifice  whose  centre  is  at  a  distance  h  below  the  surface.  The 
water  flows  in  at  top  with  a  vertical  velocity  vy,  and  the  top  area  of  cross-section  is  A.  The  water  flows  out  of 
the  orifice  with  a  velocity  v  in  any  direction,  and  the  area  of  the  orifice  at  right  angles  to  v  is  a.  If  water  flows 
in  as  fast  as  it  flows  out,  so  that  the  water-level remains  constant,  find the  theoretic  velocity  of  efflux  v,  disregard- 
ing friction  and  all  resistances. 

ANS.  The  quantity  of  water  flowing  in  per  second  is  Avy,  and  the  quantity  flowing  out  as  av.  By  the 
conditions,  for  constant  level  we  must  have 

Av3  =  av,    or    vy  =  -jv. 

The  initial  energy  &,  of  a  particle  of  mass  m  at  the  top  level  is  mgh 
potential  and  — -  kinetic,  or 

mvl 
§,  =  mgh  +  — 7. 

When  this  particle  reaches  the  orifice  its  energy  g  is  all  kinetic  and 
§  =  .     If  no  energy  is  lost,  Si  =  &,     or 


mgh  + 


mv\        mv* 


Substituting  the  value  of  vy,  we  have 


(I) 


If  a  is  very  small  compared  to  A,  we  can  practically  neglect  the  fraction  —^  relative  to  i,  and  we  then 


have 


The  theoretic  velocity  of  efflux  in  this  case  is  the  same  as  for  a  particle  falling  freely  through  the  distance  h. 

This  is  known  as  Torricelli's  principle. 

If  y  is  the  density  or  mass  of  a  cubic  unit  of  water,  in  a  very  small  lime  r  the  mass  discharged  is 

m  =  yavr. 
The  kinetic  energy  at  efflux  is  then  the  work  done  in  giving  the  mass  m  the  velocity  v,  or 

mv*       yav*r 
work=  —  =^-. 

If  the  mass  m  acquires  the  velocity  v  in  a  very  short  time  r  from  rest,  the  average  velocity  is  -   and  the 
distance  is  —  .     If  we  divide  the  work  by  this  distance,  we  have  for  the  uniform  force  /''during  the  short  time 


r  in  the  direction  of  v 
or,  from  (i), 


F  =  yav*. 


F  =    y^  \  poundals. 


For  F  in  gravitation  measure 


(3) 


CHAP.  V.] 

From  (3),  for  a  small  relative  to  A 


EXAMPLES. 


283 


F  =  iyah. 


or  the  weight  of  a  column  of  water  whose  base  is  the  area  a  of  the  orifice  and  whose  height  is  twice  the  distance  h. 
If  a  is  the  angle  of  v  with  the  horizontal,  we  have  for  the  horizontal  component  of  F 


Fx  =  2yah  .  cos  a 


(4) 


Since  action  and  reaction  are  equal,  this  is  the  horizontal  pressure  on  the  side  of  the  vessel  in  a  direction 
opposite  to  v  cos  a. 

We  can  also  obtain  (2)  directly  by  the  principle  of  impulse  (page  257).     Thus 

„  „       mv 

fv  =  mv,    or     f1  =  — . 

Substituting  m  =  yavr,  we  have  at  once 

F  =  yav\ 

(6)  In  the  preceding  example  let  the  vessel  move  horizontally  with  the  uniform  velocity  vx ,  while  the  water 
flows  in  at  the  top  with  a  vertical  velocity  vy ,  the  top  area  being  A,  and  is  discharged  with  the  velocity  z. 
making  an  angle  a.  with  the  horizontal,  the  area  of  orifice  at  right  angles  to  v  being  a. 

ANS.  We  have  for  constant  level,  as  before, 

Avy  =  av, 

and,  disregarding  the  motion  of  the  vessel,  we  have,  as  in  the 
preceding  example, 


=  mgh  + 


mv 

2 


and  therefore,  just   as  in   the  preceding  example,  we  have  for 
S,  =  8 

.        mvl        mv* 


and  for  the  velocity  of  efflux  relative  to  the  vessel. 


just  as  before. 

Now  the  absolute  velocity  at  A  is  given  by 


(1) 


Hence  the  total  energy  at  A  is 

&>  =  —    —  +  mgh  =  — (v'x  +  Vy)  +  mgh. 

The  absolute  velocity  at  a  is  given  by 

vl  =  Vx  +  v*  —  2wx  cos  a. 
Hence  the  total  energy  at  a  is 

is  = =  (7/1  +  V*  —   227Z/X  COS  a). 


(2) 


(3) 


Equation  (2)  gives  the  work  the  mass  m  at  A  can  do,  and  equation  (3)  the  work  this  mass  at  a  can  do 
after  leaving  the  vessel.  The  difference  ©i  —  §  is  then  the  work  done  upon  the  vessel.  If*  then  we  subtract 
(3)  from  (2)  and  reduce  by  (i),  we  obtain  for  the  work  done  upon  the  vessel 


work  =  mwx  cos  a. 


284  KINETICS  OF  A  PARTICLE.  [CHAP.  V. 

In  a  very  short  time  r  we  have  m  =  yavT,  where  y  is  the  density  or  mass  of  a   cubic  unit  of  water. 
Hence 

work  =  yav*vxT  cos  a (4) 

But  the  distance  in  the  time  r  is  vxr.     Hence,  dividing  (4)  by  vxr,  we  have  for  the  horizontal   force  FK 
on  the  vessel 

Fx  =  yav*  cos  a. 

This  is  the  same  result  as  obtained  in  the  preceding  example.     Thus,  if  we  put  v  —  ^zgh,  we  have 

FK  =  2yagh .  cos  a  poundals, 

or,  in  gravitation  measures, 

Fx  =  2yah  .cos. a, 
just  as  before. 

(7)  A  horizontal  stream  of  water  whose  cross-section  is  a  and  velocity  7/1  meets  a  surface  moving  in  the 
same  direction  with  a  velocity  v.     Find  the  pressure,  disregarding  friction  and  resistances. 

ANS.  Let   the   water   pass   off   the  surface  in   a   direction 
fi  '  making  the  angle  or  with  the  direction  of  motion.     In  any  small 

\a  yy  time  r  the  mass  of  water  discharged  is  m  =  y.iv^r,  where  y  is  the 
density  or  mass  of  a  cubic  unit  of  water.  The  initial  kinetic 
energy  of  m  is  then 

*        mv\*      yav*r 

>«-*  :~:     ~^~ 

The  velocity  of  the  water  as  it  leaves  the  surface  is  vt  —  v 
y>"v    relative  to  the  surface.     The  velocity  of  the  surface  is  v.     The 
resultant  velocity  is  then  given  by 

vr*  =  (t/i  —  z>)*  +  v*  +  2(vi  —  v)v  cos  a. 

The  final  kinetic  energy  of  the  mass  m  is  then 

,,,       m  yav\r  r 

&  =  — Vr '  =    ' [zv  —  2Wi  +  2V    +  2(V\  —  V)V  COS  <X], 

The  difference  §i  —  g  is  the  work  done  on  the  body, 

work  =  yavir(vi  —  v)v( i  —cos  a) (i) 

If  we  divide  this  by  the  distance  zr,  we  have  for  the  force  on  the  surface 

F  =  yavi(vi  —  v)('i  —  cos  «)  poundals (2) 

This  is  the  same  result  as  found  in  example  (3),  page  259,  by  the  principle  of  impulse. 
Equation  (2)  gives  F  in  poundals.     For  gravitation  measure  we  have 

F.  =  '- — -  (v\  —  v)  (i  —  cos  a). 
If  the  surface  moves  in  the  opposite  direction,  we  have  z/i  +  v  in  place  of  ZM  —  v,  and 


F  =  ?~±  (Vl  +  v)  (i  -  cos  a). 


If  the  surface  is  at  rest,  v  =»  o  and 


If  in  the  latter  case  E  =  90°,  this  becomes 


_ 

F  =  -  --  =  2ya 
g 


Hence  the  normal  pressure  of  a  jet  of  water  against  a  plane  surface  at  rest  is  equal  to  the  weight  of  a 
column  of  water  whose  cross-section  is  that  of  the  jet  and  whose  height  is  twice  that  due  to  the  velocity. 


CHAP.  V.]  EXAMPLES.  285 

If  a  =  i  So*  and  v  =  o,  we  have 


or  twice  as  much  as  when  <r  =  90*.  * 

The  work  done  on  the  surface  is,  from  (i),  a  maximum  when  v  =Vi  —  w  or  v  =  —  .     That  is,  the  work  is 
a  maximum  when  the  velocity  of  the  surface  is  half  that  of  the  jet.     The  maximum  work  is  then,  from  (2), 

yav*T(\  —  cos  a) 


If  «  =  180",  this  becomes         l   .  Vl*  =  —  L  t  or  a\i  the  kinetic  energy  of  the  water. 

If  a  =  90°,  it  becomes  faVlT  f  v*  =  ^L  ,  or  one-half  the  kinetic  energy  of  the  water. 

Hence  the  maximum  -work  of  a  jet  of  water  striking  a  plane  surface  at  right  angles,  disregarding 
friction,  is  only  one  half  the  kinetic  energy  of  the  jet. 


CHAPTER   VI. 

THE   POTENTIAL. 

The  Potential. — Let  a  particle  at  a  fixed  point  O  act  either  by  attraction  or  repulsion 
upon  a  particle  at  B.      Let  BA  be  any  path  of  the  particle  from  B  to  A,  the  distance  OB 
being  R,  and  the  distance    OA    being  r.      With   OA  =  r  as  a 
radius  describe  an  arc  of  a  circle  Aa. 

Then  the  force  upon  B  is  a  central  force,  and  we  have 
proved,  page  278,  that  the  work  done  by  or  against  the  central 
force  while  the  particle  moves  from  B  to  A  is  independent  of 
the  path  and  equal  to  that  necessary  to  move  it  from  B  to  a, 
when  Oa  =  r. 

The  fixed  particle  at  O  is  then  a  centre  of  force,  and  the  space  surrounding  this  par- 
ticle we  call  the  FIELD  OF  FORCE. 

If  then  we  take  any  convenient  point  of  reference,   as  C,  the, work  done  in  transferring 
a  particle  of  unit  mass  from  any  point  of  the  field  to  this  point,  or  from   this  point  to  any 
point  of  the  field,  has  a  definite  value  for  every  point  of  the  field,   no  matter  what  the  path. 
This  definite  work  for  any  given  point  of  the  field  ivhen  the  particle  moved  has  a  mass  o 
unity  is  called  the  POTENTIAL  of  the  point. 

The  unit  of  potential  is  then  the  same  as  the  unit  of  work,  as  one  foot-poundal  or  one 
foot-pound  or  one  erg. 

The  magnitude  of  the  potential  will  depend  upon  the  position  of  the  point  of  reference. 
The  sign  will  be  plus  or  minus  according  as  work  is  done  by  or  against  the  force  of  the 
field.  We  denote  the  potential  by  the  letter  77. 

Principle  of  the  Potential. — The  application  of  the  potential  rests  upon  the  following 
principle: 

Let  A  and  B  be  any  two  points  in  the  field  of  force  due  to  a  particle  at  <9,  and  let  C  be 
any  point  of  reference.  Then,  since  the  work  done  during  any 
displacement  is  independent  of  the  path,  the  work  done  by  or 
against  the  force  of  the  field  in  transferring  a  unit  mass  from  A  to 
B  is  equal  to  the  difference  of  the  works  done  in  transferring  it  from 
C  to  A  and  C  to  B. 

If  then  //„  and  TIb  are  the  potentials  of  the  points  A  and  B,  the  difference  is  the  work 
of  moving  unit  mass  from  A  to  B  or  B  to  A.  If  F  is  the  mean  force  in  the  direction 
ABt  we  have  this  work  equal  to  F  X  AB.  Hence 

£=na~  IIb,    or    F  =  g*^fl*. 

286 


CHAP.  VI.]  EQUIPOTENTIAL  SURFACE.  287 

When  A  and  B  are  indefinitely  near,  the  mean  force  F  becomes  the  instantaneous  force 

TT        TT 
in  the  direction  AB,  and     a  .  „      becomes  the  rate  of  change  of  the  potential  of  the  point  A 

per  unit  of  distance  in  the  direction  AB.     Hence 

The  rate  of  change  of  the  potential  of  any  point  per  unit  of  distance  in  any  direction  is 
equal  to  the  component  force  in  that  direction  which  acts  upon  a  particle  of  unit  mass  placed  at 
that  point. 

The  particle  possesses  potential  energy  at  whatever  point  of  the  field  of  force  it  may  be 
placed.  The  excess  of  its  potential  energy  at  one  point  over  its  potential  energy  at  another 
point  is  then  the  work  done  by  or  against  the  force  of  the  field  in  moving  from  one  point  to 
the  other.  This  is  equal  to  the  difference  of  potential.  Hence  the  appropriateness  of  the 
term  "  potential." 

The  theory  of  the  potential  is  of  great  use  in  magnetic  and  electrical  investigations. 

Equipotential  Surface. — A  surface  at  every  point  'of  which  the  potential  has  the  same 
value  is  called  an  EQUIPOTENTIAL  SURFACE. 

If  then  a  particle  is  moved  from  any  point  on  such  a  surface  to  any  other  point  on  this 
surface,  no  work  is  done  by  or  against  the  force  of  the  field.  There  is  then  no  component 
force  in  any  direction  tangential  to  such  a  surface,  and  hence  no  rate  of  change  of  potential 
per  unit  of  distance  in  that  direction.  The  resultant  force  at  any  point  of  such  a  surface  is 
then  normal  to  the  surface.  Thus  the  surface  of  water  at  rest  forms  an  equipotential 
surface  for  which  there  is  no  rate  of  change  of  potential,  and  the  resultant  force  for  every 
particle  on  the  surface  is  normal  to  the  surface.  The  work  done  by  or  against  gravity  in 
moving  a  particle  from  one  point  to  another  of  such  a  surface  is  zero. 

Lines  of  Force. — Any  line  so  drawn  in  a  field  of  force  that  its  direction  at  every  point 
is  the  direction  of  the  resultant  force  at  that  point  is  called  a  LINE  OF  FORCE.  As  the 
resultant  force  at  any  point  is  normal  to  the  equipotential  surface  passing  through  that  point, 
lines  of  force  are  normal  to  the  equipotential  forces  they  meet. 

Tubes  of  Force. — If  from  points-  in  the  boundary  of  any  portion  of  an  equipotential 
surface  lines  of  force  arc  drawn,  the  space  thus  marked  off  is  called  a  TUBE  OF  FORCE. 

Gravitational  Potential.— The  choice  of  the  point  of  reference  and  of  the  mode  of 
defining  potential  are  matters  of  convenience  and  vary  with  the  kind  of  field  of  force  under 
consideration. 

The  potential  in  a  field  of  force  due  to  the  attraction  of  gravity  is  called  GRAVITATIONAL 
POTENTIAL.  The  point  of  reference  is  taken  in  this  case  at  an  infinite  distance,  and  since  it 
is  convenient  to  have  the  potential  for  all  points  of  a  gravitational  field  positive,  and  the 
force  of  the  field  is  always  attractive,  we  define  gravitational  potential  of  a  point  as  the  work 
done  BY  the  force  of  the  field  in  moving  unit  mass  from  a  point  at  an  infinite  distance  TO  the 
given  point.  Or,  since  there  is  thus  a  loss  of  potential  energy,  the  work  done  by  the  force 
of  the  field  must  equal  the  gain  of  kinetic  energy,  and  hence  we  may  also  define  gravitational 
potential  of  a  point  as  the  kinetic  energy  acquired  by  unit  mass  in  falling  from  infinity  under 
the  attraction  of  a  given  mass  to  that  point. 

The  force  of  gravity  varies  inversely  as  the  square  of  the  distance, 

and  we  have  seen  (page  280)  that  the  work  of  such  a  force  when  a  /*^ 

particle  moves  from  a  distance  R  to  a  distance  r  from  the  centre  of  p 

force  is  given  by 

'T        Tx  f0- 


where  FQ  is  the  force  at  a  given  distance  r0. 


288  KINETICS  OF  A  PARTICLE.  [CHAP.  VI. 

If  we  take  rx  infinite,  —  is  zero  and  this  becomes 


If  the  mass  of  the  attracting  particle  at  O  is  m  and  of  the  moving  particle  M,  we  have 
for  the  force  of  attraction  at  any  distance  rn  (page  203) 

_  ,Mm 
ro" 
where  k  (page  205)  is  given  by 


where  g  is  the  acceleration  of  gravity,   m0  the  mass  of  the  earth  and  r0  the  radius  of  the 
earth.     We  have  then  for  the  work  of  moving  a  particle  M  from  infinity  to  the  distance  r 


If  we  adopt  the  astronomical  unit  of  mass  (page  206),  this  becomes  W  =  M  —  ,  and  if 
we  take  the  mass  J/as  unity  we  have  W=  \M]—  ,  where  [M~]  is  the  unit  of  mass,  or  the 

numerical  equation  W  —  —  . 

If  the  field  of  force  is  due  to  any  number  of  particles  of  masses  ml  ,  mz,  mz,  etc.,  at 
distances  rlt  r2,  r3,  etc.,  from  M,  we  have  the  numeric  equation  when  M  is  unity 


The  expression  2—  is  the  gravitational  potential  of  the  point  at  which  the  attracted 
particle  of  unit  mass  is  placed.      We  have  then 

zr  =  ^,  .............    (0 

or,  mathematically  defined,  the  gravitational  potential  of  any  point  due  to  the  attraction  of 
any  mass  is  the  sum  of  the  quotients  of  all  the  elementary  attracting  masses  divided  by  their 
distances  from  the  point,  provided  we  adopt  the  astronomical  unit  of  mass. 

Since  equation  (i)  gives  the  work  done  by  the  force  of  the  field  in  moving  unit  mass 
from  a  point  at  an  infinite  distance  to  the  given  point,  the  work  done  in  moving  a  massy)/  is 

......     (2) 


CHAP.  VI.]  GRAVITATIONAL  POTENTIAL.  289 

if  we  use  the  astronomical  unit  of  mass  (page  206),  or 


if  we  use  the  ordinary  unit  of  mass,  where  k  is  given  by 


where  g  is  the  acceleration  of  gravity,  m0  the  mass  and  r0  the  radius  of  the  earth. 

[Differential  Equations.  —  We  have  then  for  the  gravitational  potential  of  any  point  of  a  field  of  force 
due  to  the  attraction  of  any  number  of  particles  mi,  ;«a,  tn»,  etc.,  at  distances  r\,  r*,  r»  from  that  point, 


From  the  principle  of  the  potential  (page  286),  if  we  take  the  point  as  an  origin  of  co-ordinates,  we  have 
for  the  component  force  in  the  directions  of  the  axes  of  X,  Y,  Z,  for  a  unit  mass  at  the  point, 


dU 

"W 


(2) 


where  the  astronomical  unit  of  mass  (page  206)  is  to  be  used.     For  the  ordinary  unit  of  mass  we  multiply  by 

*=£•  ............  ••;•••« 

where  g  is  the  acceleration  of  gravity,  m0  the  mass  and  ra  the  radius  of  the  earth  (page  205). 
For  a  mass  M,  then,  at  the  point  we  multiply  by  kM. 
For  the  resultant  force  on  unit  of  mass  in  the  direction  of  any  radius  vector  from  the  point  we  have 


where  the  astronomical  unit  of  mass  (page  206)  is  to  be  used.     For  ordinary  unit  of  mass  we  multiply  by  k, 
and  for  any  mass  M  at  the  point  by  kM. 

If  ds  is  an  element  of  the  path  of  the  attracted  particle  of  unit  mass,  making  an  angle  0  with  r,  then 

ds  =  -  f.  ,  and  we  have  for  the  component  of  the  force  tangent  to  the  path,  upon  unit  mass, 

dIL      dU 


where  the  astronomical  unit  of  mass  (page  206)  is  to  be  used. 

For  ordinary  unit  of  mass  we  multiply  by  k,  and  for  any  mass  M  by  kM. 

We  have  from  (4),  II  =   /  Rdr  ;  and  since  for  an  equipotential  surface  the  potential  has  the  same  value 
at  every  point,  the  condition  for  an  equipotential  surface  is 


(6) 


where  C  is  a  constant. 

A  surface  which  fulfils  for  each  of  its  points  this  condition  is  an  equipotential  surface  for  the  system  of 
attractions.  As  any  value  can  be  given  to  C  between  its  greatest  and  least  values,  there  will  be  an  indefinitely 
great  number  of  equipotential  surfaces  corresponding  to  any  given  system  of  attractions. 


KINETICS  OF  A  PARTICLE. 


[CHAP.  Vt. 


From  equation  (5)  we  see  that  Ft  is  zero  when  0  =  90",  and  becomes  equal  to  the  resultant  force  A', 
equation  (4),  when  6  =  0.     That  is,  the  resultant  attraction  R  is  at  right  angles  to  the  equipotential  surface. 

The  direction  of  A"  is  then  a  line  of  force. 

If  dv  is  an  element  of  volume,  and  <5  its  density,  we  have  for 
its  mass  m  =  8dv.     Hence  for  rectangular  co-ordinates 


n  =  f—  —  r8 


If  we  use   polar  co-ordinates,  we   have   for  the   elementary 
volume  dv  =  r*dr  cos  8  </6  d<f>,  and  hence 


=  J  J     i  Sr  dr  cos  Q  dB  d<f>.     .    .    .. 


(8, 


Examples.— (I)  Particles  of  masses 3.928,  39.28  and  392.8  kilograms  are  situated  at  three  of  the  corners 
of  a  square  -whose  side  is  I  metre.     Find  the  potential  at  the  fourth  corner. 

ANS.   n  =  J2— ,  and  the  astronomical  unit  of  mass  is  3928  grams  (page  206).     Hence  II  =  1.087  ergs. 

(2)  Find  the  potential  and  attraction  of  a  homogeneous  circular  ring  of  radius  r  upon  a  point  C  on  the 
Perpendicular  to  its  plane  through  its  centre  O. 

ANS.   Let  the  distance  of  the  point  C  from  the  centre  O  be  x.     Then  the  distance  Cm  for  any  particle  of 
the  ring  is  fV*-+  x\     If  the  linear  density  of  the  ring  is  S,  the  mass  is  2itrS, 

and  therefore  the  potential  H  =  —        ==. 
fV1  +  x* 

The  attraction  upon  a  unit  mass  at  C  parallel  to  the  plane  of  the  ring  is 
then   -— -,  taking  the  astronomical  unit  of  mass  (page  206).  But  r  is  constant 

and  hence  • ,    =  o.     That  is  the  sum  of  the  component  attractions  of  the 

elements  of  the  ring  in  the  plane  of  the  ring  is  zero.     The  attraction  in  the  direction  CO  upon  a  unit  mass  at 

C  taking  the  astronomical  unit  of  mass,  is  Fx  =  —j-  = — - ,  the  minus  sign  denoting  attraction  or 

force  towards  the  centre  O. 

If  we  multiply  the  value  of  II  and  Fx  by  kM,  where  M  is  the  mass  of  any  particle  at  O,  and  k  is    %— 

(page  205),   we  have  the  result  for  any  mass  M  at  C,  using  the  ordinary  unit  of  mass. 
When  x  —  o.  the  potential  at  the  centre  of  the  ring  is  n  =  2*8. 

(3)  Find  the  potential  and  attraction  of  a  circular  arc  at  its  centre. 

C  ANS.   Let  6  be  the  angle  subtended  by  any  portion  of  the  arc  estimated  from 

its  middle  point  D. 

The   length  of  any  element  is  rdQ,  its  mass  is  rS  dQ,  where  5  is  the  linear 
0\\  density,  and  the  potential  is 


f+^ 

J-a         ' 


where  a  is  the  angle  A  CD. 

D  This  is  independent  of  the  radius  r  of  the  arc. 

The  attraction  of  any  element  whose  mass  is  rSdQ  for  a  unit  mass  at  C,  using  the  astronomical  unit  of  mass 

(page  206).  is  — j-.     The  component  of  this  at  right  angles  to  CD  is — ;— sin  0,  and  along  CD, 5— cos  0 

We  have  then  for  the  resultant  attraction  at  right  angles  to  CD 


S     f** 
7*  =  -     I  <&  sin  6  = 

T  J  -  a 


CHAP.  VI.] 


EXAMPLES. 


291 


and  for  the  resultant  attraction  along  CD 


-if*' 

%y  —  a 


Fy=-~-    I  JQcosO  =  -^sin«, 


the  minus  sign  denoting  attraction. 

For  any  mass  M  at  C,  using  the  ordinary  unit  of  mass,  we  multiply  by  kM,  where  k  =  —  —  (page  205). 

(4)  Find  the  potential  and  attraction  of  a  straight  line  upon  an  external  point. 

ANS.  Let  AB  be  the  line  and  C  the  point.  Drop  the  perpendicular  CD,  take 
D  as  origin,  and  let  CD  =  y.  Then  for  any  point  P  of  the  line  distant  DP  =  x 
we  have  CP  =  r  =  Vf  +  **•  Let  5  be  the  linear  density.  Then  the  mass  of 
any  element  is  $dx,  and  the  potential  is 


n= 


Taking  this  between  the  limits  of  x  —  DA  =  +  a  and  x  =  DB  =  +  b,  we  have 


n  =  s  log    +  y -?  ~L— . 


The  component  attraction  upon  unit  mass  at  C  in  the  direction  of  the  line  is 

dTL  8 


Introducing  the  limits  +  a  and  +  ^. 


For  the  component  attraction  upon  the  unit  mass  at  C  perpendicular  to  the  line  we  have 


dy 


Introducing  the  limits  +  a  and  +  b,  we  have 


and 


-.-L\ 

CBr 


CB'-~CD~'       ~CA=~CD'      «*Crf-8b«,      £=CZ?sin# 
F*  =  ^^fcos  /*  ~  c°s  A      />  =  -^  (sin  a  -  sii 


,Vd 

Let  the  angle  Z>C4  =  a,     DCB  =  ft,     ACB  =  ft—a  =  r.     Then 

I  COS 


klNETlCS  OF  A  PARTICLE. 
The  resultant  force  upon  unit  mass  at  6' is  then 


[CHAP.  VI. 


The  tangent  of  the  angle  which  this  resultant  force  makes  with  the  vertical  is 

Ft  _  cos  ft  —cos  a  _        a  +  ft 
F9       sin  a  —  sin  ft  2 

Therefore  the  resultant  attraction  bisects  the  angle  ACB. 

The  results  are  all  for  unit  mass  at  C  and  astronomical  unit  of  mass  (page  206).     For  mass  M  at  Cand 

zrS 
ordinary  unit  of  mass  we  have  only  to  multiply  Fx,  Fy,  R  by  kM,  where  k  =  - —  (page  205). 

(5)  Find  the  potential  and  attraction  for  a  circular  disc  at  a  point  on  the  perpendicular  to  its  plane  through 
its  centre, 

ANS.  Let^  be  the  radius  and  dy  the  thickness  of  an  elementary  ring,  and  $  the  surface  density.     Then 
the  mass  of  the  elementary  ring  is  2nSydy.     If  the  distance  OC  is  x,  we  have  for  the  potential  of  the  disc 

n  = 


which  for  the  limits^  =  R  =  radius  of  disk,and^  =  o  becomes 

For  the  centre  of  the  disc  this  becomes  2itSR. 

The  potential,  then,  is  constant  for  x  constant.   The  com- 
ponent force  upon  unit  mass  at  C  parallel  to  the  disc  is  then 

-jj^  —  o.      For  the   component   force   along   OC  we  have 
the  minus  sign  denoting  attraction. 

For  mass  M  at  <9and  ordinary  unit  of  mass  we  have  only  to  multiply  Fx  by  kM,  where  k  =  ^-  (page  205). 

(6)  Find  the  potential  and  attraction  at  the  vertex  for  a  right  tvne  with  circular  base. 

ANS.  Let  the  half  angle  at  the  vertex,  OCB,  of  the  preceding  figure  be  0.     Then  —    *          =  cos  6. 

Hence,  from  the  preceding  example,  we  see  that  the  attraction  of  all  circular  elementary  slices  for  a  particle 
at  C  is  the  same,  and  equal  to 

—  2*8ax(i  —  cos  6). 
The  total  attraction  is  then 

F3t=^=-  2it8x(i  -  cos  0), 

which  for  the  limits  h  and  o  becomes 

FM=  -  2itSh(\  -  cos  6). 

For  mass  M  at  C  and  ordinary  units  of  mass  we  have  only  to  multiply  by  kM,  where  k  =  ^~  (page  25). 
We  have  then 

H  =  -  *6x*(\  -  cos  S), 
or  for  limits  o  and  h 

n  *=  Tt8h\  i  -  cos  6). 


CHAP.  VI.] 


EXAMPLES, 


293 


(7)  Find  the  potential  and  attraction  of  a  spherical  shell  at  any  point. 

ANS.  Let  r  be  the  radius  of  the  shell,  /  its  thickness,  p  the  distance  of  the  point  B  from  the  centre  C, 
AB  =  a  =  the  distance  of  any  point  of  the  shell  from  the  given  point  B.     Take  the 
origin  at  C,  and  let  BC  coincide  with  the  axis  of  Y. 

Then  the  elementary  volume  is 


dv  =  r*t  sin 


and 


a  =   4/r5  +  p'  —  2rp  cos  0. 


Hence,  if  $  is  the  density, 
H  =  <5/r* 


'/V 

I       yr* 


sin  QdQ  d<f> 


—  2rp  cos  0  + 


Integrating  first  with  respect  to  0,  we  have 


and  then  with  respect  to  9, 


H  = 


-  (r*  -  2rp  + 


When  the  point  B  is  within  the  shell  p  <  r,  and  when  it  is  outside  of  the  shell  p  >  r. 
In  the  first  case,  when  B  is  within  the  shell,  we  have 


P)-  (r-  p)]  = 


where   m  is  the  mass  of  the  shell.     The  resultant  force  of  attraction  is  then  R  =   -j—  =o.     This  is  the  same 

dp 

result  as  in  example  (i),  page  207. 

In  the  second  case,  when  B  is  outside  the  shell,  we  have 


where  m  is  the  mass  of  the  shell.     The  resultant  force  of  attraction  then   is  R  =  —  =  —  -^ ,  where  the 

dp  p1 

minus  sign  denotes  attraction. 

If  we  take  the  mass  M  at  B  and  use  the  ordinary  unit  of  mass,  we  have  R  =  —  k — — .     This  is  the  same 

P 
result  as  already  obtained,  page  205. 

( 8)  Find  the  potential  and  attraction  of  a  thick  homogeneous  spherical  shell  at  any  point. 

ANS.  Let  the  external  radius  be  r\  ,  and  the  internal  radius  ry.     Then  in  the  preceding  example  we  can 
out  /  =  dr,  and  we  have  for  the  potential  of  that  part  of  the  shell  outside  of  the  spherical  surface  containing 

the  point    /    14jt$rdr,  and  for  the  potential  of  that  part  of  the  shell  inside  of  the  spherical  surface  contain- 

JP 

ln« -he,point     /     2=21-=-.     Hence 


n  = 


rdr 


294  KINETICS  OF  A  PARTICLE.  [C  HAP.  VI 

The  mass  of  the  shell  is  m  =  ^(r,1  -  r,f). 
If  the  point  is  wholly  within  the  shell, 


H 


and  if  the  thickness  is  very  small,  r\  —  r*  =  t  and  r(  +  r»  =  2r,  and  n  =  4««5/r,  as  found  in  the  preceding 

nple.    Also,  the  attraction  is  R  =  -^-  = 

dp 

If  the  point  is  wholly  without  the  shell, 


example.   Also,  the  attraction  is  fi  =  —  =  o,  as  found  in  the  preceding  example. 


and  R  = j,  as  found  in  the  preceding  example. 

If  the  shell  becomes  a  sphere,  ra  =  o  and  rt  =  r,  and  we  have  for  an  interior  point 

n  =  27T<5ra 


where  m  =    -nr*S.     For  an  exterior  point 


For  p  =  r  we  have  in  both  cases 


_ 

3" 


Hence  we  see  that  for  a  homogeneous  sphere  we  may  take  the  potential  and  attraction  at  any  external 
point  as  though  the  whole  mass  were  concentrated  at  the  centre,  while  the  attraction  at  an  interior  point  is 
directly  proportional  to  the  distance  from  the  centre.  The  first  result  has  been  proved,  page  205  ;  the 
second  in  example  (2),  page  207. 

(9)  Find  the  potential  and  attraction  for  a  cylinder  of  length  J^and  radius-  R  for  a  point  on  the  axis  at  a 
distance  d  from  the  nearest  end, 

ANS.    We   have    found,   example   (5).   for   the   component   force    along    the   axis   of   a   circular   disk 

—  2ird(  i  --       *  —  -J  ;  the  component  at  right  angles  to  the  axis  being  zero.     If  the  disk  has  a  thickness 
dx,  we  have  for  a  cylinder 


^=  =  Fx  =  -  2x8 
ujc  ^ 

or,  taking  the  limits  </+/and  d, 
Hence 

/»  r-,1  _^ ^»  x 

II  =    /   —  2w5rtjr(,T  —  yjr'  +  .A'*)  =  —  2»T<5      —  —  —  yx  +  JP log  V.J 

J  \_2  2  2 

The  value  of  II  is  obtained  by  taking  the  limits  d+l  and  d.     For  d  =  o  we  have  for  the  attraction  upon 
unit  mass  at  the  end  surface 

f*  =  —  2jTi$(/—   f//s  +  /(•*  +  A1) , 

and 

r~  / n       ot 

U  =  2K8\    ±\fl*  +  I?  —  l-—{L 
2          2 


CHAP.  VI.]  EXAMPLES.  295 

For  a  mass  M,  using  the  ordinary  unit  of  mass,  we  multiply  Fx  by  kM,  where  k  =£H±  (page  205). 

ma 

(10)  If  the  radius  of  the  earth  is  4000  miles,  find  the  potential  for  a  point  on  the  surface. 

ANS.   From  example  (8)  we  have  for  astronomical  unit  of  mass  II  =  — .     For  ordinary  unit  of  mass  we 

ro 

multiply  by  £~t.     Hence  II  =  gr0  ft.-poundals,  or  r0  ft.-pounds  =  4000  x  5280  =  21  120000  ft.-pounds. 
;«0 

(u)  At  the  distance  of  the  moon,  240  ooo  miles  from  the  centre  of  the  earth,  find  the  shortest  distance 
through  which  I  Ib.  must  be  moved  to  do  one  ft. -pound  of  work. 
ANS.  We  have  the  work  given  by 


W  =  mgr*(*-  -  l-\  ft.-poundals 


for  a  mass  m  at  a  distance  r\  moved  to  a  distance  r.      In  the  present  case  m  =  i  Ib.     Hence  for  work 
in  ft.-pounds 


If  W—\  ft.-pound, 

_    — rri  _     —     ri 

ri 

Taking  r0  =  4000  miles  and  r\  =  240000  miles,  we  have 

240  ooo  x  5280 

r\  —  r  =  —     — - — 5-^-,  =  3600  ft. 
(4000  x  5280) 

240  ooo  x  5280 

Value  of  g  above  Sea-level. — Suppose  a  mountain  of  uniform  density  8  and  cylindrical  in  shape,  and  a 
particle  of  mass  m  at  the  centre  of  its  upper  surface.  Then,  from  example  (9),  we  have  for  the  force  of 
attraction,  using  the  astronomrcal  unit  of  mass, 


If  we  divide  by  m,  we  have  for  the  acceleration  due  to  the  attraction  of  the  mountain 

where  R  is  the  radius  of  the  cylinder  and  /  its  length,  or  the  height  of  the  mountain. 

If  R  is  so  large  compared  to  /  that  ^5  can  be  neglected,  this  reduces  to  2it8l.     For  the  ordinary  unit  of 

mass  we  have,  multiplying  by  k  =  *— -  (page  205),  for  the  acceleration  due  to  the  attraction  of  the  mountain 


Let  50  be  the  mean  density  of  the  earth,  so  that  m0  =  -7r<V0s.     Then  the  acceleration  due  to  the 

attraction  of  the  mountain  is 

3  */ 
zVo'^ 

The  acceleration  due  to  the  attraction  of  the  earth  is 


We  have  then  for  the  acceleration  at  the  distance  /above  sea-level 

Q=. 


2g6  KINETICS   OF  A  PARTICLE.  [CHAP.  VI. 

The  value  of  -r  from  what  we  know  of  the  density  of  matter  at  the  earth's  surface,  may  be  taken  equal 
to  -•     Also  we  may  write,  approximately, 


Hence  we  have,  approximately, 


where  /  is  the  height  above  sea-level,  rt  is  the  mean  radius  of  the  earth,  and^-  the  corresponding  acceleration 
at  sea-level. 

The  assumptions  made  in  this  determination  are  more  applicable  to  elevated  table-land  than  to  a 
mountain.  The  equation  obtained  is  the  accepted  formula  for  estimating  the  value  of  g  at  two  places  so  far 
as  dependent  on  the  height  above  sea-level. 


KINETICS   OF   A  MATERIAL   SYSTEM. 


CHAPTER   I. 

GENERAL  PRINCIPLES. 

Material  System. — A  material  system  consists  of  an  indefinitely  large  number  of  par- 
ticles which  act  and  react  upon  each  other.  Such  a  system  may  constitute  a  rigid  body  or  a 
system  of  rigid  bodies,  an  elastic  body  or  a  system  of  elastic  bodies,  a  liquid  or  a  gaseous  mass. 

Internal  and  External  Forces. — The  forces  acting  between  the  bodies  or  particles  of  a 
material  system  are  internal  forces  or  STRESSES  (page  174).  The  forces  exerted  upon  the 
system  by  other  bodies  or  particles  outside  of  the  given  system  are  external  forces. 

When  one  body  or  particle  acts  upon  another  with  a  certain  force,  it  is  itself  acted  upon 
by  the  other  with  an  equal  and  opposite  force  (page  174).  Hence  the  internal  forces  or 
stresses  between  any  two  bodies  or  particles  of  a  material  system  consist  of  two  equal  and 
opposite  forces  in  the  same  straight  line. 

It  follows  that  the  internal  forces  or  stresses  existing  between  any  system  of  bodies  or 
particles  must  constitute  a  system  of  forces  in  equilibrium. 

Impressed  and  Effective  Forces. — The  external  forces  acting  upon  any  material  system 
we  call  IMPRESSED  forces. 

Any  particle  of  the  system  of  mass  m  has ^  at  any  instant  an  acceleration  f  in  a  certain 
direction.  By  the  equation  of  force  (page  170)  the  force  acting  upon  this  particle  is  ;;//. 
This  we  call  the  EFFECTIVE  force  on  the  particle. 

It  is  t  lie  force  which,  acting  upon  an  isolated  particle  of  the  system  at  any  instant,  would 
make  if  move  at  that  instant  precisely  as  it  really  does  move  as  part  of  the  system. 

D'Alembert's  Principle. — Let  m  be  the  mass  of  a  particle  of  any  material  system,  and 
f  its  acceleration.  Then  its  effective  force  as  just  defined  is  mf.  Let  F  be  the  resultant 
impressed  force  on  this  particle,  and  .S  the  resultant  internal  force  or  stress  acting  on  the 
particle. 

Then  mf  must  be  the  resultant  of  F  and  S.  Hence  if  mfbe  reversed  in  direction,  we 
should  have  a  system  of  forces  F,  S,  and  mf  reversed,  in  equilibrium. 

The  same  would  hold  for  every  other  particle  of  the  system.  But  we  have  just  seen 
that  for  all  the  particles  the  internal  forces  5  form  by  themselves  a  system  in  equilibrium. 

Hence  it  follows  that  all  the  impressed  forces  F  and  reversed  effective  fotcts  mf  must 
also  form  a  system  of  forces  in  equilibrium. 

We  have  then  the  following  general  principle  which  holds  for  any  material  system  or 
body  whether  solid,  liquid  or  gaseous: 

297 


*98  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  I. 

Consider  all  the  effective  forces  as  reversed  in  direction.  Then  the  impressed  forces  acting 
upon  the  system,  and  the  reversed  effective  forces,  constitute  a  system  of  forces  in  equilibrium. 

This  is  known  as  D'ALEMBERT's  PRINCIPLE.  It  reduces  any  dynamic  problem  to  one 
of  equilibrium  between  actual  ("impressed")  forces  and  fictitious  ("reversed  effective") 
forces. 

Velocity  of  Centre  of  Mass — Momentum  of  a  System. — Let  ;«ir  m2,  tn3,  etc.,  be  the 
masses  of  the  particles  of  any  material  system,  sl ,  s2,  s3,  etc.,  the  distances  of  these  particles 
from  any  given  plane,  J  the  distance  of  the  centre  of  mass  of  the  system  from  that  plane, 
and  m  =  2m  the  mass  of  the  system.  Then  we  have  (page  21) 

_        ^,^4-  myS2  -f-  *%r3-{-  etc. 

s  = — — 

m 

At  the  end  of  an  indefinitely  small  time  T,  let  the  distances  be  s^,  s2,  s3' ,  etc.    Then 

_,__  m^i  -j-  myS2  -j-  w3$3'  -j-  etc. 
m 

If  we  subtract  the  first  equation  from  the  second  and  divide  by  r,  we  have 

***l(V   —  5j)  m2(S2      ~    S2)     . 


J'-J 


m 


_ 
But  if  T  is  indefinitely  small,  —       -   is  the  velocity  'v  of  the  centre  of  mass  in  a  direction 


s  i  _  s      s  '  _  s 
perpendicular  to  the  given  plane,  and  —       —  >  —      —  -  ,  etc.,  are  the  velocities  vl  ,   v2,  etc. 

of  the  particles  in  this  direction.      Hence 

_       WjZ'j  -f-  in2v2  -\-  w3^3-f-  etc. 

~^~  ^^ 

or 


But  »*,*>,,  m2v2,  etc.,  are  the  momentums  of  the  particles  in  any  given  direction,  and 
2mi>  i$  then  the  momentum  of  the  system  in  that  direction. 

Hence,  for  any  material  system,  the  momentum  in  a,ny  direction  is  the  same  as  for  a  par- 
ticle of  mass  equal  to  the  mass  of  the  system  moving  with  the  velocity  of  the  centre  of  mass  in 
that  direction. 

Also,  the  velocity  of  the  centre  of  mass  in  any  direction  is  equal  to  the  momentum  of  the 
system  in  that  direction  divided  by  the  mass  of  the  system. 

From  (i)  we  have 

2mv  —  ~v2m  =  o  = 

But  2m(v  —  v)  is  the  momentum  of  the  system  in  any  direction  relative  to  the  centre  of 
mass. 

Hence  the  momentum  of  a  material  system  in  any  direction  relative  to  the  centre  of  mass 
is  zero. 


CHAP-  L]  ACCELERATION  OF  THE  CENTRE  OF  M4SS. 


299 


Acceleration  of  the  Centre  of  Mass.  —  We  find  the  acceleration  of  the  centre  of  mass  in 
precisely  the  same  way.  Thus  if  vl  is  the  initial  and  v  the  final  velocity  of  the  centre  of 
mass  of  any  material  system  in  any  given  direction  during  an  indefinitely  small  time  r,  we 
have,  from  (i), 


Subtracting  the  first  from  the  second  and  dividing  by  r,  we  obtain 


If  r  is  indefinitely  small,  —    — ~  is  the  acceleration/  of  the  centre  of  mass  in  the  given 
direction,  and   = is  the  sum  2mf  of  all  the  effective  forces  in  that  direction.    Hence 


(2) 


That  is,  for  any  material  system  the  effective  force  in  any  direction  is  the  same  as  for  a 
particle  of  mass  equal  to  that  of  the  system  moving  witli  the  acceleration  of  the  centre  of  mass 
in  that  direction. 

Also,  the  acceleration  of  the  centre  of  mass  in  any  direction  is  equal  to  the  acceleration  of 
the  system  in  that  direction  divided  by  the  mass  of  the  system. 

From  (2)  we  have 

2mf  —  f2m  —  o  =  2m(f  —  /). 

But  ^m(f  —  /)  is  the  effective  force  of  the  system  relative  to  the  centre  of  mass. 

Hence  the  effective  force  of  a  system  in  any  direction  relative  to  the  centre  of  mass  is  zero. 

Motion  of  Centre  of  Mass.  —  Let  Flt  F2,  F3,  etc.,  be  the  components  in  any  direction 
of  all  the  impressed  or  external  forces  acting  upon  any  material  system,  and  m^f^  m^f2,' 
>«3/3,  etc.,  the  components  in  the  same  direction  of  all  the  effective  forces. 

Then,  by  D'Alembert's  principle  (page  297),  we  have 

2F—2mf=o,      or     2F=2mf=mf^      ......      (3) 

Hence  the  motion  of  the  centre  of  mass  of  any  material  system  is  not  affected  by  the 
internal  forces  or  stresses  between  the  particles  of  the  system,  but  only  by  the  external  or 
impressed  forces. 

Also,  the  acceleration  of  the  centre  of  mass  is  the  same  as  for  a  particle  of  mass  equal  to 
the  mass  of  the  system,  all  the  impressed  forces  being  transferred  to  this  particle  without 
change  of  magnitude  or  direction.  This  principle  we  have  already  deduced  for  a  rigid  body 
(page  175).  We  see  that  it  holds  for  any  material  system,  rigid  or  not. 

Conservation  of  Centre  of  Mass.  —  If,  then,  there  are  no  external  or  impressed  forces, 
2F  =  o,  f=o,  and  the  motion  of  the  centre  of  mass  cannot  change.  If  at  rest,  it  remains 
at  rest.  If  in  motion,  it  moves  with  uniform  speed  in  a  straight  line,  no  matter  what  the 
internal  forces  or  stresses  may  be. 


3°°  KINETICS  OF  A  MATERIAL  SYSTEM.  J('HAP.  I. 

That  is,  no  material  system  can  of  itself  and  ivithout  the  action  of  external  forces  change 
the  motion  of  its  centre  of  mass. 

This  is  called  the  principle  of  conservation  of  centre  of  mass. 

Conservation  of  Momentum.—  If  z;,  is  the  initial  and  v  the  final  velocity  of  a  particle 
in  any  direction  during  an  indefinitely  small  timer,  the  acceleration  /  in  that  direction  is 
v  —  if  m(v  _  v  "\ 

—  -  —  and  the  effective  force  mf  in  that  direction  is  --  -.     We  have  then,  from  (3), 


We  see,  then,  that  the  change  of  momentum  2m(v  —  v^)  in  any  direction  of  a  material 
system  is  not  affected  by  the  internal  forces  or  stresses,  but  only  by  the  impressed  forces. 

If  then  the  impressed   forces  are  zero,  2m(v  —  v^)  is  zero,  and  the  momentum  of  the 
system  in  any  direction  does  not  change. 

That  is,  no  material  system  can  of  itself  and  without  the  action  of  external  forces  change 
its  momentum  in  any  direction. 

This  is  called  the  principle  of  conservation  of  momentum. 

Moment  of  Momentum.  —  The  moment  of  momentum  of  a  particle   relative   to  a  given 
point  or  axis  is  the  product  of  its  mass  by  the  moment  of  its  velocity 
'V    (PaSe  89)  relative  to  that  point  or  that  axis. 

Thus  let  m  be  the  mass  of  a  particle,  v  its  velocity,  /  the  lever- 
arm  or  perpendicular  from  any  point  O'  upon  the  direction  of  v. 
Then  the  moment  of  the  velocity  (page  89)  is  vl.  Or  if  r  .is  the 
radius  vector  of  m,  e  the  angle  of  v  with  r,  and  o>  the  angular  velocity 

relative  to  O',  rco  =  v  sin  e  and  -v  =  -r-          But  /=  r  sin  e.      Hence 

sm  e. 

vl  =  vr  sin  e  =  r?co. 
The  moment  of  momentum  of  the  particle  relative  to  O'  is  then 

mvl  = 


Its  line  representative  is  a  straight  line  O'  M  at  right  angles  to  the  plane  of  v  and  /, 
whose  magnitude  is  mvl  =  mr^aa,  and  whose  direction  is  such  that  looking  along  it  in  its 
direction,  as  shown  by  the  arrow  in  the  figure,  the  rotation  is  seen  clockwise. 

By  "  direction  "  of  moment  of  momentum  we  always  mean  the  direction  of  its  line 
representative. 

We  see,  then,  that  the  moment  of  momentum  of  a  particle  relative  to  a  point  is  repre- 
sented in  magnitude  and  direction  by  a  straight  line  of  definite  magnitude  and  direction 
through  that  point,  just  like  moment  of  velocity  (page  89),  and  the  same  principles  apply. 
The  same  holds  for  moment  of  momentum  relative  to  an  axis. 

We  can  therefore  combine  and  resolve  moment  of  momentum  just  like  moment  of 
velocity,  or  just  like  velocity,  acceleration  or  force,  and  can  find  the  resultant  for  any  num- 
ber of  concurring  line  representatives  just  as  for  force. 

Moment  of  Momentum  for  a  System.  —  For  a  material  system  each  particle  has  its 
own  velocity,  and  for  any  point  O'  its  own  moment  of  momentum  of  definite  magnitude  and 
direction,  given  by  its  line  representative  through  O',  as  in  the  preceding  figure.  For  the 
entire  system  we  have  then  a  number  of  line  representatives  concurring  at  O't  and  the 
resultant  of  all  these  is  the  moment  of  momentum  of  the  system  relative  to  0  '. 


CHAP.  I.] 


ACCELERATION  OF  MOMENT  OF  MOMENTUM. 


301 


The  moment  of  momentum  of  a  material  system  relative  to  any  point  reduces,  then,  to  a 
resultant  moment  of  momentum  of  definite  magnitude  about  a  definite  axis  through  that  point. 

Let  this  axis  be  the  axis  of  Z',  and  let  O'X',  O'  Y'  be  the 
other  two  rectangular  axes.  Let  ^,  y,  ~z  be  the  co-ordinates 
of  the  centre  of  mass  O.  With  O  as  origin  take  parallel  axes 
OX,  OY,  OZ,  and  let  x,  y,  z  be  the  co-ordinates  of  any 
particle  P  relative  to  O.  Then  the  co-ordinates  of  P  for 
origin  O'  are  ^  -(-  x,  ~y  -f-  y,  ~z  -j-  z. 

Let  vx,  vy,  vz  be  the  components  of  the  velocity  of  the 
centre  of  mass  O.  Then  the  velocity  components  of  P  are 

vx—V*—  y°°z  »  vy=i~Vy~^~  Xa)" 

where  GO,,  is  the  angular  velocity  relative  to  the  axis  OZ  or  O'Z'  . 
The  moment  of  momentum  for  a  particle  of  mass  m  at 
P  is  then  given  by 

mvy(  *  -f-  x)  —  mvx(  y+y). 

Substituting  the  values   of  vx,a.nd  vy,  we   have  for  the  entire  system  the  moment  of 
momentum 


M  =  2mvy(  x  +  x}  —  2mvx(y  -j-  y) 


,  +  2m(  xx  -j-  yy)oo,. 


But  2m  =  m  =  mass  of  the  system,  vx  and  vy  are  constant,  and  2mx  =  o,  2my  =  o, 
since  O  is  the  centre  of  mass.      Hence 

M  =  mv     — 


But  2m(x*  -{-y^oo,  is  the  moment  of  momentum  of  the  system  about  an  axis  parallel  to 
the  given  axis,  through  the  centre  of  mass  O,  and  mzyir  —  mivxy  is  the  moment  of  momentum 
of  a  particle  of  mass  m  at  the  centre  of  mass. 

Hence  the  moment  of  momentum  of  a  material  system  about  a  given  axis  is  equal  to  the 
moment  of  momentum  about  a  parallel  axis  through  the  centre  of  mass  plus  the  moment  of 
momentum  of  a  particle  of  mass  equal  to  t/ie  mass  of  the  system,  situated  at  the  centre  of  mass. 

Acceleration  of  Moment  of  Momentum.  —  If  vl  is  the  initial  and  v  the  final  velocity  of  a 

particle  of  mass  m,  in  any  given  direction,  during  an  indefinitely  small  time  r,  then  - 
is  the  acceleration  /*«  that  direction,  and 


is  the  effective  force  in  that  direction. 

The  moment  of  this  force  relative  to  any  point  O'  is  then 

_  m(v-vjt 


where  r  is  the  radius  vector,  GO^  and  GO  are  the  initial  and  final  angular  velocities  and  a  is  the 
angular  acceleration.  The  line  representative  is  a  straight  line  O' M  at  right  angles  to  the 
plane  of /"and  /,  whose  magnitude  is  mfl  and  whose  direction  is  such  that,  looking  along  it  in 
its  direction  as  shown  by  its  arrow  in  the  figure,  the  rotation  is  seen  clockwise.  By  "  direc- 
tion "  of  moment  of  a  force  we  always  mean  the  direction  of  its  line  representative. 


302  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  I. 

But  —  —  is  the  acceleration  of  the  moment  of  momentum  in  the  direction  O'M 

of  the  particle  relative  to  O'. 

For  the  entire  system  we  have  for  the  acceleration  of  moment  of  momentum   in  this 
direction 


But,  by  D'Alembert's  principle  (page  297),  the  external  or  impressed  forces  and  the 
reversed  effective  forces  form  a  system  of  forces  in  equilibrium.     Hence 


2FI  =  2mfl, 
or 


(4) 


That  is,  for  a  particle  or  for  a  material  system  of  particles  the  acceleration  of  the  moment 
of  momentum  in  any  direction  relative  to  any  point  is  equal  to  the  moment  in  that  direction  of 
all  the  impressed  forces  relative  to  that  point. 

Conservation  of  Moment  of  Momentum.  —  If  now  2F/=  o,  we  have 


or  the  change  of  moment  of  momentum  in  any  direction  relative  to  any  point  is  zero.  But 
we  have  seen  (page  301)  that  the  moment  of  momentum  of  a  particle  or  of  a  system  relative 
to  any  point  has  a  definite  magnitude  and  direction.  Its  projection,  then,  in  any  direction 
does  not  change  when  2F/  =  o,  and  it  is  therefore  invariable  in  magnitude  and  direction. 

This  is  called  the  principle  of  conservation  of  moment  of  momentum,  and  it  may  be 
expressed  generally  as  follows  : 

When  the  resultant  moment  2FI  of  all  the  impressed  forces  acting  upon  a  particle  or  upon 
a  material  system,  relative  to  any  point  or  any  axis,  is  always  zero,  the  moment  of  momentum 
relative  to  tliat  point  or  axis  does  not  change  either  in  magnitude  or  direction,  and  its  line  repre- 
sentative is  then  a  straight  line  of  constant  magnitude  and  invariable  direction,  passing  through 
that  point  or  coinciding  with  that  axis. 

Now  2FI  is  always  zero  for  all  points  when  there  are  no  impressed  forces  ;  also  when 
all  the  impressed  forces  are  always  in  equilibrium.  In  both  cases  the  principle  of  conserva- 
tion of  moment  of  momentum  holds  for  any  point,  and  we  have  an  invariable  axis  for  any 
point. 

Also,  2FI  is  always  zero  for  any  one  given  point  when  all  the  impressed  forces  always  pass 
through  that  point  or  when  they  always  reduce  to  a  single  resultant  through  that  point.  In 
this  case  the  principle  holds  for  that  point,  and  we  have  an  invariable  axis  through  that  point. 

Also,  2FI  is  always  zero  for  any  axis  when  all  the  impressed  forces  always  pass  through 
that  axis  or  are  always  parallel  to  that  axis.  In  this  case  the  principle  holds  for  this  axis. 

Invariable  Axis  and  Plane.  —  In  all  these  cases  we  have  an  invariable  axis  either 
through  any  point,  or  through  a  given  point,  or  coinciding  with  a  given  axis.  But  from  the 
principle  of  page  301  we  see  that  in  all  cases  we  have  a  parallel  axis  through  the  centre  of 
mass,  whether  the  centre  of  mass  is  fixed  or  not.  ' 

This  axis  through  the  centre  of  mass,  invariable  in  direction,  is  called  the  INVARIABLE 
AXIS  of  the  system,  and  the  plane  through  the  centre  of  mass  at  right  angles  to  it  is  the 
INVARIABLE  PLANE  of  the  system. 


CHAP.  I.]  CONSERVATION  OF  AREAS.  3°3 

Thus,  in  the  solar  system,  if  there  are  no  forces  external  to  the  system,  we  have  an 
invariable  axis  through  the  centre  of  mass  of  the  system  and  an  invariable  plane  at  right 
angles  to  this  axis,  through  the  centre  of  mass.  Also,  if  for  each  planet  and  satellite  the 
force  of  attraction  on  every  particle  always  passes  through  the  centre  of  mass  of  the  system, 
we  have  for  each  planet  and  satellite  an  invariable  axis  through  that  point.  The  moment  of 
momentum  of  each  does  not  change  in  magnitude,  and  lies  along  its  oVn  invariable  axis. 
The  resultant  of  all  is  the  moment  of  momentum  of  the  system,  and  lies  along  its  invariable 
axis. 

If  we  take  the  axis  of  Z  as  the  invariable  axis  of  a  system  for  which  the  moment  of 
momentum  is  constant  in  magnitude,  the  projection  on  the  axes  of  X  and  Y  will  be  zero. 
Hence,  for  the  invariable  axis,  the  moment  of  momentum  of  a  system  is  a  maximum. 

Conservation  of  Areas.  —  The  moment  vl  of  a  velocity  is  equal  to  twice  the  areal 
velocity  of  the  radius  vector  (page  89).  The  moment  of  momentum  mvl  of  a  particle  is  then 
proportional  to  twice  the  areal  velocity  of  the  radius  vector. 

Hence,  from  the  principle  of  conservation  of  moment  of  momentum, 

When  thet  resultant  moment  2  Fl  of  all  the  impressed  forces  acting  upon  a  particle,  relative 
to  any  point  or  any  axis,  is  always  zero,  the  areal  velocity  of  the  radius  vector  of  the  particle 
does  not  change  either  in  magnitude  or  direction. 

For  a  system  the  areal  velocity  of  any  particle  relative  to  the  invariable  axis  does  not 
change. 

This  is  called  the  principle  of  conservation  of  areas.  Thus,  in  the  solar  system,  the  areal 
velocity  of  the  radius  vector  of  any  planet  does  not  change.  This  is  Kepler's  second  law 
(page  120). 

Kinetic  Energy  of  a  System.  —  The  kinetic  energy  of  a  material  system  is  the  sum  of. 
the  kinetic  energy  of  all  its  particles,  or 


Let  vx,  ~vy,  vf  be  the  component  velocities  of  the  centre  of  mass  O  relative  to  the  origin 
O'  ,  and  vx,  vy  ,  vz  the  component  velocities  of  any  particle  relative  to  the  centre  of  mass  O. 
Then  the  component  velocities  of  the  particle  relative  to  O'  are 


and  the  kinetic  energy  of  the  system  is 

K  =  ~ 
Expanding,  we  can  write 


But  (page  298)  2mvx  =  o,  2mvy  =  o,  2mvt  =  o.  Also,  -  2tn(v^  +  v}  -|-  z/«)  is  the  kinetic 
energy  of  all  the  particles  moving  with  velocities  equal  to  their  velocities  relative  to  the 
centre  of  mass  O,  and-^w(  v*x-\-  ~v*y  +  z£  )  =-m(z^-f-  V*,  -\-vfy  is  the  kinetic  energy  of  a 

particle  of  mass  m  equal  to  the  mass  of  the  system  moving  with  the  velocity  of  the  centre 
of  mass. 


304  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  I. 

Hence  the  kinetic  energy  of  a  system  is  equal  to  the  sum  of  the  kinetic  energy  of  the 
particles  relative  to  the  centre  of  mass  plus  the  kinetic  energy  of  a  particle  of  mass  equal  to  the 
mass  of  the  system  moving  with  the  velocity  of  the  centre  of  mass. 

_m  _  !!'<"  ^  then  /  is  the  lever-arm   of  the  velocity  v  relative  to  O, 

and  co  is  the  corresponding  angular  velocity,  we  have  v  =  loo,  and 

the  kinetic  energy  ^ml^tf.     The  kinetic  energy  of  all  the  particles 
relative  to  O  is  then  -2ml*(*P.      If  v  is  the  velocity  of  the  centre 

\j  2 

of  mass,  and  m  the  mass  of  the  system,  we  have  then  for  the  kinetic  energy  of  the  system 


5C  =    mv2  -f    SmlW  ..........      (5) 

Potential  Energy  of  a  System.  —  The  potential  energy  of  a  system  is  the  sum  of  the 
potential  energies  of  its  particles.  Since  the  potential  energy  of  each  particle  depends  upon 
its  position  (page  274),  the  potential  energy  of  the  system  depends  upon  its  configuration  and 
position.  If  no  external  forces  act  upon  the  system,  its  potential  energy  will  depend  upon 
its  configuration  only. 

Law  of  Energy.  —  Since  the  law  of  energy  (page  275)  holds  for  every  particle  of  a  system, 
it  holds  for  the  entire  system. 

Therefore,  if  &,  is  the  total  initial  and  &  the  total  final  energy,  potential  and  kinetic, 
of  a  system,  and  if  W'\s  the  work  done  against  non-conservative  forces,  we  have 


or  the  total  loss  of  energy  is  equal  to  the  work  done  against  non-conservative  forces.      Such 
a  system  is  called  a  non-conservative  system. 

Conservation  of  Energy.  —  If  there  are  no  non-conservative  forces  W  is  zero,  and  we 
have 


or  the  total  energy  does  not  change.      Such  a  system  is  called  a  CONSERVATIVE  system. 

Perpetual  Motion.  —  When  a  material  system  after  a  series  of  changes  returns  to  its 
original  position  and  configuration,  its  potential  energy  is  not  changed  ;  and  if  the  system  is 
conservative,  the  kinetic  energy  is  unchanged.  Such  a  system  would  then  go  through  its 
cycle  of  changes  for  ever  if  set  in  motion. 

If  the  bodies  of  the  solar  system  encounter  no  resistance,  we  have  a  system  of  this  kind. 
If  we  could  abolish  friction  and  all  other  non-conservative  forces,  a  machine  could  be 
constructed  which,  once  started,  would  run  forever,  provided  no  work  is  done  by  it. 

Since  friction  and  other  non-conservative  forces  cannot  be  thus  abolished,  such  a 
machine  is  not  possible. 

Law  of  Conservation  of  Energy  General.  —  The  law  of  conservation  of  energy  just 
enunciated 

£,  —  &  =  o 

appears  as  a  special  case  of  the  law  of  energy 


CHAP.  L]  EXAMPLES.  305 

when  W  the  work  against  non-conservative  forces,  such  as  friction,  is  zero.  In  general, 
then,  when  work  is  done  against  such  forces  we  say  that  energy  is  "  lost."  Experiment 
shows  that  when  energy  is  thus  "  lost"  heat  is  always  developed,  and  that  heat  is  a  form  of 
energy  dependent  upon  kinetic  energy  of  particles.  This  heat  is  the  exact  equivalent  of  the 
energy  "lost."  If  then  we  take  into  account  heat  energy,  the  law  of  conservation  of 
energy  becomes  general,  and  we  have  in  ajl  cases,  whether  the  forces  are  conservative  or 
non-conservative, 


That  is,  the  entire  energy  of  an  isolated  system,  including  kinetic,  potential,  and  thermal, 
is  constant. 

Examples.— (i)    What  effect  has  the  bursting  of  a  bomb  upon  the  motion  of  its  centre  of  mass? 
ANS.   Neglecting  air  resistance,  none  whatever.     By  the  principle  of  conservation  of  centre  of  mass* 
internal  forces  cannot  affect  the  motion  of  the  centre  of  mass. 

(2)  Show  that  the  centre  of  mass  of  the  universe  must  either  be  fixed  in  space  or  move  in  a  straight  line 
with  uniform  speed. 

ANS.  By  the  principle  of  conservation  of  centre  of  mass  no  material  system  can  of  itself  and  without 
the  action  of  external  force  change  the  motion  of  its  centre  of  mass. 

(3)  Two  particles  of  mass  nti  and  mi  are  moving  in  the  same  straight  line  with  velocities  z/i  and  v*. 
Find  the  velocity  of  their  centre  of  mass. 

ANS.  From  equation  (i),  page  298,  "v  =     _   . 

m 

We  have  m  =  m\  +  mi  and  2mv  =  m\v\  +  miVi  if  the  velocities  are  in  the   same  direction.  *  Hence 

_  _  mivi oz/a  ^  vejocjtjes  m  tne  same  direction  and 

ffti-r  mi 

_       m\v\  —  m^Vi  .t 

v  =  —  it  Vi  is  opposite  to  v\. 

nt\  -rnti 

(4)  Two  particles  of  mass  mi  and  mi  at  a  distance  s'  are  at  rest  on  a  smooth  horizontal  plane  and  are 
drawn  together  by  a  uniform  force  F.     After  a  time  t  the  mass  mi  has  a  velocity  Vi.     Find  the  final  velocity 
Vi  of  the  mass  mi ,  the  internal  force  F,  the  distance  s  apart  at  the  end  of  the  time  t,  and  the  distance  of  each 
mass  from  the  centre  of  mass. 

ANS.  Let  v  be  the  velocity  of  the  centre  of  mass  O.    Then,  since  Vi  and  Vi  are  opposite  in  direction,  we 
have,  by  the  principle  of  velocity  of  , 

centre  of  mass,  page  298, 


(mi  +  mi)v=mivi-mivi  ™f         F  J  °  Y*  & 

Hence  v  =  **1^~^7/a. 

But  the  centre  of  mass  is  initially  at  rest,  or  v  =  o,  and  by  the  principle  of  conservation  of  centre  of 
mass  (page  300)  it  must  remain  at  rest.     Hence 


(i) 


We  also  obtain  this  result  directly  from  the  principle  of  conservation  of  momentum  (page  300). 

Since  the  force  F  on  mi  is  uniform,  the  acceleration  is  uniform  and  the  distance  passed  over  by  mi  in 

the  time  /,  starting  from  rest,  is  —  /.    The  distance  passed  over  by  mi  is  — /.      The  distance  s'  —  s  is  then 
given  by 

(»!_+  ft),     = 

2*2i 


306  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  1. 

By  the  principle  of  impulse  (page  257)  the  uniform  force  Fon  mi  is  ttL^l%  and  on  mt  ,  f"y*.     Since  these 
forces  are  equal  and  opposite,  we  have 


.......  .        .  ..  .  . 

from  which  we  obtain  again  (i). 

Again,  the  initial  energy  is  Si  =  Fs',  potential.     The  final  energy  &  is  Fs  potential  and  —  m\v*  ^  -- 
kinetic,  or 

&  =  fr  ~s  +  —miVi*  H 


By  the  principle  of  conservation  of  energy,  &i  =  §,  or 

Ft  =  Fs  +  1  mivS  +  ~  m,vS, 
or 

F  —  tftlVl*  +  m*v? 

2(sf  -  JT~ 

Inserting  the  value  of  s'  -  s  from  (2),  and  of  z/,  from  (i),  we  obtain  again 

,..  _  m\v\  _  mtVt 
~T~'        t     ' 

The  distance  of  m\  from  the  centre  of  mass  C  at  the  start  is 


mi  +  m\  ' 
and  at  the  end  of  the  time  / 


The  distance  of  «,  from  the  centre  of  mass  C  at  the  start  is 


mi  + 
and  at  the  end  of  the  time  / 


rt  = 


If  mi  =  50  Ibs.,  mt  =  loo  Ibs.,  v,  =  10  ft.  per  sec.,  /  =  ~  sec.,  /  =  3  ft.,  we  have 

Vi  =  20  ft.  per  sec. ;    s  =  2.25  ft. ;     F  =  20000  poundals  = pounds; 

r\  =  2  ft. ;    ri  =  1.5  ft. ;     r,'  =  i  ft. ;    r,  =  0.75  ft. 

(5)  In  the  preceding  example  suppose  the  particles  revolve  about  the  centre  of  mass  with  the  initial  angular 
velocity  o>'. 

ANS.  Take  the  same  notation  as  before  and  let  a>i  be  the  angular  velocity  of  mi  at  the  distance  ri,  and 
o»t  of  m^  at  the  distance  rt.  We  have  the  same  values  for  vt,  /  —  s,  ri,<.t\,  rY.and  r»,as  before.  It  is 
therefore  only  required  to  find  F. 

We  have,  by  the  principle  of  conservation  of  areas, 

rjV  =  r.'oj,     and     rjV  =  r,1*,, 
hence 


CHAP.  L] 


EXAMPLES. 


307 


or  K)I  and  GO*  are  equal,  as  they  should  be,  since,  by  the  principle  of  conservation  of  centre  of  mass,  the  centre 
of  mass  is  fixed. 

The  initial  energy  of  the  system  is 

&  =  FS  +  ~  mirfu*  +  ~  mir!*P. 
The  final  energy  of  the  system  is 


r-  *  l     .     '  I 

=.r.y  ^ m\Vi    H wazv  -f-  — 

222 


By  the  conservation  of  energy,  §a  =  $,  and  inserting  the  values  of  -z/a,  aoi,  o>a,  r\,  r\,  r\,  r*,  we  have 


_    \  _  *»i»»i       ^»^»   ,   nu 
.         2m, 


Hence 


_  _  miVi       mimes'* GO'* (s'  +  s) 
/  2(mi  +  m^s* 


(6)  /«  a  «/^<?/  #«</  a.r/<?  M<?  radius  of  the  wheel  is  a  =  3  ft.,  of  the  axle  b  =  2  ft.  Let  r  =  i  inch  be  the 
radius  of  the  journal,  and  //  =  0.07  be  the  coefficient  of  kinetic  friction.  Let  the  moving  mass  P '  =  10  Ibs.  and 
the  mass  lifted  Q=  5  Ibs.  Let  P  start  from  rest  and  fall  for  a  time  t  =  5  sec.  Disregarding  rigidity  of 
the  rope  and  mass  of  rope,  wheel  and  axle,  discuss  the  apparatus,  {g  = 

ANS.  The  impressed  forces  are  Pg  and  Qg  down,  the  upward 
pressure  of  the  axle  R  and  the  forces  of  friction  which  form  a  couple 
+  F,  —  F  with  lever-arm  r  as  shown  in  the  figure.  The  effective  forces 

are  Pf  down  and  Q—f  up.  Reversing  these,  we  have,  by  D'Alembert's 
principle  (page  297), 


R-Pg- 


R  =  (P 


if  we  take  masses  in  Ibs.,  distances  in  ft.  and^,/"  in  ft.-per-sec.  per  sec. 
The  friction  for  new  bearing  (page  229)   is  then,  if  ft  is  the  angle 
of  bearing, 


Taking  moments  about  the  centre,  we  have,  by  D'Alembert's  principle, 
or  inserting  the  value  for  F  just  found  and  solving  for/,  we  have 


/  = 


If  we  disregard  friction  fi  =  o,  and  take  a  =  b,  we  have  the  same  value  for  /  as  already  found  in  example 
(2),  page  243. 


KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  I. 

If  the  angle  of  bearing  ft  is  small,  Tsin  ft  =  ft,  and  we  have 


/=  i  -  -si-  —  j-  --  K  .f  =  o.S4«-  =  17.49  ft.-p«r-sec.  per  sec. 


Again,  let  the  centres  of  mass  of  P  and  Q  be  initially  at  the  same  distance  h  above  any  plane  AB. 
The  initial  energy  is  then 


At  the  end  of  the  time  /suppose  /Mias  fallen  a  distance  j  and  g  risen  a  distance—  ,  while  the  mass  /'has  the 

F  ^> 

velocity  v  and  the  mass  0  the  velocity  -v.     Then  the  final  energy  of  P  is  Pg(h  —  s)   potential  and  —  V* 
kinetic,  and  of   Q,  Qg(h  +  -J  +  ~i»*-     The  total  final  energy  is  then 

t~Pf(k  -s)  +  P-  +  +  &  (h  +  £)  +  g>. 

The  work  consumed  by  friction  is  F  —  s.     By  the  law  of  energy  the  loss  of  energy  is  equal  to  the  work 
consumed  by  friction,  or 


Substituting  the  values  of  §i,  &  and  /"already  found,  we  obtain 


That  is,  the  loss  of  potential  energy,  Pgs  —  Q  -s,  equals  the  gain  of  kinetic  energy,  — z/»  +  =-  .  —v*,  plus  the 
work  of  overcoming  friction.  Also,  the  loss  of  potential  energy  of  P  or  Pgs  equals  the  gain  of  potential 
energy  of  Q  or  Qg  -  s,  plus  the  gain  of  kinetic  energy  of  the  system,  plus  the  work  of  overcoming  friction. 

If  we  substitute  s  =  -//*,  v  =//  and  solve  for/,  we  obtain  the  same  value  for/ as  already  obtained. 
The  acceleration  of  Q  is  then 

-/  =  0.363^  =  11.66  ft.-per-sec.  per  sec. 

The  velocity  of  P  at  the  end  of  /  =  5  sec.  is 

v  =//  =  8744  ft.  per  sec., 
and  the  velocity  of  Q 

—v  =  58.29  ft.  per  sec. 

The  distance  passed  through  by  P  is 

*  =  £//*  =  218.59  ft., 
and  by  Q 

-s  =  145-73  ft- 


CHAP.  I.]  .  EXAMPLES.  309 

Tension  in  string  for  P  is  P(g  —  /)      =  4-S^  poundals  or  4.56  pounds; 


work  of  P  =  4.56.1  =  996.77  ft.-lbs.  =  /**  --  ; 


"    onQ  =  6.82  x  i=.  993.88       "       =  Q-s  +    -^- 


The  difference  of  these  works,  or  2.89  ft.-lbs.,  is  the  work  consumed  by  friction. 
The  power  of  P 
The  efficiency  is 


The  power  of  P  is  "  *77  =  199.35  ft.-lbs.  per  sec.,   or     I99'35  =  0.36  horse-power. 


CHAPTER    II. 

EQUILIBRIUM  OF  A   MATERIAL  SYSTEM. 

Equilibrium  of  a  Material  System. — We  have  seen  (page  299)  that  the  motion  of  the 
centre  of  mass  of  a  material  system  is  not  affected  by  the  internal  forces,  but  only  by  the 
external  forces.  Also,  the  acceleration  of  the  centre  of  mass  is  the  same  as  for  a  particle  of 
mass  equal  to  the  mass  of  the  system,  all  the  external  forces  being  transferred  to  this  particle 
without  change  of  magnitude  or  direction. 

If  all  these  forces  constitute  a  system  of  forces  in  equilibrium,  and  every  particle  of  the 
material  system  is  at  rest,  it  is  said  to  be  in  static  equilibrium. 

If  not  at  rest,  since  the  external  forces  constitute  a  system  of  forces  in  equilibrium,  the 
centre  of  mass  has  no  acceleration,  and  the  resultant  moment  2F!  of  all  these  forces  relative 
to  any  point  is  zero.  But  we  have  seen  (page  302)  that  when  2FI  is  zero  the  moment 
of  momentum  does  not  change,  and  we  have  then  uniform  angular  velocity  of  every  particle 
about  an  invariable  axis  through  the  centre  of  mass,  and  uniform  velocity  of  translation  of 
this  axis.  A  system  in  this  condition  is  said  to  be  in  molar  equilibrium. 

If  there  is  no  rotation  about  an  axis,  but  simply  uniform  translation  in  a  straight  line, 
then  the  forces  acting  upon  every  particle  are  in  equilibrium,  and  the  system  is  said  to  be  in 
molecular  equilibrium. 

The  necessary  and  sufficient  conditions  of  molar  equilibrium  are  then 

Fx  =o,          Fy=  o,          F,  =  o, 
Mx  =  o,        My  =  o,        Mt  —  o. 

That  is,  the  external  forces  acting  upon  the  system  form  a  system  of  forces  in  equilib- 
rium, or  the  algebraic  sum  of  the  components  of  all  the  external  forces  in  any  direction  is 
zero,  and  the  algebraic  sum  of  the  moments  of  these  forces  about  any  axis  is  zero. 

For  molecular  equilibrium,  we  have,  in  addition  to  these  conditions,  the  condition  that 
all  the  forces  external  and  internal  on  every  particle  form  a  system  of  forces  in  equilibrium. 

For  static  equilibrium  we  have  still  the  added  condition  that  every  particle   is  at  rest. 

In  both  molecular  and  static  equilibrium,  then,  the  work  for  an  indefinitely  small  change 
of  position  or  configuration  must  be  zero.  Let  2zv  be  the  work  of  all  the  external  and  2iu' 
of  all  the  internal  forces  for  an  indefinitely  small  change  of  position  or  configuration.  Then 
we  have 

2w  -f-  ^w'  =  o,     or     2-w  =  —  ^v/. 

If  the  system  is  rigid,  2w'  =  o,  and  hence  2w  =  o. 

ILLUSTRATIONS. — Thus  a  billiard-ball  at  rest  on  a  table  is  in  static  equilibrium.  All  the  external  forces 
acting  upon  it  form  a  system  of  forces  in  equilibrium,  and  every  particle  is  at  rest.  Hence  the  forces  acting 
on  every  particle  must  form  a  system  in  equilibrium.  For  an  indefinitely  small  change  of  position  or  con- 
figuration the  work  is  zero. 

310 


CHAP.  II.  ] 


STABLE  EQUILIBRIUM. 


If  the  ball  rotates  about  a  fixed  vertical  axis  with  uniform  angular  velocity,  it  is  in  molar  equilibrium. 
All  the  external  forces  acting  upon  it  form  a  system  of  forces  in  equilibrium,  but  the  forces  acting  upon  every 
particle  do  not.  Since  the  moment  of  momentum  cannot  change,  the  centre  of  mass  is  at  rest,  and  the  ball 
must  always  rotate  about  the  vertical  axis  with  uniform  angular  velocity.  For  an  indefinitely  small  change 
of  position  or  configuration  the  work  is  not  zero. 

If  the  ball  rolls  so  that  the  centre  of  mass  moves  with  uniform  speed  in  a  straight  line,  it  is  also  in  molar 
equilibrium. 

If  the  ball  is  projected  through  the  air  without  rotation,  so  that  every  particle  has  uniform  motion  of 
translation  only  in  a  straight  line,  the  ball  is  in  molecular  equilibrium.  All  the  external  forces  form  a  system 
in  equilibrium,  and  the  internal  forces  on  every  particle  also  form  a  system  in  equilibrium.  For  any 
indefinitely  small  change  of  position  or  configuration  the  work  is  zero. 

If  the  particles  of  a  mass  of  water  in  a  horizontal  circular  dish  are  rotating  about  a  vertical  axis  through 
the  centre  of  mass  and  are  acted  upon  only  by  gravity  the  mass  is  in  molar  equilibrium.  The  centre  of  mass,, 
is  fixed,  and  no  particle,  disregarding  friction,  can  change  its  moment  of  momentum  about  the  vertical  axis. 

A  beam  or  spring  bent  by  a  load  and  at  rest  is  in  static  equilibrium.  If  every  particle  has  uniform 
motion  of  translation  only  in  a  straight  line,  it  is  in  molecular  equilibrium.  In  both  cases,  for  an  indefinitely 
small  change  of  configuration,  the  work  of  the  load  is  equal  and  opposite  to  the  work  of  the  internal  forces. 

Stable  Equilibrium. — The  same  conditions  must  hold  for  stable  equilibrium  of  a  material 
system  as  for  each  particle  (page  277). 

Hence  for  stable  equilibrium  the  potential  energy  is  a  minimum  and  the  kinetic 
energy,  if  any,  is  a  maximum.  If  the  potential  energy  is  a  maximum,  the  equilibrium  is 
unstable  and  the  kinetic  energy,  if  any,  a  minimum.  If  the  potential  energy  is  neither  a 
maximum  nor  a  minimum,  the  equilibrium  is  stable  for  displacements  for  which  the  poten- 
tial energy  increases,  and  unstable  for  displacements  for  which  the  potential  energy  decreases. 
If  the  potential  energy  is  constant,  the  equilibrium  is  neutral.  If  for  all  possible  displacements, 
large  or  small,  the  potential  energy  is  constant,  the  equilibrium  is  indifferent. 

ILLUSTRATIONS. — Let  a  prism  stand  on  a  level  base.   For  equilibrium  the  weight  W  acting  at  the  centre 
of  mass  C,  and  the  resultant  upward  pressure  R  on   the  base  at 
P,  must  be  equal  and  opposite  and  in  the  same  straight  line. 

If  the  prism  is  so  constrained  that  it  can  have  displacement  of 
translation    only   along   a  horizontal  plane,  the   potential  energy 
W  x  CP  does  not  change  for  any  displacement  large  or  small,  and    — 
the  equilibrium  is  indifferent. 

If  it  is  possible  to  rotate  the  prism  about  an  edge  at  Bor  A,  the 
potential  energy  is  increased,  so  long  as  the  weight  W  falls  inside  the  base,  and  the  equilibrium  is  stable. 

If,  however,  the  weight  W  cuts  the  base  at  an  edge  B,  then  for  rotation 
about  A  the  potential  energy  increases  and  the  equilibrium  is  stable,  while  for 
rotation  about  B  the  potential  energy  decreases  and  the  equilibrium  is 
unstable. 

Again,  if  a  pendulum  hangs  vertically  with  the  bob  below  the  point  of 
suspension  P.  the  potential  energy  W  x  CO  is  a  minimum  for  all  possible 
displacements  and  the  equilibrium  is  stable.  If  the  pendulum  is  reversed,  the 
potential  energy  is  a  maximum  and  the  equilibrium  is  unstable.  If  the  pendu- 
lum swings,  the  kinetic  energy  is  a  maximum  in  the  first  case  and  a  minimum 
in  the  second  case. 

Principle  of  Least  Work. — The  work  done  by  external  forces 

I 1 in  changing  the  configuration  of  an  elastic  body  can  be  given  back 

by  the  body  when  the  external  forces  are  removed.      It  is  therefore 

potential  energy.  If  such  a  body  is  in  static  or  molecular  equilibrium,  the  forces  external 
and  internal  form  a  system  of  forces  in  equilibrium.  If  the  equilibrium  is  stable,  the  work 
done  in  causing  the  change  of  configuration  must  be  a  minimum  consistent  with  the  conditions 
of  equilibrium  of  the  forces. 

This  is  called  the  "  principle  of  least  work  "  for  elastic  bodies. 


P  B 


W 


EQUILIBRIUM  OF  A   MATERIAL  SYSTEM. 


[CHAP.  II. 


Stability  in  Rolling  Contact. — As  an  application  of  the  preceding,  let  us  investigate 
the  equilibrium  of  a  body  with  a  curved  surface  resting 
and  rolling  upon  a  curved  surface. 

Let  O  be  the  centre  of  curvature  of  the  fixed  surface 
APB,  and  o  the  centre  of  curvature  of  the  body  aPc  resting 
on  it  at  P. 

The  reaction  R  at  P  is  for  equilibrium,  equal  and 
opposite  to  the  resultant  R'  acting  at  A,  of  all  the  other 
forces,  and  in  the  same  straight  line. 

Let  aPc  be  displaced  by  rolling  through  an  indefinitely 
small  angle  so  that  it  comes  into  the  position  a' PC',  A'  and 
o'  being  the  new  positions  of  A  and  o.  Let  the  radius 
of  curvature  oP =  p  and  OP=  pl ,  and  the  angle  a'o'P'  =  ft 
and  POP'=  a.  Let  the  distance  PA  =  h. 


We  have  then  PP'  =  a'P',  or 


pjt  =  p/3. 


Take  a  plane  OX  at  right  angles  to  R' .  The  equilibrium  will  be  stable  when  the 
potential  energy  is  a  minimum,  or  when  AO  is  less  than  A'O',  and  unstable  when  the 
potential  energy  is  a  maximum,  or  when  AO  is  greater  than  A'O'. 

The  distance  A  O  =  pl  -f-  //. 

We  have  then  for  stable  equilibrium 


for  unstable  equilibrium 

and  for  neutral  equilibrium 
Now  we  have 


pl+h<A'0, 

pl  +  h>  A'O', 


A'O'  =  (p  +  ft)  cos  a  -  (p  —  //)  cos  (a  -f  /?). 

v 
But,  by  Trigonometry, 


and 


cos  (a  -f-  ft)  =  cos  a  cos  ft  —  sin  a  sin  /?, 


cos  a  =  i  —  2  sin2-,          cos  ft  =  i  —  2  sin2-. 


Substituting,  we  have,  after  reduction, 
=  pl  +  A  —  2pl  sin2 


a  sin 


—  h  sin  a  sin  ft  —  2/t  sin2  -  —  2/1  sin2  -(\  —  2  sin2  -J. 

Now  when  a  is  very  small,  2  sin2  —  can  be  neglected  relative  to  I,  and  we  can  take  the 
arc  «  in  place  of  sin  a.      Hence,  since  ft  =  — lar,  we  can  take 

.      *       PI  .   ,«       «2  .   ,0 

sin  a=  a,         sin  p  =  -a,         smz-  =  — ,         sur-  = 

r  2         4  ^ 


CHAP.  II.]  STABILITY  IN  ROLLING   CONTACT. 

Making  these  substitutions,  we  have 


We  have  then  for  stable  equilibrium 


Pi  +  h<A'0',      or     k<^^t      or     \>j  +  j>'. 


for  unstable  equilibrium 


while  for  neutral  equilibrium  we  have 


.       pl          =  A,     or     >fr  = 

If  the  concavity  of  either  surface  is  turned  the  other  way,  we  have  the  same  result 
except  that  the  sign  of  the  corresponding  radius  will  be  changed.  If  either  surface  is  plane, 
its  radius  is  infinite. 

The  same  results  can  be  obtained  more  simply  as  follows  :  The  equilibrium  will  be 
stable,  unstable  or  neutral  according  as  A'  lies  to  the  left,  right  or  directly  over  P'  ,  that  is, 
according  as  the  horizontal  distance  of  A'  from  P  is  less,  greater  than  or  equal  to  the 
horizontal  distance  of  P'  from  P. 

The  horizontal  distance  of  P'  from  P  is  pl  sin  a.      The  horizontal  distance  of  A'  from  j 
is  the  same  as  from  a'  ,  or  h  sin  (a  -f-  /?). 
Hence  we  have 

h  sin  (or  -j-  /?)  =  pl  sin  OL. 


— 
the  same  conditions  as  before. 


Inserting  fi  =  —  V,  we  have  for  a  very  small,  so  that  we  can  take  the  arc    for  the  sine, 


Examples.— (i)  A  body  made  up  of  a  cone  and  a  hemisphere  having  a  common  base  rests  with  the  axis 
vertical  on  a  horizontal  plane.  Find  the  greatest  height  of  the  cone  for  stable  equilibrium. 

ANS.  Let  h  be  the  height  of  the  cone,  r  the  radius  of  the  hemisphere,  and  C  the  centre  of  mass.  The 
height  required  is  that  height  for  which  PC  =  r. 

•->  j 

The  volume  of  the  hemisphere  is  -xr3.     The  volume  of  the  cone  is  —itr^h. 

The  centre  of  mass  of  the  hemisphere  is  at  a  distance  above  P  equal  to  gV.    The 

o 

centre  of  mass  of  the  cone  is  at  a  distance  above  P  equal  to  r  +  — .      We  have 
then 

2  5          Ttr^h       I 

-Ttr3  x  f-r  + x  [r  +  - 

PC=* L 3__A ±,=rt     or     A  =  ,TJ. 

3    ,    ~  ~3~ 

(2)  A  prolate  spheroid  rests  with  its  axis  horizontal  on  a  roi'gh  horizontal  plane.  Show  that  for  rolling 
displacement  in  its  equatorial  plane  the  equilibrium  is  indifferent,  and  for  rolling  displacement  in  the  vertical 
plane  through  the  axis  it  is  stable. 


3*4 


EQUILIBRIUM  OF  A  MATERIAL  SYSTEM. 


[CHAP.  II. 


(3)  A  right  circular  cylinder  of  radius  r  rests  with  its  axis  horizontal  on  a  fixed  sphere  of  radius  R 
greattr  than  r.  Show  that  for  rolling  displacement  the  equilibrium  is  stable  or  unstable  according  us  the  plane 
of  displacement  makes  an  angle  with  the  vertical  plane  through  the  axis  of  the  cylinder  whose  sine  is  less  or 


greater  than  y  J  —  jr . 


ANS.  Let  p  be  the  radius  of  curvature  of  the  rolling  curve  at  the  point  of  contact.     Then  the  condition 
for  stable  equilibrium  is 


-s,  4.  _ 

r      R        ' 


Let  the  plane  of  displacement  make  the  angle  0  with  the  vertical  plane  through  the  axis  of  the  cylinder. 


The  rolling  curve  is  then  an  ellipse  whose  semi-minor  axis  is  r  and  whose  semi-major  axis  is 
radius  of  curvature  at  the  point  of  contact,  that  is,  at  the  vertex  of  the  minor  axis,  is 


The 


P  = 


sln~e 


Hence  for  stable  equilibrium 


sin'0 


or    sin  9  <  i/  i  —  -. 


(4)  A  prolate  hemispheroid  rests  with  its  vertex  on  a  horizontal  plane.     Show  that  for  rolling  displace- 
ment the  equilibrium  is  stable  or  unstable  according  as  the  eccentricity  of  the  generating  ellipse  is  less  or  greater 


than 


ANS.  Let  a  be  the  semi-major  and  b  the  semi-minor  axis.   Then  the  distance  OCto  the  centre  of  mass  is 

50  /~    \i 

4      $a  —  a         8 

The  distance  PC  then  is  ^-a.     The  radius  of  curvature  at  P  is  p  =  — .     We  have 

a 


then  for  stable  equilibrium 

i 


But  the  eccentricity  of  the  generating  ellipse  is 


Hence  for  stable  equilibrium  e  < 


CHAPTER    III. 

ROTATION   ABOUT   A    FIXED   AXIS.* 

Rotation  about  a  Fixed  Axis  —  Effective  Forces.  —  We  have  seen  that    D'Alembert's' 

principle  (page  297)  reduces  any  kinetic  problem  to  one  of  equilibrium  between  actual 
(impressed]  forces  and  fictitious  (reversed  effective]  forces.  In  order  to  apply  it,  then,  we  must 
be  able  to  find  in  any  given  case  the  effective  forces  and  the  moments  of  the  effective  forces. 

Let  us  consider  first  the  case  of  rotation  about  a  fixed  axis.  Let  O  be  the  centre  of 
mass,  and  let  us  take  the  origin  of  co-ordinates  O'  at  the  point  of  intersection  of  the  axis  of 
rotation  with  a  plane  through  the  centre  of  mass  O  at  right  angles  to  this  axis.  Let  ~x,  y  ',  Is 
be  the  co-ordinates  of  the  centre  of  mass  O  for  this  origin  O'  and  any  co-ordinate  axes 
O'X',  O'Y',  O'Z'. 

Equations  (i),  page  158,  give  the  components  of  the  tangential  acceleration  for  any 
particle  of  a  rotating  body.  If  we  multiply  ea,ch  term  by  the  mass  m  of  the  particle  and 
sum  up  for  all  the  particles,  we  shall  have  the  components  of  the  effective  tangential  forces  of 
the  -body.  These  forces  cause  change  of  magnitude  of  the  velocity  of  each  particle  in  its 
plane  of  rotation.  In  summing  up,  since  x,  y,  z  are  taken  from  the  centre  of  mass  O  as 
origin,  we  have  ~2,mx  =  o,  ~2my  =  o,  ~2.mz  =  o. 

We  have  then  for  the  components  of  the  effective  tangential  forces  for  all  the  particles 
of  the  body,  since  2m  =  m  the  mass  of  the  body, 


where  ax,  «y,  az  are  the  components  of  the  angular  acceleration  a  about  the  axis  of  rotation. 

These  forces  cause  change  of  magnitude  of  the  velocity  of  each  particle  in  its  plane  of 
rotation. 

Equations  (3),  page  159,  give  the  components  of  the  deflecting  acceleration  for  any 
particle  of  a  rotating  body.  Here  again,  multiplying  each  term  by  the  mass  m  of  the  particle 
and  summing  up  for  all  the  particles,  we  have  the  components  of  the  effective  deflecting 
forces  for  all  the  particles, 


2mfr   =  —  niyoo^  —  inyctf*2,  }• (2) 


*  Before  reading  this  and  the  following  chapters  the  student  should  be  familiar  with    the  principles  of 
Chap.  II,  page  153,  and  Chap.  Ill,  page  31. 

315 


3l6  KINETICS  OF  A  MATERIAL   SYSTEM    ROTATION—  FIXED  AXIS.  [CHAP.  III. 

where  a>x,  toyt  oo,  are  the  components  of  the  angular  velocity  a>  about  the  axis  of  rotation. 
These  forces  cause  change  of  direction  of  velocity  of  each  particle  in  its  plane  of  rotation. 

For  fixed  axis  the  plane  of  rotation  of  each  particle  does  not  change.  There  are 
therefore  no  deviating  accelerations. 

Adding  equations  (i)  and  (2)  we  have  then  for  the  components  of  the  effective  forces 
for  a  body  rotating  about  a  fixed  axis,  for  any  co-ordinate  axes  we  please. 


<»!  + 

'  ^ 


If  we  take  Jhe  fixed  axis  coinciding  with  one  of  the  co-ordinate  axes,  as,  for  instance, 
with  O'X',  we  have  «,  =  o,  GO,  —  o,  ay  =  o,  at  =  o,  and  equations  (3)  become 

2mfx=  o, 

(4) 


If  we  take  distance  in  feet  and  mass  in  Ibs.,  all  these  equations  give  force  in  pou-ndals. 
For  force  in  pounds  we  divide  by  g  (page  171). 

We  see  from  these  equations  that  the  effective  force  in  any  direction  is  the  same  as  for 
a  particle  of  mass  equal  to  that  of  the  body  moving  with  the  acceleration  of  the  centre  of 
mass  in  that  direction,  as  already  proved  on  page  299. 

If  the  axis  of  rotation  passes  through  the  centre  of  mass,  we  have  !*•  =  o,  y  =  o,  z  =  o, 
and  hence  2»t/x  =  o,  2m  fy  =  o,  2mft  =  o. 

That  is,  when  a  body  rotates  about  an  axis  through  the  centre  of  mass  the  effective 
forces  in  any  direction  are  zero. 

Moments  of  the  Effective  Forces.  —  Equations  (10),  page  160,  give  the  component 
moments  of  the  acceleration  for  any  particle  of  a  rotating  and  translating  body.  For  a  fixed 
axis  the  deviating  acceleration  is  zero,  and  therefore,  as  we  see  from  equations  (4),  page  159, 
we  have  ooxoay  =  o,  coxcot  =  o,  <*>y<*)t  =  O,  ooyoox  =  o,  ootcax  =  O,  oatoay  =  o.  We  can  also  put 
•^',  y  ',  z  in  place  oi~x-\-x,  ~y  -\-  y,  ~z  -\-  z,  and  for  rotation  only  we  have  fx  =  o,fj,  =  o, 
/.  =  o. 

If  we  make  these  changes  in  equations  (ib),  page  160,  multiply  each  term  by  the  mass 
m  of  the  particle  and  sum  up  for  all  the  particles,  we  shall  have  the  components  of  the 
moments  of  the  effective  forces  for  all  the  particles  of  the  body.  In  summing  up  we  have 


where  I'x  ,  Ij,  /.'  are  the  moments  of  inertia  of  the  body  for  the  axes  O'X',  O'Y',  O'Z'. 

We  have  then,  from  equations  (10),   page    160,  for  the  component   moments  of   the 
effective  forces, 


.      .      .      (2) 


M'fx  =  - 

M'fy  =  —  at2*nc'y  —  a^mx'y  -\-  (a?*  —  QC%)2mxJ  +  Iyay 


M'ft  =  — 


CHAP.  III.]  ORIGIN  OF  THE   TERM  "MOMENT  OF  INERTIA."  317 

If  the  fixed  axis  coincides  with  one  of  the  co-ordinate  axes,  as,  for  instance,  O '  X\  we 
have  Goy  =  O,  GO,  =  o,  ay  =  o,  a,  =  o,  and  hence 


=  comx'z'  — 


(5) 


Since  we  can  take  any  co-ordinate  axes  we  please,  let  the  co-ordinate  axes  be  principal 
axes  at  the  point  O'.  Then  (page  35)  2mx'y'  =  o,  2my'zf  =  o,  2mz'x'  =  o,  and  equations 
(4)  become 

M'fx=I'xax,          M»  =*;«,,          M'fz  =  I'tas (6) 

If  the  axis  of  rotation  is  a  principal  axis,  let  it  coincide  with  O'X',  for  instance.  Then 
ay  —  o,  nx  =  o,  and  from  (5)  we  have  Mfx  =  I'xax ,  Mfy  —  o,  Mft  =  o. 

That  is,  when  a  body  rotates  about  a  principal  axis  with  angular  acceleration  or,  the 
moment  of  the  effective  forces  relative  to  the  axis  of  rotation  is  equal  to  the  moment  of 
inertia  /'  of  the  body  relative  to  the  axis  multiplied  by  the  angular  acceleration,  or 

M'f  =  I'a. 

If  Ix ,  Iy,  Iz  are  the  moments  of  inertia  for  principal  axes  through  the  centre  of  mass. 
O,  parallel  to  O'X',  O'Y',  O'Z' ,  we  have  (page  33) 

/;  =  Ix  +  mCy2  +  P),         /;  =  Iy  +  m(P  +  j?),        //  =  /.  +  m(^  +  5?), 

where  m  is  the  mass  of  the  body  and  ^,  y,  ~z  the  co-ordinates  of  the  centre  of  mass 
Inserting  these  values  in  (6),  we  see  that  the  moment  of  the  effective  forces  about  any  fixed 
axis  is  equal  to  the  moment  about  a  parallel  axis  through  the  centre  of  mass  plus  the 
moment  of  the  effective  force  of  a  particle  of  mass  equal  to  the  mass  of  the  body  at  the 
centre  of  mass. 

If  we  take  distance  in  feet  and  mass  in  Ibs.,  these  equations  (4),  (5),  (6)  give  moments 
in  poundal-feet.  For  pound-feet  divide  by  g  (page  171). 

Origin  of  the  Term  "  Moment  of  Inertia." — Let  F be  the  resultant  of  all  the  impressed 
forces  at  right  angles  to  the  axis  of  rotation  and  p  its  lever-arm,  so  that  Fp  is  the  resultant 
moment  of  all  the  impressed  forces  relative  to  the  axis  of  rotation.  Then,  by  D'Alembert's 
principle,  when  the  axis  of  rotation  is  a  principal  axis 

Fp—Mf—o,      or     Fp=I'a. 

The  term  "moment  of  inertia"  is  due  to  Euler.  Euler  used  the  term  "inertia"  as 
synonymous  with  what  we  call  mass.  Thus  the  equation  of  force, 

F=  mf, 
would  be  read  in  the  terminology  of  Euler 

Force  =  inertia  X  linear  acceleration. 
In  the  equation 

Fp  =  I'a  =  a^mr3 

Euler  called  the  term  ^mr*  "  moment  of  inertia"  and  thus  obtained  the  analogous  expression 
moment  of  force  =  moment  of  inertia  X  angular  acceleration. 


318  KINETICS  OF  A  MATERIAL  SYSTEM-ROTATION  -FIXED  AXIS.  [CHAP.  III. 

The  term  "  moment  of  inertia  "  in  modern  scientific  terminology  is  an  improper  expres- 
sion. Inertia  (page  169)  is  a  property  of  matter  like  color  or  hardness,  and  we  cannot 
properly  speak  of  moment  of  inertia  any  more  than  of  moment  of  color  or  hardness.  The 
term  "second  moment  "  of  mass  would  more  correctly  describe  the  product  Stnr2,  and  has 
been  used  by  some  recent  authors.  The  expression  "moment  of  inertia,"  however,  has 
beco'me  firmly  established  by  long  usage. 

The  student,  while  using  it,  should  consider  it  simply  as  a  name  for  the  quantity  ZZtnr2, 
which  occurs  so  frequently  in  dynamic  problems  that  it  is  convenient  to  give  it  a  special  name. 

Momentum  of  Rotating  Body.  —  From  equations  (2),  page  154,  we  have  tne  component 
velocities  for  any  particle  of  a  rotating  body 

vM=(s  +  £)a>,  —  (y+y)a>,,       vy=(x  +  x}™t—(z+z)c*>x,       v,  =  (J+y)cox-  (x  +  x)vy. 

If  we  multiply  each  term  by  the  mass  m  of  the  particle  and  sum  up  for  all  the  particles, 
we  shall  have  the  components  of  momentum  for  all  the  particles.  In  summing  up,  since  x, 
y,  z  are  taken  from  the  centre  of  mass,  we  have  "Sinx  =  o,  2my  =  o,  2mz  —  o. 

We  have  then  for  the  components  of  the  momentum  of  a  rotating  body,  since  2m  =  m 
the  mass  of  the  body, 


,  —  TOLZGOX, 


For  axis  of  rotation  through  the  centre  of  mass  we  have  ~x  =  o,  ~y  =  o,  ^  =  o,  and 
hence  2mvx  =  o,  2mi>y  =  o,  2mvt  =  o. 

Hence  the  momentum  of  a  body  rotating  about  an  axis  is  the  same  as  for  a  particle  of 
mass  equal  to  ttie  mass  of  the  body  at  the  centre  of  mass. 

Moment  of  Momentum  —  Rotating  Body.  —  Equations  (10),  page  155,  give  the  component 
moments  of  the  velocity  for  any  particle  of  a  rotating  and  translating  body.  For  rotation 
only  we  have  vx  =  o,  vy  =  o,  v,  =  o.  If  we  make  these  changes  in  equations  (10),  page  155, 
multiply  by  the  mass  m  of  the  particle  and  sum  up  for  all  the  particles,  we  shall  have  the 
components  of  the  moment  of  momentum  for  the  body,  for  any  co-ordinate  axes  we  please. 
Let  us  take  these  axes  principal  axes  at  O'.  Then  we  have  'Smx'y'  —  o,  ^my'z'  —  o, 
2mz'x'  =  o.  We  have  also  Zm(  /*  +  z"^  =  f'x,  2m(^  +  x'z)  -  /',,  2»t(x'*  +/a)  =  /,'  , 
and  2  mx  =  o,  2»i  y  =  o,  2ntz  =  o. 

We  have  then  from  equations  (10),  page  155,  the  component  moments  of  momentum 


(8) 


If  the  axis  of  rotation  coincides  with  a  principal  axis,  as,  for  instance,  O'X',  we  have 
a?,  =  o,  cot  =  o,  and  from  (8)  we  have  M'vx  =  Jxa>x,  M^  =  o,  M'n  =  o. 

That  is,  when  a  body  rotates  about  a  principal  axis  with  angular  velocity  a?,  the  moment 
of  momentum  relative  to  the  axis  of  rotation  is  equal  to  the  moment  of  inertia  /'  of  the  body 
relative  to  that  axis,  multiplied  by  the  angular  velocity,  or 

Ml=  Too. 

If  /,  ,  Iy  ,  It  are  the  moments  of  inertia  for  principal  axes  through  the  centre  of  mass  O, 
we  have 

/',=  /,  +  ^(?  +  **)•         7-  =  7«  + 


CHAP.  III.]  PRESSURES  ON  FIXED  AXIS. 


319 


Inserting  these  values  in  (8),  we  see  that  the  moment  of  momentum  for  a  body  rotating 
about  an  axis  is  equal  to  the  moment  of  momentum  about  a  parallel  axis  through  the  centre 
of  mass,  plus  the  moment  of  momentum  for  a  particle  of  mass  equal  to  the  mass  of  the 
body  at  the  centre  of  mass. 

Pressures  on  Fixed  Axis.  —  Equations  (4)  give  the  component  effective  forces  for  a  body 
rotating  about  any  axis  O'Xf,  and  equations  (5)  give  the  component  moments  of  the 
effective  forces  for  a  body  rotating  about  any  axis  O'  X'  . 

Let  the  axis  be  supported  at  the  ends  a  and  b  so  that 
it  cannot  change  its  direction,  and  let  the  distances 
O'a  =  llt  O'b  =  /2,  the  plane  Y'Z'  being  a  plane  through 
the  centre  of  mass  O. 

Let  the  component  pressures  at  a  and  b  be 

R'x,  R'y,  R'f     and     R'x,  Ry>  R"t.  , 

Let  the  components  of  all  the  other  impressed  forces        |    /v 
be 

2FX,     2Fy, 


and  the  component  moments  of  these  forces  about  the  co-ordinate  axes  be 

about  O'X'  =  2F     —  2F 


"      O'Z'  = 
We  have  then  from  equations  (4),  by  D'Alembert's  principle,  for  any  fixed  axis  O'X' 

+  3F,  =  -  m^G?2,  -  mj",,  ] 

[.  .....      .      .      (9) 

where  m  is  the  mass  of  the  body  ;   x,  y,  J,  the  co-ordinates  of  the  centre  of  mass. 
We  have  also  from  equations  (5),  by  D'Alembert's  principle,  for  any  axis  O'X' 


(10) 


=  Ix'ax. 


From  the  last  of  equations  (10)  we  can  find  a  x  ,  and  then  from  the  first  two  and  the  first 
two  of  equations  (9)  we  can  find  Ry'  ,  Ry'  ,  RM',  Rn"  .  This  leaves  Rx  and  Rx"  indeterminate, 
but  their  sum  is  given  by  the  last  of  equations  (9). 

If  we  take  distance  in  feet  and  mass  in  Ibs.  ,  the  pressures./?  will  be  in  poundals.  For 
pounds  divide  by  g  (page  171). 

In  equations  (9)  the  terms  —  mFo?^2  and  —  'myGox2  are  the  sums  of  the  components 
parallel  to  Z'  and  Yf  of  the  effective  deflecting  forces  (page  315)  of  all  the  particles,  and  the 
terms  —  *&yax  and  —  mzax  are  the  sums  of  the  components  parallel  to  Z'  and  Y'  of  the 
effective  tangential  forces  (page  315)  of  all  the  particles. 


320  KINETICS  OF  A  MATERIAL  SYSTEM-ROTATION-FIXED  AXIS.  [CHAP.  III. 

In  equations  (10)  the  terms  —  cs&Smy'x'  and  of^mx  '  z'  are  the  moments  about  Z'  and 
V  of  the  effective  deflecting  forces,  and  —  ax2mx'zf  ,  —  a^mx'y'  are  the  moments  about 
Z'  and  Y'  of  the  effective  tangential  forces,  and  I'xax  is  the  moment  about  the  axis  of 
rotation  X'  of  the  effective  tangential  forces. 

If  the  axis  of  rotation  passes  through  the  centre  of  mass  we  have  y  =  o,  ~z  =  o,  and  we 
see  from  equations  (9)  that  the  sums  of  the  components  of  the  effective  deflecting  and 
tangential  forces  are  zero,  or  these  forces  reduce  to  a  couple. 

If  the  axis  of  rotation  is  a  principal  axis,  then  taking  the  other  two  axes  as  principal 
axes  we  have  2my'x'  =  o,  2mx'z'  —  o,  and  we  see  from  equations  (10)  that  the  moments 
of  the  effective  deflecting  and  tangential  forces  about  Y'  and  Z'  are  zero.  There  is  then 
no  tendency  of  the  axis  of  rotation  O'X'  to  turn  about  the  other  two  co-ordinate  axes  O'  Z' 
orO'Y'. 

If,  then,  the  fixed  axis  of  rotation  is  a  principal  axis  through  the  centre  of  mass,  there 
will  be  no  pressure  on  this  axis  due  to  rotation,  and  in  such  case  we  have,  from  equations 
(9)  and  (10), 

Rxr  -f-  Rx'  +  2FX  =  o,         Ry'  +  Ry"  +  2Fy  =  o,         JR.'  +  R,"  +  2F.  =  o, 
2F,v  =  o,          -  J?,'/,  +  *."/,  +  ^Fjt  -  SF*  =  o, 


That  is,  for  a  body  rotating  about  a  fixed  principal  axis  through  the  centre  of  mass  the 
Pressures  on  the  axis  are  the  same  as  if  the  body  did  not  rotate,  and  are  given  by  the  conditions 
of  equilibrium  of  the  impressed  forces  only. 

Conservation  of  Moment  of  Momentum  —  Fixed  Axis.  —  We  have,  from  equations  (8), 
for  the  component  moments  of  momentum  for  a  body  rotating  about  a  fixed  axis,  taking  the 
co-ordinate  axes  as  principal  axes  at  O', 


and,  from  equations  (6),  for  the  component  moments  of  the  effective  forces 


If,  in  equations  (#),  we  have  ocx  •=  o,  ay=.  o,  at  =  o,  we  shall  evidently  have  oox  ,  ooy,  GO, 
constant  in  equations  (a).  But  when  ax  —  o,  ay=  o,  a,  =  o,  we  have  Mfx  =  o,  Mfy  =  o,  Mft=  o. 

Since,  by  D'Alembert's  principle,  the  moment  of  the  effective  forces  is  equal  to  the 
moment  of  the  impressed  forces,  we  have  the  moment  of  the  impressed  forces  zero. 

Hence,  if  the  moment  of  the  impressed  forces  about  a  fixed  axis  is  always  zero,  the 
moment  of  momentum  about  that  axis  is  constant. 

Kinetic  Energy  —  Fixed  Axis.  —  The  kinetic  energy  of  a  particle  of  mass  m  and  velocity 

v  is  -mv*  (page  272).      If  a  particle  has  the  component  velocities  vx  ,  vy,  vtt     we  have 


and  hence 


CHAP.  III.]  EQUATIONS  FOR  ROTATION  AND   TRANSLATION.  321 

From  page  154  we  have  for  the  component  velocities  for  any  particle  of  a  rotating  body 


where  x',  y1  ',  z'  are  taken  from  the  origin  O',  the  intersection  with  the  axis  of  rotation  of  a 
plane  through  the  centre  of  mass  O  at  right  angles  to  the  axis  of  rotation.  If  we  square  these 

component  velocities,  multiply  each  term  by  -m,  sum  up  for  all  the  particles  and  add,  we 

shall  have  the  kinetic  energy  for  a  rotating  body  for  any  co-ordinate  axes  O'X',  O'  Y'  ,  O  '  Z1  ', 
we  please.  Let  us  take  these  co-ordinate  axes  as  principal  axes  of  the  body  at  the  origin  (y. 
Then  we  have 


=  O,  myz  =  O,  mzx  =  o. 

We  thus  obtain  for  the  kinetic  energy  of  a  rotating  body 


K  =  - 

But 

2m(y*  +  z"*)  =  rm  ,         2m(z*  -f  *'2)  =  // 
Hence 

*=^+5#aJ+5/>2  .........      (II) 

But  GOX  =  a?2  cos2  a,  a$  =  a?2  cos2  /?,  a?2  =  a?2  cos2  y,  where  GO  is  the  angular  velocity 
about  the  axis  of  rotation  and  a,  /3,  y  are  the  direction  angles  of  this  axis.  Therefore 
equation  (n)  becomes  (page  35) 


(12) 


where  /'  is  the  moment  of  inertia  relative  to  the  axis  of  rotation. 

Hence  the  kinetic  energy  for  a  body  rotating  about  an  axis  is  equal  to  one  half  the  product 
of  the  moment  of  inertia  of  the  body  relative  to  that  axis  and  the  square  of  the  angular  velociy 
about  that  axis. 

Analogy  between  the  Equations  for  Rotation  and  Translation — The  student  should  not 
fail  to  note  the  analogy  between  the  equations  for  rotation  and  translation. 
Thus  for  translation 

F  -  m/  =  force, 
while  for  rotation  (principal  axis) 

Fp  =  I' a  =  moment  of  force. 
For  translation 

mv  =  momentum, 
while  for  rotation  (principal  axis) 

/'<»=  moment  of  momentum. 
For  translation 

=  uniform  force, 


322  KINETICS  OF  A  MATERIAL  SYSTEM ^ROTATION-FIXED  AXIS.  [CHAP.  ill. 

while  for  rotation  (principal  axis) 

/'(o>  —  <») 

— - —    i-  =  moment  of  uniform  force. 

For  translation 

-  ihz/*  =  kinetic  energy, 

while  for  rotation  (any  axis) 

— 7'oj2  =  kinetic  energy. 

We  see  that  if  linear  acceleration  and  velocity  are  replaced  by  angular  acceleration,  and 
velocity  and  mass  by  moment  of  inertia,  force  and  momentum  become  moment  of  force  and 
momentum,  and  kinetic  energy  is  given  in  both  cases. 

Reduction  of  Mass. — It  is  often  desirable  to  be  able  to  reduce  a  rotating  mass  to  a 
particle  of  equivalent  mass  at  any  desired  distance  from  the  axis. 

Thus  let  a  body  of  mass  m  rotate  about  a  principal  axis  YY  with  angular  velocity  co 
and  angular  acceleration  a.  Then  the  moment  of  momentum  is  I 'GO  and 

the  moment  of  the  force  is  I 'a,  and  the  kinetic  energy  is  -I'd. 


JNow  suppose  a  particle  of  mass  m  at    any  desired  distance  d  from  the 
axis    to  have  the  same  angular  velocity  and  acceleration.       Then  its  mo- 


ment of    momentum  is  md*(*>,  the  moment  of  the  force  is  md*a,  and  the 
"Y  kinetic  energy  is  —md*coz. 

If  then  we  have 

mdz  =  I't     or     w  =  — 8, 

the  particle  m  at  d  will  have  the  same  moment   of  momentum,  moment  of  force  and  kinetic 
energy  as  the  body  itself.      It  is  therefore  equivalent  to  the  body. 

Hence  to  reduce  a  rotating  mass  to  a  particle  of  equivalent  mass  at  any  desired  dis- 
tance from  the  axis  we  divide  the  moment  of  inertia  by  the  square  of  the  distance. 

Examples. — (i)  A  driving-wheel  of  a  locomotive  has  a  arank-pin  and  cross-head,  etc.,  of  mass  m\  =2600 
Ibs.,  the  centre  of  mass  being  at  a  distance  r\  =  6  inches  from  centre  of  wheel.  If  the  wheel  has  a  radius  of 
r  =  j  feet  and  is  running  60  miles  an  hour,  find  the  pressure  on  the  rail  when  the  crank-pin  has  its  lowest  and 
highest  position.  (Take^  =  32  ft.-per-sec.  per  sec.) 

ANS.   Let  m  be  the  mass  of  the  wheel  alone.     A  speed  of  60  miles  an  hour  is  88  feet  per  sec.     The 

88 
angular  speed  is  then  given  by  roo  =  88,  or  oo  =  —  radians  per  sec. 

The  impressed  forces  are  the  upward  pressure  fi  of  the  rail  and  the  weight  (m.  +  mi)g  of  the  wheel  and 
crank-pin  and  cross-head.  The  effective  force  is  nttrt<o*  acting  towards  the  centre,  for  the  crank.  The 
sum  of  the  effective  forces  for  the  wheel  is  zero  (page  316). 

By  D'Alembert's  principle,  for  crank  at  its  highest  position  we  have 

R  -  (m-f  mj)g  +  mirioo*  =  o,     or     R  =  (m  +  m^g  —  m^r^oo*  poundals. 
Inserting  numerical  values  and  dividing  by^-,  we  have 

K  =  (m  -  32355)  pounds. 
For  crank  at  lowest  position  we  have 

R  —  (m  +  >n\\g  —  m,riao*  =  o,     or     R  =  (mi  +  37555)  pounds. 
The  effect  is  to  cause  a  blow  upon  the  rail  every  time  the  crank  passes  the  lowest  point. 


CHAP.  III.]  EXAMPLES.—  ROTATION.  323 

(2)  In  the  preceding  example  let  the  thickness  of  the  wheel  be  t  =  2  inches,  and  density  $  =  480  Ibs.  per 
cubic  foot.  It  is  required  to  balance  the  crank  by  filling  in  between  the  spokes.  Find  the  inner  radius  r0  for 
any  given  angle  6  =  60°. 

ANS.  Let  mi  be  the  mass  of  the  ring,  and  J  the  distance  of  its  centre  of  mass.  Then 
we  should  have 


< 
We  have  (page  29)  w»  =  —  (r1  —  r0»)    and 


0  - 


Substituting  and  solving,  we  have  for  r0 


3*«in 
2<5/sin  — 


Inserting  numerical  values,  we  have  r0  =  2.79  feet. 

3.  A  sphere  of  mass  m  and  radius  ri  has  an  angular  velocity  <»,,  and  contracts  until  its  radius  —r\.       Find 
the  final  angular  velocity  GO  if  there  are  no  external  forces. 

ANS.  The  moment  of  momentum  cannot  change  (page  320).     Hence /i<»i  =  loo,    or    «?  — -*'001 

2  _  2  _  r  5 

We  have  (page  49)   7i  =  -mr,s     and     I  =  —  m—,     Hence 


The  angular  velocity  increases  as  the  sphere  contracts. 

Find  the  gain  of  kinetic  energy. — The  initial  kinetic  energy  is  (page  321) 


and  the  final  kinetic  energy  is 

The  gain  of  kinetic  energy  is  then 


This  gain  of  kinetic  energy  must  be  at  the  expense  of  potential  energy  (page  304). 
(4)  In  the  preceding  example  find  the  loss  of  potential  energy  due  to  contraction. 

ANS.  Let  m  be  the  mass  of  a  particle  on  the  surface  of  the  sphere.     The  attraction  between  the  sphere 
and  this  particle  is  (page  203) 

kmtn  _  m.mr*g 
r\*  mQr^ 

where  r0  is  the  radius  of  the  earth  and  ma  the  mass  of  the  earth,  and  g  the  acceleration  of  gravity  at  the 
earth's  surface. 

The  attraction  for  any  point  within  the  sphere  varies  directly  as  the  distance  from  the  centre.      Hence 
at  a  distance  pi  from  the  centre  the  attraction  is 


During  contraction  the  attraction  is  inversely  as  the  square  of  the  distance.     Hence  the  attraction  at 
a  distance  x  of  a  particle  originally  at  a  distance  pi  is 


324  KINETICS  OF  A  MATERIAL  SYSTKM-ROT AT  ION-FIXED  AXIS.  [CHAP.  III. 

The  loss  of  potential  energy  of  the  particle  is  then 


The  mass  of  a  unit  of  volume  of  the  sphere  is     m     .     The  volume  of  a  spherical  shell  of  radius  p,  is 

l*r,' 

Hence  the  mass  of  an  elementary  shell  is 


Substituting  this  for  m,  we  have  for  the  loss  of  potential  energy  of  an  elementary  shell 


The  total  loss  of  potential  energy  is  then 


«-i)   jpsfa  ^  3SV«!f(«  ~  ') 
i*       J  $m*r\ 


This  loss  of  potential  energy  must  be  converted  into  kinetic  energy  (page  304). 
We  have  just  seen  that  the  gain  of  kinetic  energy  is 


(2) 


Hence  if  (i)  is  greater  than  (2)  the  difference  must  be  converted  into  heat.    The  energy  converted  into 
heat  is  then 


If  we  divide  by^  we  have  this  energy  in  ft.-lbs.      If  we  then  divide  by  the  "mechanical  equivalent  of 
heat  "/,  we  obtain  the  number  of  heat-units.      We  have  then  for  the  number  of  heat-units  generated 

No.  of  heat-units  =  m(n  ~  '*  [3mr«V  -  m*ri\n+  i  )<»,']. 


If  S  is  the  density  of  the  sphere  and  r  is  the  density  of  water,  the  mass  of  a  volume  of  water  equal  to 
the  sphere  is  -?.  If  <r  is  the  specific  heat  of  the  sphere  and  T  the  number  of  degrees  rise  of  temperature, 
we  have 


No.  of  heat-units  =  — m7*. 


Hence 


CHAP.  III.] 


EXAMPLES.— RO  TA  TION. 


325 


(5)  A  disc  of  mass  m  has  a  motion  of  translation  v  and  of  rotation  <»i  in  its  own  plane.     If  any  point  of 
the  disc  is  suddenly  fixed,  find  the  angular  velocity  GO. 

ANS.  Let^  be  the  perpendicular  distance  between  the  fixed  point  P  and  the 
direction  of  motion  of  translation  v,  and  d  the  distance  between  P  and  O. 
Let  /=  mx-2  be  the  moment  of  inertia  for  the  axis  at  O,  and  /'  =  m(x-2  +  d3)  for 
the  axis  at  P. 

The  moment  of  momentum  about  axis  at  P  before  stoppage  is 


or    00 


and  after  stoppage  /'<».     Hence 


/'GO  = 


*9  + 


(6)  A  disc  of  mass  m  =  8  Ibs.  and  radius  r  =  2  feet  rotates  about  a  fixed 
horizontal  axis  AB.  A  perfectly  flexible  string  wound  on  the  disc  has  a  mass 
P  —  16  Ibs.  attached  to  its  lower  end.  Find  the  distance  described  by  P  in 
t  =  2  seconds,  neglecting  friction  and  mass  of  the  string.  ( Take  g  =  j^). 


ANS.     The  moment  of  inertia  for  axis  AB  is  /  = 


The    impressed 


forces  are  the  upward  reaction  R  at  O  and  the  weight  rag  of  the  disc  and  the 
weight  Pg  ol  P.     Let /be  the  acceleration  of    P,  then  Pf  downward  is  the 
P0  T  effective  force  on  P. 

The  moment  of  the  effective  forces  of  the  particles  of  the  disc  is  la  (page  317),  and  the  sum  of  the 
components  of  the  effective  forces  of  the  particles  of  the  disc  in  any  direction  is  zero  (page  316). 
Reversing  the  effective  forces,  we  have,  by  D'Alembert's  principle, 

R-mg-Pg  +  Pf  =  o, 
Pgr  —  Pfr  —  la  =  o. 

But  /  =  — r*  and  rot-  =/.     Hence  we  obtain  from  the  second  of  these  equations 


and  from  the  first  equation 


/  =  — -£—  =  —g  =  25.6  ft.-per-sec.  per  sec., 


R  =  mg  +  Pg  —  —Pg  —  (m+  —P)g  poundals, 


or  .#  =  m  +  —  P  =  ii.2  pounds. 

We  see  that  the  reaction  R  is  less  than  the  weight  m  +  P  =  24  pounds  of  the  apparatus. 
Since/  is  uniform,  we  have  for  the  distance  described  in  /  seconds  starting  from  rest 


NOTE.  —  The  mass  reduced  to  the  circumference  (page  322)  is  P  -\  —  5  =  P  +  —  .     We  have  then  directly 

(reduced  mass)  x  acceleration  =  moving  force,     or     f  P  +  -  \f  =  Pg. 

(7)  A  horizontal  uniform  disc  is  free  to  revolve  about  a  vertical  axis  through  its  centre.  A  man  walks 
around  on  the  outer  edge.  Find  the  angular  distance  described  by  the  man  and  the  disc  when  he  has  walked 
once  roztnd. 


ANS.  Let  M  be  the  mass  of  the  man.  and  D  the  mass  of  the  disc,  and  r  its  radius.    Then  /=  —r*. 

2 


3a6  KINETICS  OF  A  MATERIAL  SYSTEM-ROTATION-FIXED  AXIS.  [CHAP.  III. 

Let  a  be  the  angular  acceleration  of  the  disc,  and  F  the  force  exerted  on  its  circumference.     Then 
(page  317) 

Fr      2F 
Fr  =  la.,    or    a  —  —  =  — . 

If  a\  is  the  angular  acceleration  of  the  man,  we  have 

Fr  =  Mr*al ,    or    F  =  Mrav. 
Hence 


The  angular  distance  described  by  the  disc  is  then  0  =  —  at*,  and  by  the  man  61  =  —  a,/*,  and  when 
the  man  has  walked  once  round  we  have 

~2.a        2   ' 
Inserting  the  value  of  en,  we  have  for  the  angular  distance  of  the  man 

I         ^  _       21fD 
2 

and  for  the  angular  distance  of  the  disc 


2  D  +  2M' 

(8)  Let  a  body  of  mass  m  on  the  horizontal  arm  AB  be  free  to  rotate  about  the  vertical  axis  ED.  Let 
the  body  be  acted  upon  by  a  horizontal  force  F  of  constant  magnitude  always  at  right  angles  to  AB  at  the  distance 
AB  —  r.  Let  the  distance  AO  of  the  centre  of  mass  O  from  the  axis  be  d.  Find  the  number  of  turns  which  the 
body  will  make  about  the  axis  in  the  time  t,  neglecting  the  mass  of  the  arm. 

ANS.    Let  K  be  the  radius  of  gyration  of  the  body  for  an  axis  through  O 
parallel  to  ED.     Then  the  moment  of  inertia  of  the  body  (or  axis  ED  is 


and  we  have  (page  317) 

Fr 


If  6  is  the  angular  distance,  we  have     6=  !«/»    or 

Frt* 


The  number  of  complete  revolutions  will  then  be 

0  Fr? 


If  the  body  is  a  sphere  *  feet  in  diameter,  "weighing  100  Ibs.,  the  centre  5  feet  from  the  axis,  and  F  is  a 
force  of  25  pounds  at  the  end  of  a  lever  8  feet  long,  find  the  number  of  turns  in  5  minutes,      (g  =  32.) 

AHS.  .  =  -?Sr->«J  '  *  1°°'     .22=?  =  ,845jSV.urn, 

4*  x  ico 


(-  +  25 


CHAP.  III.] 


EXAMPLES.— RO  TA  TION. 


327 


The  time  necessary  to  make  one  turn  is 


x  ioo  I-  +  25 

2CP-    X    8 


=  2.23  sec. 


(9)  A  sphere  whose  mass  is  m  rests  upon  the  rim  of  a  horizontal  disc  of  mass  D.  A  perfectly  flexible 
string  passes  round  the  disc  and  over  a  pulley  and  has  a  mass  P  at  its  lower  end.  Disregarding  friction  and 
mass  of  pulley  and  string,  find  the  distance  described  by  P  in  the  time  t. 

ANS.   Let  R  be  the  radius  of  the  disc  and  r  the  radius  of  the  sphere.    If  the  sphere  moves  with  the  disc  as 
if  it  were  part  of  it,  i.e.,  rotates  about  the  axis  ab  in  the  same  time  that 
it  rotates  about  the  parallel  axis  AB,  we  have  the  moment  of  inertia 
of  the  sphere  relative  to  the  axis  AB  ,a 

2  

?mJ 

In  this  case  we  have  for  the  angular  acceleration  about  AB 


2  + 


+  £j 


D 


Hence  the  acceleration /of  P  is 


The  distance  described  by  P  is  then 


If  the  sphere  does  not  rotate  about  the  axis  ab,  as  when,  for  instance,  it  is  hung  from  the  rim,  we  may 
consider  it  as  a  particle,  and  its  moment  of  inertia  is  then  m/v*.     We  have  then 


PgP 


(2) 


Hence  we  have 


2/>  +  2m  +  D' 


If  the  sphere  has  an  angular  acceleration  at  ,  not  equal  to  a,  about  ad  in  same  direction  as  the  disc,  we 
have  for  the  moment  of  the  force  causing  this  acceleration  -mr"a,  .     Hence  by  D'Alembert's  principle 


P  +  m  +  -- 


(  10)  A  hollow  circular  disc  whose  outer  radius  is  a\  and  inner  radius  b\  and  thickness  t\  revolves  about  an 
axis  perpendicular  to  its  plane.  Find  the  thickness  (*  of  an  equivalent  disc  whose  outer  radius  is  at  and  inner 
radius  b^. 

ANS.  If  the  discs  are  equivalent,  the  same  force  moment  should  give  the  same  angular  acceleration  a  to 
each,  or 

I\a  =  fr,a,         or         I\  =  It  . 


But  7i  =  mi(ai2  +  of) 

In  the  same  way  7»  =  8itt?(ay*  —  £,4).     Hence 

'»  =  ?V^JK-'>- 


3*8 


KINETICS  OF  A  MATERIAL  SYSTEM— ROTATION-FIXED  AXIS. 


[CHAP.  III. 


(i  i)  A  sphtre  of  radius  r\  rotates  about  the  axis  YY  at  a  distance  a.     Find  the  height  h  of  an  equivalent 
cylinder  of  base  radius  r\ ,  whose  axis  is  parallel  to  YY  at  a  distance  b. 
v  ANS.  The  moment  of  inertia  of  the  sphere  of  mass  OTI  relative  to  YY  is 


The  moment  of  inertia  for  a  cylinder  of  mass  m«  is 


We   have  for  equivalence  //  =  /»'.     Hence  if  the  cylinder  and  sphere 
'jfi     rotate  about  their  own  axes  parallel   to  YY  in  the  same  time    they  rotate 
about  YY, 


But  if  Si  is  the  density  of  the  sphere  and  St  of  the  cylinder, 


Hence 


n  — 


If  the  cylinder  and  sphere  rotate  about  the  axis  YY  without  turning  on  their  own  axes,  they  can  be 
treated  as  particles  and  we  have 


If  the  cylinder  and  sphere  have  the  angular  velocity  or  acceleration  oa,  or  a,  about  their  own  axes  and 
o>  or  a  about  YY  in  the  same  or  opposite  directions,  we  have 

— fiir»*»i  ±  m,aa<»  =  m2  —  <»,  ±  m,6*ta, 


Hence 


•2  _ 

5 


j,  ± 


or     h 


(12)  Upon  a  vertical  hollow  axle  whose  outer  radius  isr\  and  inner  radius  rt ,  and  whose  length  is  I,  there  is 
fixed  a  circular  disc  of  radius  a  at  right  angles  to  the  axle.  Under  the  action  of  a  force  F  of  constant  magnitude 
acting  with  the  lever-arm  a  the  angular  velocity  oo,  is  attained.  Find  the  time  t\  of  attaining  this  velocity. 
If  now  the  force  F  ceases  to  act,  find  the  time  t  of  coming  to  rest  and' the  number  of  revolutions  in  that  time. 

ANS.  Let  the  mass  of  the  axle  be  A,  and  of  the  disc  D.  Then  the  moment  of  inertia  of  the  axle  is 
(page  46) 


and  the  moment  of  inertia  of  the  disc  is 


The  total  moment  of  inertia  for  the  centre  line  of  the  axle  is  then 


and  this  axis  is  a  principal  axis. 

The  pressure  on  the  foot  of  the  axle  is  (I)  +  A)g  poundals.     The  moment  of  the  friction  for  hollow  flat 
pivot  (page  225)  is 


where  n  is  the  coefficient  o(  kinetic  friction. 


CHAP.  III.] 


EXAMPLES.— RO  TA  TION. 


329 


If  ai  is  the  angular  acceleration,  we  have  the  moment  of  the  effective  forces  Ia\  (page  317).     Hence,  by 
UAlembert's  principle, 

Fa  —  M 
Fa  —  M —  7«i  =  o,     or     ai  = . 

Hence  the  angular  velocity  at  the  end  of  the  time  A  is 


a,       Fa  —  M     '  _ 

If  the  force  F  now  ceases,  we  have  the  angular  retardation 

M 


and  the  angular  velocity  at  the  end  of  any  time  /  from  the  instant  the  force  F  ceases  is 

CD  =  oo,  +  at  =  oo,  -  —. 

When  the  apparatus  comes  to  rest  oo  =  o,  and  the  time  t  of  coming  to  rest  is 
t  =  ~Jf 


¥ 

The  number  of  radians  described  in  this  time  /  is 


The  number  of  revolutions  is  then 


D  -] 

,-)  +  -<«-  +  „•>} 


Again,  the  kinetic  energy  at  the  instant  the  force  .F  ceases  is  (page  321)  -<»,*.      The  work  of  the  friction 
for  one  revolution  is  2itM.     In  n  revolutions  the  work  is  2itn.\f.     We  have  then,  in  coming  to  rest, 


as  before. 

(13)  Suppose  a  wheel  and  axle  composed  of  hollow  discs  for  the  wheel  and  axle  and  a  solid  cylinder  for  the 
journal.      The  radius  of  the  wheel  is  a  =  3  ft.,  of  the  axle  b  =  2  ft.,  of 
the  journal  r  =  i  inch.     Let  the  mass  of  the  wheel  be  W  —  3  Ibs.,  of 
the  axle  A  =  3  Ibs.,  of  the  journal  J  =  2  Ibs.     Let  the  moving  mass  be 
P  =  10  Ibs.  and  the  mass  lifted  Q=  5  Ibs.    Let  the  string  be  perfectly 
flexible,  and  disregard  its  mass.     Let  P  start  from  rest  and  fall  for 
t  =  5  sec.     Discuss  the  motion  of  the  apparatus,  taking  into  account  the 
mass  of  the  wheel,  axle  and  journal,  and  the  friction  ;  the  coefficient  of 
friction  being  u  =  0.07.      Take  g  =  32\ft.-per-sec.  per  sec.     (Compare  _p 
example  (6),  page  307.) 

ANS.  We  have  for  principal  axis  through  the  centre  of  mass  O 

W 
Moment  of  inertia  of  wheel     =  —  (a*  +  P)  =  32.5  lb.-ft.2 


"   axle        =  -(P  +  r*)  = 
"  journal  =  — r2 


i 

144 


33°  KINETICS  OF  A  MATERIAL   SYSTEM—  ROTATION—  FIXED  AXIS.  [CHAP.  III. 

Hence  moment  of  inertia  of  wheel,  axle  and  journal  is 


The  impressed  forces  are  the  upward  reaction  R  at  the  centre  O,  the  downward  weights  Pg  ,  Qg,  Wg, 
Agt  Jg,  the  friction  4-  F  and  the  equal  and  opposite  reaction  —Foi  the  bearing.  The  moment  of  the  fric- 
tion is  then  the  moment  of  a  couple,  and  is  Fr  at  any  point  in  its  plane  (page  185). 

The  effective  forces  are  Pf  down  and  g—  /up,  and  the  effective  forces  of  the  particles  of  the  wheel, 

journal  and  axle.     The  sum  of  these  in  any  direction  is  zero  (page  316),  and  their  moment  about  O  is  la 
(page  317). 

If  we  reverse  the  effective  forces  and  then  apply  D'Alembert's  principle,  we  have 


(i) 


Pfa  +Q-f=o  ...........     (3) 


From  (2)  we  have  the  pressure  on  the  bearing, 


R  =  (P  +  Q+  W  +  A  +  f)g—  \P  -  Q~jf  •  •  •  poundals. 


We  see  that  the  pressure  on  the  bearing  is  less  than  the  weight  of  the  apparatus. 
The  friction  for  new  bearing  is  then  (page  229) 


=  - 

sin  ft  sin  ft 


+J)g  -(P-  Q-}/~]  poundals, 
\  a/ 


where  u  is  the  coefficient  of  kinetic  friction  and  ft  is  the  bearing  angle.     We  have  also   aa  =/,  or  a  =  /. 

a 

Inserting  this  value  for  a  and  Fin  (3),  we  have  for  the  acceleration  /at  the  circumference  of  the  wheel 


...     (4) 


If  we  disregard  mass  of  wheel,  axle  and  journal,  we  have  the  same  value  for  /as  already  found    in 
example  (6),  page  307. 

If  ft  is  small,  sin  ft  =  ft,  and  we  have  for  the  given  numerical  values 

/  =  0.4013^  =  12.912  ft.-per-sec.  per  sec.,  and  a  =  /  —  4t,      radians 


4-  3°4 
a      *****     Sec. 


The  acceleration  of  Q  is  then 

=  8.608  ft.-per-sec.  per  sec. 


a 


CHAP.  III.]  EXAMPLES.— ROTATION.  331 

The  velocity  of  P  at  the  end  of  the  time  t  =  5  sec.  is 

v  =  ft  =  64.56  ft.  per  sec., 
and  the  angular  velocity  of  the  wheel  is 

oo  =  —  =  21.52  radians  per  sec. 
a, 

The  velocity  of  Q  at  the  end  of  the  time  /  =  5  sec.  is 

— v  —  43.04  ft.  per  sec. 
a 

The  pressure  R  on  the  bearing  is 

R  =  22.32426^-  poundals  =  22. 32426  pounds, 

whereas  the  weight  of  the  apparatus  is  25  pounds. 
The  friction  is 

F  =  nR  =  1.5627^-  poundals  =  1.5627  pounds. 

The  moment  of  the  friction  is 

Fr  =  0.13022^-  poundal-ft.  =  0.13022  pound-ft. 
The  moment  of  the  effective  forces  of  the  particles  of  the  wheel,  axle  and  journal  is 

fa  =  5.1531^- poundal-ft.  =  5.1531  pound-ft. 
The  distance  s  described  by  P  is 

s  =  -//"  =  161.4  ft. 
The  distance  described  by  Q  is 

— s  =  107.6  ft. 
a 

The  tension  Tr  on  right-hand  string  is 

Tr  —  P(g  —  J)  —  5.9864^  poundals  =  5.9864  pounds. 
The  tension  TI  on  left-hand  string  is 

TI  =  Q(g  +  -/)  =  6.337%-  poundals  =  6.3378  pounds. 

Moment  of  tension  on  right     =  17.9592  pound-ft., 

"    left       =  12.6756 
Difference     =     5.2836 

and  this  difference  we  see  is  equal  to  Fr  +  fa,  the  moment  of  friction  and  effective  forces  of  wheel,  axle  and 
journal,  as  should  be. 

Work  of  P  =  5.9864  x   161.4  =  966.205  ft.-lbs.  =  Ps ^~, 


"     on  Q  =  6.3378  x  107.6  =  681.947       -      =  o-s  + 


Difference     =  284.258 


332  KINETICS  OF  A  MATERIAL  SYSTEM— ROTATION-FIXED  AXIS.  [CHAP.  III. 

This  difference  must  be  work  on  wheel  and  work  of  friction. 
Now  work  of  friction  is  F^—s  =  7.005  ft.-lbs. 

Hence  work  on  wheel  =  277.253  ft.-lbs..  and  is  equal  to  — of  (page  321),  as  should  be. 
The  power  of  P  (page  262)  =  9^'2°5  =  i93-24i  ft.-lbs.  per  sec.,  or 

—  =0.35  horse-power. 

The  efficiency  of  the  apparatus  (page  267)  is 

681.947 

e  =  —,,          =  0.70. 
966. 205 

Note.—  The  total  mass  reduced  to  the  circumference  (page  322)  is 
Ma  = 


The  weight   Qg  reduced  to  the  circumference  is  Q-  g,  and  the   force  of   friction   F  reduced   to  the 

circumference  is  F-. 
a 

The  moving  force  at  the  circumference  is  then 

moving  force  =  I P  —  Q—\g F* 

We  have  then  directly,  reduced  mass  x  acceleration  =  moving  force,'or 


-^k- 

If  we  substitute  the  values  of  Ma  and  F,  we  obtain  (4). 

(14)  Suppose  a  wheel  and  axle  composed  of  hollow  discs  for  the  rim  or  outer  circumference  C,  the  hub  H, 
and  the  axle  A  ;  of  a  solid  cylinder  for  the  journal  f,  and  of  four  spokes,  each  spoke  S  being  a  bar  of  uniform 
cross-section.  Let  the  outer  radius  of  C  be  a  =  20  inches,  and  the  inner  radius  r\  =  19  inches  ;  the  outer 
radius  of  H  be  r\  =  8  inches  and  the  inner  radius  b  =  b  inches  ;  the  radius  of  the*  journal  J  is  r  =  /  inch. 
Let  the  mass  of  the  rim  or  outer  circumference  be  C  =  40  Ids. ,  of  the  hub  H  =  12  Ibs. ,  of  the  axle  A  =  10  Ibs. , 
of  the  journal  J  =  2  Ibs.,  and  of  each  spoke  S  =  jf  Ibs.  Let  the  moving  mass  P  =  bo  Ibs.  and  the  mass  lifted 
Q  =  1 60  Ibs.  Let  the  string  be  perfectly  flexible,  and  disregard  its  mass.  Let  P  start  from  rest  and  full  for  a 
time  t  =  3  sec.  Discuss  the  motion  of  the  apparatus,  taking  into  account  the  mass  of  the  wheel,  axle,  journal 
and  spokes,  and  the  friction  ;  the  coefficient  of  kinetic  friction  being  M  =  0.07.  Take  g  =  j2^ft.-per-sec.  per 
sec. 

ANS.  The  moment  of  inertia  of  a  spoke  for  axis  through  its  centre  of  mass  at  right  angles  to  plane  of 
wheel  is  (page  38) 


For  parallel  axis  through  centre  of  wheel  it  is  then  (page  33) 


CHAP.  III.]  EXAMPLES.— ROTATION. 

For  four  spokes  we  have  then 


333 


The  moment  of  inertia  of  the  rim  is  (page  46) 


of  the  hub, 


of  the  axle, 


of  the  journal, 


144 


The  total  moment  of  inertia  for  the  axis  through  the  centre  of  the  wheel  at  right  angles  to  its  plane  is 
then 

7181 

/  =    — ; Ib.-ft.*, 

64 

and  this  axis  is  a  principal  axis. 

The  impressed  forces  are  the  upward  reaction  R,  the  weights  Pg,  Qg,  Cg,  Hg,  Ag,Jg,  ^Sg,  the  friction 
+  F  and  the  equal  and  opposite  reaction  —  F  of  the  bearing. 

The  effective  forces  are  Pf  down  and  Q  —f  up  and  the  effective  forces  of  the  particles  of  the  wheel.     If 
we  reverse  these  forces  and  apply  D'Alembert's  principle,  we  have 


.       +  R  -  Pg  -  Qg  -  Cg  -  Hg  -  Ag  -  Jg  -  *Sg  - 
-  Pga  +  Qgb  +  Fr  -  la  +  Pfa  +  Q  — f  =  o. 
From  (i)  we  have  the  pressure  on  the  bearing, 


Pf  -Q-f=°, 


(i) 
(2) 


We  see  that  the  pressure  on  the  bearing  is  less  than  the  weight  of  the  apparatus.     Let  the  mass  of  the 
apparatus  be  M,  so  that 

M  =  C  +  H+  A+J  +  4S+Q  +  P. 
Then  we  can  write 


The  friction  for  new  bearing  is  then  (page  229) 


where  J*  is  the  coefficient  of  kinetic  friction  and  ft  is  the  bearing  angle.     We  have  also  aa.  =f,     or     a  =  +-. 
Inserting  these  values  for  a  and  F  \n  (2),  we  have  for  the  acceleration  /at  the  circumference 


/  = 


P 
P 


n*\ 
~Qa) 


334  KINETICS  OF  A  MATERIAL  SYSTEM  -ROTATION-FIXED  AXIS.  [CHAP.  III. 

If  ft  is  small,  sin  ft  =  ft,  and  we  have  for  the  given  numerical  values 


/  =  o.og£-  =  2.895  ft.-per-sec.  per  sec.,    and    a  =  *-  =  1.737 
The  acceleration  of  O  is  then 

—  f  =  0.8685  ft.-per-sec.  per  sec. 
The  velocity  of  P  at  the  end  of  the  time  /  =  3  sec.  is 

v  =//  =  8.685  ft-  Per  sec'» 
and  the  angular  velocity  of  the  wheel  is 

o>=  —  =  5.211  radians  per  sec, 
a 

The  velocity  of  Q  at  the  end  of  /  =  3  sec.  is 

—  v  =  2.6055  ft.  per  sec. 
a 

The  pressure  R  on  the  bearing  is 

R  —  297.92^  poundals  =  297.92  pounds, 

Whereas  the  weight  of  the  apparatus  is  299  pounds. 
The  friction  is 

F  =  fiR  =  20.8544^-  poundals  =  20.8544  pounds. 
The  moment  of  the  friction  is 


Fr  =  i.73787ir  poundal-ft.  =  1.737875  pound-ft. 
The  moment  of  the  effective  forces  for  all  rotating  particles  is 

la  =  227.8666  poundal-ft.  =  227.866  pound-ft. 
The  distance  /  described  by  P  is 

*  =  -ff  =  13-0275  ft. 
The  distance  described  by  Q  is 

-t  =  3-90825  ft. 

The  tension  Tr  on  the  right-hand  string  is 

Tr  =  P(g  —  /)  =  54-6^  poundals  =  54.6  pounds. 
The  tension  Tt  on  the  left-hand  string  is 

Tl  =  Q(#  +  ~^f)=  l64-33f  poundals  =  164.32  pounds. 

Moment  of  tension  on  right  =  91.00  pound-ft.. 

"     left     =  82.16 
Difference  =    8.84 

and  this   difference   we   see  is  equal  to  Fr  +  la,  the  moment  of  friction  and  effective  forces  of  rotating 
particles,  as  should  be. 

Work  of  P  =    54.6    x  13.0275    =  711.3015  ft.-lbs.  =  Ps  -  —  , 

3f 
•     on  Q  =  164.32  x     3.90825  =  642.2036      "  b_          QPv* 

Difference  =    69.0979       •«       ~  *  Jf  "     2^r- 
This  difference  must  be  work  on  wheel  and  work  of  friction.     Now,  work  of  friction  is 

F=  '3-5841  ft.-lbs. 


CHAP.  III.]  EXAMPLES.— ROTATION.  335 

Hence  work  on  wheel  =  55.5138  fi.-lbs.,  and  this  is  equal  to  —GO*  (page  321),  as  should  be. 

,  „       711.3015 
Ihe  power  of  P  = — -  =  237.1005  ft.-4bs.  per  sec.,  or 


The  efficiency  of  the  apparatus  is 


237.1005 

— =  0.43  horse-power. 

550 


642.2036 

— —  =  0.93. 

711.3015 


Note. — Just  as  in  example  (13),  the  moving  force  reduced  to  the  circumference  is  \P  —  Q  —\g F. 

If  then  the  reduced  mass  (page  322)  is  Ma,  we  have,  just  as  before, 


Mt 

In  the  present  case  the  reduced  mass  is 


~   +  ~i 

a        a 


If  we  substitute  this  and  the  value  for  F,  we  have/  directly. 


CHAPTER   IV. 

COMPOUND  PENDULUM.  CENTRE  OF  OSCILLATION.  CENTRE  OF  PERCUSSION. 
EXPERIMENTAL  DETERMINATION  OF  MOMENT  OF  INERTIA.  EXPERIMENTAL 
DETERMINATION  OF  g. 

Simple  Pendulum. — A  particle  suspended  from  a  fixed  point  O'  by  a  perfectly  flexible 
inextensible  string  without  mass,  and  swinging  under  the  action  of  gravity,  is  called  a  simple 
pendulum.  It  is  then  a  purely  ideal  conception. 

Let  the  mass  of  the  particle  be  m,  the  length  of  string  /,  the  angle  with  the  vertical  at 
,any  instant  6,  the  angular  acceleration  a,  and  the  angular  velocity  co. 

The  impressed  forces  are  the  tension  T  in    the  string  and  the  weight .  mg. 
The  effective  forces  are  mla  at  right  angles  to  /,  and  mla?  along  /towards  O'. 

By  D'Alembert's  principle,   reversing  the  effective  forces  and    taking    mo- 
ments about  O', 

•  £\ 

™g  X  /  sin  6  —  ml*a  =  o,      or     a  =  ^  Sin — (i) 

-mla. 

Instead  of  a  particle  of  mass  m  suppose  we  have  a  body  of  mass  in.  Let 
this  body  have  the  angular  acceleration  u'  about  a  principal  axis  through  its  centre  of  mass 
O,  and  the  centre  of  mass  the  acceleration  a  about  a  parallel  axis  at  O'.  Let 
the  distance  OO'  be  d.  Then,  by  the  principle  of  page  317,  the  moment  of 
the  effective  forces  of  rotation  about  axis  at  O'  is 

md*a  -f  la', 

where  /is  the  moment  of  inertia  of  the  body  for  axis  at  O,  and,  by  D'Alem- 
bert's principle, 

ni£-  X  d  sin  0  —  mW2«  —  la'  =  O (2) 

Now  if  a'  =  o,  that  is,  if  the  body  has  no  angular  acceleration  about  the  axis  through 
the  centre  of  mass  O,  we  have,  as  before, 


and  we  still  have  a  simple  pendulum  of  length  O'O  =  d,  and  can  treat  the  body  as  if  it  were 
a  particle  of  equal  mass  at  O. 

But  if  the  body  has  an  angular  acceleration  about  the  axis  through  O,  we  can  no  longer 
treat  the  body  as  a  particle  and  we  have  no  longer  a  simple  pendulum. 

336 


CHAP.  IV.] 


COMPOUND  PENDULUM. 


337 


Compound  Pendulum.—  If   a'  =  a,  then  the  body  has  the  same  angular  acceleration 
about  the  axis  through  O  that  it  has  about  the  parallel  axis  through  O'  '.     This 
is  the  case  of  a  rigid  body  swinging  under  the  action  of  gravity  about  a  fixed 
axis  at  O'  .     Such  a  body  is  called  a  compound  or  physical  pendulum.      It  is  an 
actual  pendulum  and  not  a  purely  ideal  conception. 

We  have  in  such  case,  from  (2), 

_  mgd  sin  0  /0j 


or,  since  /  =  m/c2,  where  K  is  the  radius  of  gyration  for  the  ax  s  through  the 

centre  of  mass  O, 


a  = 


gd  sin  0 


(3) 


EQUIVALENT  SIMPLE  PENDULUM. — If  now  we  equate  (i)  and  (3),  we  shall  obtain  the 
length  /  of  the  equivalent  simple  pendulum,  that  is,  the  length  of  a  simple  pendulum  which 
has  at  any  instant  the  same  angular  acceleration  as  the  actual  pendulum,  and  which  there- 
fore swings  in  the  same  time.  We  have  then,  since  m(/c2-j-  d*)  = /' ==  the  moment  of 
inertia  relative  to  the  axis  at  Ort 


**  +  <**_  I 


(4) 


That  is,  the  length  of  the  equivalent  simple  pendulum  is  equal  to  the  -moment  of  inertia 
relative  to  the  axis  at  O'  divided  by  the  moment  of  the  mass  relative  to  this  axis. 

CENTRE  OF  OSCILLATION. — If  we  lay  off  this  distance  /  given  by  (4)  along  O'O,  we 
obtain  a  point  G.     This  point  is  called  the  centre  of  oscillation  or  centre  of  gyration,  because 
it  is  the  point  at  which,  if  the  whole  mass  were  concentrated,  we  should  have  ^r 
equivalent  simple  pendulum,  which  would  vibrate  in  the  same  time.  .     (5) 

Let  OG  =  p  be  the  distance  of  this  point  from  the  centre  of  mass  O. 
we  have  /  —  /  —  d,  or,  from  (4), 


=-r,     or 


or 


__ 

*  ' 


(5) 


That  is,  the  radius  of  gyration  K  is  a  mean  proportional  between  the  distances  p  and  d  ojr 
the  centres  of  oscillation  and  suspension  from  the  centre  of  mass.  A /so,  the  distance  p  is  equal 
to  the  moment  of  inertia  I  relative  to  a  parallel  axis  at  the  centre  of  mass  O  divided  by  the 
moment  of  the  mass  relative  to  the  axis  of  suspension  at  O' . 

Suppose  now  the  body  turned  end  for  end  and  suspended  from  the  point  G  instead  of  dX. 
Then,  from  (4),  the  length  of  the  equivalent  simple  pendulum  will  be 


V;X  KINETICS  OF  A  MATERIAL  SYSTEM— ROTATION— FIXED  AXIS.  [CHAP.  IV. 

or,  inserting  the  value  of/  from  (5), 

/  == 

or  just  the  same  as  before. 

Hence  the  centre  of  suspension  and  oscillation  can  be  interchanged  without  changing  the 
time  of  vibration. 

TIME  OF  VIBRATION. — The  time  of  vibration  of  the  simple  pendulum  (page  138)  is 


If  we  put  for  /  its  value  from  (4),  we  have  for  the  time  of  vibration  of  the  compound 
pendulum 


I  tf  +  d*  I~T 

=  7t  \  I  -         —  =7rA/=--/ 

V      gd  V  m^ 


(7) 


Example.—  The  bob  of  a  heavy  pendulum  contains  a  spherical  cavity  filled  with  water.   Determine  the  motion. 

ANS.  Let  m  be  the  mass  of  the  pendulum,  ~i<  its  radius  of  gyration  for  axis  through  centre  of  mass  O  of 
the  pendulum,  </the  distance  &0.  Let  m  be  the  mass  of  the  water,  and  s  the  distance  of  its  centre  of  mass 
from  O1. 

The  force  between  the  water  and  the  cavity  is  always  normal  to  the  spherical  boundary.  There  is  then 
no  force  tending  to  cause  angular  acceleration  of  the  water  about  its  centre  of  mass.  We  can  therefore  treat 
the  water  mass  as  a  particle. 

The  moment  of  inertia  of  the  pendulum  and  water  relative  to  the  axis  at  (7  is  then 


The  mass  moment  is  md  +  ms.     We  have  then  for  the  length  of  equivalent  simple  pendulum,  from  (4), 


md  +  ms 


*'ie  time  of  vibration  is  then,  from  (6), 

th-  cfl 


(md 


If  the  water  were  to  become  solid  and  rigidly  connected  with  the  cavity,  let  tc  be  its  radius  of  gyration 
for  a  diameter.     Then 

/'  =  mOc"  +  </")  +  m(K*  +  s'), 

-a 

_  ,f|/"(**   +    **') 

(ntf  +  ms)g 
This  latter  time  of  vibration  is  evidently  greater  than  the  first. 

Significance  of  the  Term  <  Radius  of  Gyration.'—  We  see  from  (5)  that  if  we  make  the 
distance  O'  O  =  d  equal  to  *,  we  have  the  distance  OG  =p  also  equal  to  K. 

That  is,  if  the  distance  from  the  centre  of  mass  O  to  the  axis  of  suspension  O'  is  made 
equal  to  K,  the  distance  from  the  centre  of  mass  O  to  the  centre  of  oscillation  or  gyration  G 
is  also  K. 

If  then  we  describe  a  sphere  about  O  with  the  radius  AT,  the  body,  if  suspended  from 
any  point  on  the  surface  of  this  sphere,  will  vibrate  in  the  same  time. 

Hence  the  distance  K  is  called  the  "  the  radius  of  gyration"  (page  32). 


CHAP.  IV.] 


CENTRE  OF  PERCUSSION. 


339 


Centre  of  Percussion. — Suppose  a  forced  to  act  at  some  point  G  in  the  line  O'O  at 
right  angles  to  the  plane  of  O'O  and  the  axis  of  suspension  at  Or.     Then, 
by  D'Alembert's  principle,  we  have 


Hence 


a- 
~ 


where  «  is  the  angular  acceleration  about  the  axis  at  O'. 

Now  the  pressure  on  the  fixed  axis  at  O'  is  R  parallel  to  F.  Since  the  centre  of  mass 
moves  as  if  all  the  mass  and  all  the  impressed  forces  were  collected  at  the  centre  of  mass, 
if  f  is  the  acceleration  of  the  centre  of  mass  we  have 

m/~=  F+  R. 
But  /  =  da.     Hence 

mda=F+R,      or     a  =  F 


(2) 


Equating  (i)  and  (2),  we  have  for  the  reaction  R  of  the  axis  at  O' 


If  pd  >  A:2,  we  see,  from  (3),  that  R  is  positive,  or  in  the  same  direction  as  F. 
\{pd  <  KZ,  we  have  R  negative,  or  opposite  in  direction  to  F. 
\ipd=«*)  or 


the  reaction  R  of  the  axis  at  O  is  zero.     In  this  latter  case  we  have  then 


The  point  G  given  by  (5)  is  called  the  CENTRE  OF  PERCUSSION,  because  if  a  body  is 
struck  at  this  point  so  that  the  force  F  is  at  right  angles  to  the  plane  of  O'O  and  the  axis  of 
suspension,  there  will  be  no  reaction  of  the  axis. 

We  see,  from  page  337,  that  the  centre  of  percussion  is  the  same  as  the  centre  of 
oscillation. 

We  can  find  equation  (4)  directly  from  our  equations  (15),  page  161,  for  the  position  of 
the  instantaneous  axis  of  acceleration. 

Thus  taking  origin  at  O,  and  the  plane  of  rotation  as  the  plane  of  XY,    we  have 

/=/«»  7ny    =    O,  /~    =    O, 

ax  =  o,  ay  =  o,  ag  =  a. 

Therefore,  from  equations  (15),  page  161,  we  have  for  the  position  of  the  instantaneous 
axis  of  acceleration 


P.  =  o.        A  =  ~'       A  =  o. 


340  KINETICS  OF  A  MATERIAL  SYSTEM-ROTATION-FIXED  AXIS.  [(HAP.  IV. 

But  In/  =  F,  if  there  is  no  reaction  R.     Hence /=  —=.     Also,  by  D'Alembert's  prin- 

Fp 
ciple,  Fp  =  /or,  or  a  =  -j~.     Substituting,  we  have 


which  is  the  same  as  equation  (4). 

To  prevent  a  hammer  from  "jarring  "  the  hand,  or  reacting  upon  a  fixed  axis  about  which  it  turns,  the 
direction  of  the  force  at  the  moment  of  striking  should  pass  through  the  centre  of  percussion  G  at  right 
angles  to  the  line  O"O. 

Suppose  we  grasp  a  prismatic  rod  O1  A  of  length  /at  the  end  Cf  and  strike  an  obstacle  with  a  force  F  at 
right  angles  to  the  line  O"O,  where  O  is  the  centre  of  mass. 

,  Then,  from  (3),  the  force  R  at  O  is 

A  G     P     O  d  O 


In  the  case  of  the  rod  /  =  —  (page  38),  and  hence  *•*  =      .  and  d  =  -.     Hence 


If  then  the  obstacle  is  struck  by  O,  p  —  o  and  R  =  —  —,  that  is,  the  force  on  the  hand  is  one  fourth  of 

that  on  the  obstacle  and  opposite  in  direction. 

/  F 

If  the  obstacle  is  struck  by  A,  p  =  —and  R  =  +  —  ,  or  the  force  on  the  hand  is  one  half  of  that  on  the 

obstacle  and  in  the  same  direction. 

If  the  obstacle  is  struck  by  G,  so  that  p  =  -7,  R  —  o,  and  there  will  be  no  force  on  the  hand.     G  is  the 
centre  of  percussion. 

Experimental  Determination  of  Moment  of  Inertia.  —  From  the  principles  of  page  337 
we  can  determine  experimentally  the  moment  of  inertia  of  a  body  relative  to  any  axis. 
ist  METHOD.  —  Thus  we  have,  from  equation  (7),  page  338, 


(i) 


We  must  first,  then,  determine  the  mass  m  of  the  body  and  locate  its  centre  of  mass  O. 
Then  suspend  the  body  from  an  axis  at  O'  and  measure  the  distance  O'  O  =  d.  Then  swing 
the  body  about  this  axis  and  note  the  time  /  of  vibration.  The  moment  of  inertia  /' 
relative  to  the  axis  at  O  is  then  given  by  (i).  The  moment  of  inertia  /  relative  to  a  parallel 
axis  through  the  centre  of  mass  is  then  given  by 

1=1'—  mW2. 

2d  METHOD.  —  From  equation  (4),  page  337,  we  have  for  the  distance  O'G  from  the  axis 
of  suspension  to  the  centre  of  oscillation 


(2) 


We  must  first,  then,  determine  the  mass  m  of  the  body  and  locate  its  centre  of  mass  O 
as  before.     Then  suspend  the  body  from  an  >  axis  at  O'  and  measure  the  distance  O'O  =  d. 


CHAP.  IV.]  EXPERIMENTAL   DETERMINATION  OF  MOMENT  OF  INERTIA.  341 

Then  swing  the  body  about  this  axis  and  note  the  time  of  vibration.  Then  turn  the  body 
over  and  suspend  it  from  another  point  G  in  the  line  O'O  at  such  a  distance  from  O,  found 
by  trial,  that  the  time  of  vibration  about  a  parallel  axis  at  G  is  unchanged,  and  measure  O'G. 
The  point  G  is  the  centre  of  oscillation,  and  the  moment  of  inertia  I'  relative  to  the  axis  at 
O  is  then  given  by  (2).  The  moment  of  inertia  for  a  parallel  axis  through  the  centre  of 
mass  is  then,  as  before, 

1  =  1'  -mdz. 

This  method  does  not  involve  knowing  the  value  of  g. 

Experimental  Determination  of  g, — Having  thus  found  m,  s  and  I'  and  the  time  of 
vibration  t,  we  have,  from  equation  (7),  page  338, 

7T2/' 


We  can  thus  determine  g  by  pendulum  observations. 

The  quantity  -=-j  is  the  pendulum    constant    which 
m^ 

Once  known,  the  observation  of  /  at  any  locality  gives  at  once  the  value  of  g. 


The  quantity  -=-j  is  the  pendulum    constant    which    must   be  accurately  determined. 


CHAPTER  V. 


IMPACT. 

Impact.  —  When  two  moving  bodies  come  into  collision  the  straight  line  normal  to  the 
compressed  surfaces  at  the  point  of  contact  is  the  line  of  impact.  If  the  centre  of  mass  of 
the  two  bodies  is  upon  this  line,  the  impact  is  central  ;  if  not,  we  have  eccentric  impact. 

When  we  consider  the  direction  of  motion,  we  can  distinguish  direct  impact  when  the 
line  of  impact  coincides  with  the  direction  of  motion,  and  oblique  impact  when  the  line  of 

impact  does  not  coincide  with  the  direction  of  motion. 

Thus  in  Fig.  I  if  the  two  bodies  move  in  the  direc- 
tions ul  and  «2,  we  have  oblique  central  impact,  and  in 
Fig.  2  we  have  oblique  eccentric  impact.  If  in  Fig.  I 
the  directions  of  motion  ul  and  ;/2  coincided  with  Ofl^  , 
we  should  have  direct  central  impact.  If  in  Fig.  2  the 
direction  of  motion  ?/,  coincided  with  O^N,  and  #2  were 
parallel,  we  should  have  direct  eccentric  impact. 

Direct  Central  Impact  —  Non-Elastic.  —  We  can  evidently  consider  the  bodies  in  direct 
central  impact  as  particles.  Let  ml  and  u^  be  the  mass  and  initial  velocity  of  one  particle 
before  impact,  and  m2,  u2  of  the  other  before  impact.  Let  ul  be  greater  than  HI  and  in  the 
same  straight  line.  Let  the  direction  of  ul  be  positive,  the  opposite  direction  negative. 

When  the  bodies  meet  there  is  a  short  interval  of  compression,  at  the  end  of  which 
both  masses  have  a  common  velocity  v.  If  the 

bodies  are  non-elastic,  they  remain  in  contact  with  the      / 

_  y  ( 

common  velocity  v. 

By  the  principle  of  conservation  of  the  centre  of 
mass  (page  300),  if  there  are  no  external  forces  the 
motion  of  the  centre  of  mass  is  unaffected  by  the  impact. 

We  have  then 

_  =  mft 


' 


Or,  by  the  principle  of  conservation  of  momentum  (page  300),  we  have 

mlul  -f-  mtuz  =  (ml  +  m^v, 


(i) 


or  the  momentum  before  equals  the  momentum  after  impact. 

In  equation  (i)  a  velocity  opposite  to  wt  is  to  be  taken  as  negative. 
The  energy  before  impact  is 


342 


CHAP.  V.]  DIRECT  CENTRAL  IMPACT—  NON-ELASTIC.  343 

and  the  energy  after  impact  is 

&  =  2  K  +  "*£»*• 
The  loss  of  energy  is  then 


This  work  causes  deformation  and  rise  of  temperature. 

If  we  take  mass  in  Ibs.  and  velocity  in  ft.  per  sec.,  equation  (2)  gives  loss  of  energy  in 
ft.-poundals.      If  we  wish  foot-pounds,  we  must  divide  by^*,  or  in  foot-pounds 


We  call  — ~T^~  ^e  harmon*c  mean  between  m^  and  m2,  and    *-3 —  is  the  height 

due  to  the  difference  of  the  velocities. 

Hence  the  loss  of  energy  in  non-elastic  impact  is  equal  to  the  product  of  the  harmonic 
mean  of  the  masses  and  the  height  due  to  the  difference  of  their  velocities. 

If  the  mass  ;«2  is  at  rest,  we  have  «2  =  o,  and 


_  = 


If  the  bodies  move  toward  each  other,  «2  is  negative,  and 


In  this  case  if  the  momentums  of  the  bodies  are  equal,  or  mji^  =  mjiv  v  =  o,  or  the 
bodies  come  to  rest.      If,  on  the  contrary,  the  masses  are  equal,  we  have 


2  2    '  2g 

If  the  bodies  move  in  the  same  direction  and  the  mass  of  the  one  in  advance,  m2,  is 
infinite,  we  have 


or  the  velocity  of  the  infinite  mass  is  not  changed  by  the  impact.      If  the  infinite  mass  is  at 
rest,  or  #2  =  o,  we  have 


Examples.  —  (i)  A  non-elastic  body  of  mass  ;«i  =  5°  Ibs.  moving  with  a  velocity  u>  =  7  ft.  per  sec. 
impinges  centrally  upon  another  of  mass  m,  —  3°  Ibs.  moving  in  the  same  direction  with  a  velocity  of  u*  =  j 
feet  per  sec.  Find  the  velocity  with  which  the  two  move  on  together  after  the  collision,  and  the  work  expended 
in  deformation  and  heating. 

ANS.  v  =  5£  ft.  per  sec.  in  the  direction  of  u\  \  —  ft.-pounds. 


344 


KINETICS   OF  A  MATERIAL  SYSTEM-IMPACT.  [CHAP.  V. 


(2)  In  order  to  cause  a  non-elastic  body  weighing  no  Ibs.  to  change  its  velocity  from  t,  to  2  feet  per  sec., 
we  let  a  non-elastic  body  weighing  5°  Ibs.  strike  it.     Find  the  velocity  of  the  latter  body,  and  the  work  expended 
in  deformation  and  heating. 

ANS.  3.2  ft.-per-sec. ;  — -  ft.-pounds. 

(3)  Two  non-elastic  masses  of  3  and  5  tons  impinge  centrally  with  velocities  of  4  and  5-5  ft-  per  sec. 
respectively.     Find  their  final  velocity  when  moving  in  the  same  and  opposite  directions. 

ANS.  4lf  ft.  per  sec.;  i^|  ft.  per  sec.  in  the  direction  of  the  larger  velocity. 

(4)  Two  non-elastic  bodies  of  3  Ibs.  and  i  oz.  are  moving  in  opposite  directions  and  impinge  centrally .     The 
first  has  a  velocity  of  3\  and  the  latter  of  9  ft.  per  sec.     In  what  direction  do  they  move  after  impact? 

ANS.   In  the  direction  of  the  first  with  a  velocity  3jj  ft.  per  sec. 

(5)  A  non-elastic  body  whose  mass  is  it>  Ibs.  moving  with  a  velocity  of  23  miles  an  hour  impinges  centrally 
on  another  moving  in  the  opposite  direction.     The  two  come  to  rest.     If  the  mass  of  the  latter  were  28  Ibs. ,  find 
its  velocity.     If  the  velocity  of  the  latter  were  66ft.  per  sec. ,  find  its  mass. 

ANS.  14^  miles  per  hour;  8|  Ibs. 

(6)  A  number  of  non-elastic  balls  of  masses  mi,mt.m mn  lie  on  a  straight  line  at  rest.     If  the  first 

have  a  velocity  of  «i  toward  the  others,  what  will  be  the  ultimate  velocity  of  all? 


ANS.  vn  — 


(7)  If  in  the  preceding  example  the  initial  velocities  are  «,,«,,*•   .  .  .  uH,find  the  ultimate  velocity. 

miUi   +  m*ui  +  .  .  .  mnun 
ANS-  "•  =        mi  +  mt  +  ...  ,„,      • 

(8)  If  in  a  machine   ib   impacts  per  minute  occur  between  two  non-elastic  masses  mi  =  too  Ibs.  and 
my  =  1200  Ibs.  moving  in  the  velocities  u\  =  5  and  ut  =  2  ft.  per  sec.,  find  the  loss  of  energy. 

16     (5  —  2)*     1000  x  1  200 

ANS.  ^-.—       —  .  —  —  =  20.3  ft.-pounds  per  sec. 

60          ig  2200 

(9)  If  two  trains  m\  =  120000  Ibs.  and  m*  =  1  60000  Ibs.  come  into  collision  with  the  opposite  velocities 
u\  =  20  and  ut  =  15  ft.  per  sec.,  find  the  loss  of  energy  which  is  expended  in  the  destruction  of  the  cars, 
considering  them  as  non-elastic. 

(20   +    I?)*     120000   X    160000 

ANS.  ~~~  28oooo—      =  i  302000  ft.-pounds. 

Direct  Central  Impact—  Perfect  Elasticity.—  When  two  bodies  collide,  and  the  impact 
is  direct  central,  there  is  a  short  interval  of  compression  at  the  end  of  which  both  masses 
have  a  common  velocity  v.  If  the  bodies  are  non-elastic,  they  remain  in  contact.  If, 
however,  they  are  elastic,  there  is  another  short  interval  of  expansion  at  the  end  of  which  ml 
has  the  final  velocity  Z'P  and  m^  the  final  velocity  v2.  If  the  bodies  are  perfectly  elastic,  the 
works  of  compression  and  of  expansion  are  equal,  and  hence  no  energy  is  lost. 

'By  the  principle  of  conservation  of  the  centre  of  mass  (page  300)  we  have  the  velocity 
F  of  the  centre  of  mass, 

_  _     ?«,«!  -j- 


or 

MlUl  +  MaUs=tttlVl  +  9Hft  .........        (I) 

That  is,  by  the  principle  of  conservation  of  momentum  the  momentum  before  equals  the 
momentum  after  impact. 

Also,  since  for  perfectly  elastic  impact  no  energy  is  lost,  the  energy  before  must  equal 
the  energy  after  impact,  or 

*l**+1**f*S*>l+m*v»  ........      (2) 


CHAP.  V.]  DIRECT  CENTRAL  IMPACT— PERFECT  ELASTICITY.  345 

From  (i)  and  (2)  we  have  at  once 


=          _ 


,  + 


If  we  subtract  the  first  of  these  from  the  second,  we  have 


That  is,  the  velocity  of  approach  (ul  —  «2)  equals  the  velocity  of  separation  (v2  —  vj. 
The  loss  and  gain  of  velocity  are 


_ 


or  /WHY-  <w  w#c/&  as  for  non-elastic  impact. 

We  take  velocities  in  the  direction  of  ul  positive,  in  the  opposite  direction  negative. 
If  the  mass  ;«2  is  at  rest,  we  have  u2  =  o,  and 


2ml 

; "7^  •  u* 


If  the  bodies  move  toward  each  other,  uz  is  negative,  and 


In  this  case,  if  the  momentums  of  the  bodies  are  equal,  or    mlul  —  mzuz,  we  have 


That  is,  the  bodies  after  impact  move  in  opposite  directions  with  the  same   speeds   they 
originally  had.      If,  on  the  other  hand,  the  masses  are  equal,  we  have 


That  is,  each  body  returns  with  the  velocity  the  other  body  had  before  impact. 

If  the  bodi( 
infinite,  we  have 


If  the  bodies  move  in  the  same  direction  and  the  mass  mz  of  the  one  in  advance  is 


or  the  velocity  of  the  infinite  mass  is  not  changed  by  the  impact.      If  the  infinite  mass  is  at 
rest,  or  u2  =  o,  we  have 

*>!  =    —   «!  ,  ^2  =  0. 

That  is,  the  velocity  of  the  impinging  body  is  reversed. 

Examples. — (i)  Two  perfectly  elastic  balls  weighing  to  Ibs.  and  16  Ibs.  collide  centrally  with  the  •velocities 
12  and  6  feet  per  sec.  Find  the  loss  and  gain  of  velocity  and  the  final  velocities  after  collision,  if  the  initial 
velocities  are  in  the  same  and  in  opposite  direction. 

ANS.  In  the  first  case  the  final  velocities  are  Vi  —  +  4^-  and  v,  =  +  IOT\  ft.  per  sec.  The  first  body 
loses,  then,  7^  and  the  other  gains  4T85  ft.  per  sec. 

In  the  second  case  the  final  velocities  are  Vi  =  —  10^  and  v9  —  +  7H  ft.  per  sec.  Each  body  then 
rebounds  with  these  velocities.  The  first  body  loses  22T23  and  the  other  gains  13/5  ft.  per  sec. 

(2)  A  number  of  perfectly  elastic  balls  of  masses  nti,  m-t ,  ms  .  .  .  mn  lie  in  a  straight  line  at  rest.  If  the 
first  have  a  velocity  of  u\  toward  the  others,  find  their  velocities  after  impact. 


346  KINETICS  OF  A  MATERIAL  SYSTEM— IMPACT.  [CHAP.  V. 

ANS.  The  velocity  of  the  first  is 


(mi  —  mt)ui 

v\  =  . 

Ml    +  Mt 


The  velocity  of  any  intermediate  ball  is 

ti  .  mt  . 


VH  = 


The  velocity  of  the  last  ball  is 


(mi  +  m,)(mt  +  mt)  .  .  .  (mn  +  MH+I) 


(3)  In  the  preceding  example  let  there  be  four  balls,  the  mass  of  the  first  mi,  of  the  second  mt  =  ami ,  of  the 
third  mt  =  amt  =  a*Mi  ,  of  the  fourth  m4  =  am*  =  a"/«i. 


'  I  +a  l  +  a  (i  +  a)"' 

If,  for  example,  the  mass  of  each  ball  is  one  half  that  of  the  preceding, 

i  4  16  64 

Vl=-Ui,  V*  =  ~Ui,  V*  =  —Ui,  V.=    ~Ui. 

Coefficient  of  Elasticity.  —  No  body  is  perfectly  elastic  or  absolutely  non-elastic.     We 
deal  with  bodies  more  or  less  elastic,  that  is,  ^  imperfectly  elastic. 

Let  a  prismatic  body  of  length  /  and  area  of  cross-section  A  be  acted  upon  by  a  stress  5 
in  its  axis,  and  let  the  elongation  or  compression  or,  in  general,  the  strain  be  denoted  by  A. 

We  know  by  experiment  that  within  a  certain  limit, 

3<  [  —  ^]  -  ^5    twice,  three  times,   or  four  times,   etc.,  the  stress  S  will 

cause  a  strain  of  2  A,  3A,  4!,  etc.     The  limit  up  to  which 

this  law  of  proportionality  of  stress  to  strain  holds  true,  for  any  material,  is  called  the  elastic 
limit  for  that  material  for  the  kind  of  stress  under  consideration. 

Thus  if  we  lay  off  the  successive  stresses  to  any  convenient  scale  horizontally,  and  lay 
off  the  corresponding  observed  strains  to  any  convenient  scale  vertically,  we  obtain  within 
the  elastic  limit  a  straight  line  AB.  The  co-ordinates  AD  and  DP  of  any  point  P  of  this 
line  give  the  stress  and  corresponding  strain.  The  point  B  at 
which  the  straight  line  begins  to  curve  gives  the  stress  AL,  and 
this  is  the  elastic  limit  stress. 

We  see,  then,  that  within   the  elastic  limit  the  ratio  ^-  of     A 

A  S        2S       D         L 

stress    to  strain  is  constant.      Now  if  A    is  the  area  of    cross-section    of   the  prism,   then 

-  is  the  unit  stress,  or  stress  per  square  inch.   Also,  if  /  is  the  original  length,  then  y  is  the 
A  * 

unit  strain,  or  strain  per  unit  of  length.      If  then   the  experiment  were  made  upon  a  prism 
of  one  unit  area  and  one  unit   length,  we  should  have   within  the  elastic  limit  the  ratio 

-  +  -t     or     -jj-    constant.      This  constant   for  any  material   is  called     the    coefficient  of 

A  /  An. 

elasticity  for  that  material,  for  the  kind  of  stress  under   consideration.     We  denote  it  by  E. 
We  have  then  within  the  elastic  limit 


and  we  can  define  the   coefficient  of  elasticity  in  any  case  as  the  unit  stress  divided  by  the 
unit  strain. 

Values  of  E  for  various  materials,  as  determined  by  experiment,  are  given  on  page  478. 


CHAP.  V.]  MODULUS   OF  ELASTICITY.  347 

Examples.  —  (i)  A  wrought-iron  rod  30  feet  long  and  4  square  inches  in  cross-section  is  subjected  to  a  tensile 
stress  of  4000  pounds.  The  elongation  is  found  to  be  o.oi  ft.  Find  E. 

ANS.   The  unit   stress  is   —  =  400°    =  1000  pounds  per  square  inch.     The  unit  strain  is  —  =  —  -  —  ft 
A  4  *         3000 

per  foot  of  length.     We  have  then 

E  =  —  ry  =  30  oooooo  pounds  per  square  inch. 
Ah- 

(2)  Taking  E  for  -wrought  iron  as  thus  determined,  find  the  compression  of  a  wrought-iron  prism  2  ft. 
long  and  12  square  inches  cross-section  under  a  stress  of  1  50000  pounds. 

SI  150000  x  24  i     . 

ANS.          A  =  —  —  —^—  =  —  inch. 

AE       12  x  30000000       loo 

Modulus  of  Elasticity.  —  Let  Fc  be  the  force  of  compression  when  two  bodies  collide  at 
the  end  of  the  period  of  compression,  and  Fr  the  force  of  expansion  or  force  of  restitution 
at  the  beginning  of  the  period  of  expansion.  If  a  body  is  non-elastic,  Fr  is  zero.  If  a  body 
is  perfectly  elastic,  Fr  =  Fc.  Strictly  speaking,  no  bodies  are  absolutely  non-elastic  or 
perfectly  elastic.  We  deal  .with  bodies  imperfectly  elastic  where  Fr  is  less  than  Fc.  The 

ratio  ~  of  the  force  of  restitution  to  the  force  of  compression  is  found  by  experiment  to  be  a 

constant  for  any  body  so  long  as  the  limit  of  elasticity  is  not  exceeded.     This  ratio  we 
denote  by  e  and  call  it  the  modulus  of  elasticity.     We  have  then 


If  we  let  a  sphere  of  mass  m  fall  from  a  height  h  upon  a  -relatively  very  large  rigidly 
supported  mass  of  the  same  material,  it  will  rebound  to  a  height  ft  ,  less  than  the  original 
height.  The  velocity  with  which  it  strikes  is  ux  =  M2gh,  and  the  velocity  of  rebound  is 
vl  —  ^2gli'  .  If  the  interval  of  compression  /  is  very  short,  we  have  the  force  of  compression 


and  the  force  of  restitution 

mv,       m 


t 
We  have  then 


We  can  thus  experimentally  determine  the  modulus  of  elasticity  e  for  various  materials. 
For  perfect  elasticity  e—\t  and  for  non-elastic  bodies  e  —  o. 

The  following  average  values  have  been  thus  determined  by  experiment : 

Cast  iron,  e  =  I  nearly.          Glass,  e  =  ^.          Steel,  e  =  --.          Clay,  wood,  e  =  o  nearly. 

Direct  Central  Impact— Imperfect  Elasticity. — Let  Fe  be  the  force  at  the  end  of  the 
period  of  compression  when  both  bodies  are  moving  with  the  common  velocity 


348  KINETICS  OF  A  MATERIAL  SYSTEM— IMPACT.  [CHAP.  V. 

Let  A,  be  the  compression  of  ml ,  and  A3  of    mr     Then  the  work  of  compression  is, 
since  -Ft  is  the  average  force, 


This  work  should  equal  the  energy  lost  during  compression,  or 

-Fe(X,  +  A2)  =  -mp*  +  l-m^  -  l-(m,  +  m2)v*, 

or  putting  for  v  its  value,  we  obtain  for  Fc  in  pounds,  if  we  take  mass  in  Ibs.  and  distance  in 
feet  and  velocity  in  ft.  per  sec., 


Let  Fr'  be  the  force  of  restitution  at  the  beginning  of  the  period  of  expansion  of  ml  , 
and  A,'  its  expansion.  Let  Fr"  be  the  force  of  restitution  at  the  beginning  of  the  period  of 
expansion  of  mz  ,  and  A2'  its  expansion.  Then  the  work  of  expansion  is 


-2r  r. 

The  total  energy  lost  then  is,  in  foot-pounds, 

L/r(A  -f  AJ  -  -Fr'\'  -  -Fr"\'  =  —m,u*  +  —*»,«»  --  -*n.v*  —  —mj>*.          (2) 

2     A    i  T      t)         2     r     i  2     »       2          2g     i    i    -T  2g     2   2          2g     i   i          2g     *  * 

Now,  from  page  346,  we  have 

J  ^'  H       J  ^ 

Al  =  4^    and    A2==^"2' 

where  /j  and  /2  are  the  lengths  of  the  masses  ;«t  and   mz  ,  ^4a  and  ^42  their  areas  of  cross- 
section,  and  EI  and  E2  their  coefficients  of  elasticity. 
For  the  sake  of  simplicity  we  put  the  quantities 

f  =  ^  =  ",     and      %  =  *&=*,  .......     (3) 

Aj  ^  A2  12 

AE 
The  quantity  //  =  —j-  in  general  we  call  the  hardness  of  a  body. 

The  hardness  of  a  body  is  then  the  ratio  of  the  stress  to  strain. 
We  can  write,  then, 

F'  Fe 

Aj  =  —       and     A2  =  -^  ..........     (4) 

where  Hl  and  fft  are  given  by  (3). 

We  also  have  from  page  347,  since  stress  and  strain  are  proportional, 

Fr  V  Fr"          V 

yiaBir    *>   and    :F~:=r  =  '2' 

•V  *|  •*  f  A2 

where  ^  and  ez  are  the  moduli  of  elasticity  for  the  masses  ;«,  and  ;«2.      We  have  then 

Fr'  =  flFet         F,"  =  f2Fe,         V  =  ^.         V  =  ^r       ....     (5) 


CHAP.  V.]  DIRECT  CENTRAL  IMPACT— IMPERFECT  ELASTICITY.  349 

If  we  substitute  the  values  for  \,  A2,  F/,  Fr" ,  A/  and  A2',  given  by  equations  (4)  and 
(5),  in  equations  (i)  and  (2),  we  obtain 


2     e  L     HI  Hz     J         2g  '    2£-  2^ 

We  also  have,  from  the  principle  of  conservation  of  momentum  (page  300), 


From  these  three  equations  we  obtain 


+ 


Equations  (7)  are  general  and  give  the  final  velocities  vv  and  v2  of  wx  and  m^.      If  both 
masses  are  of  the  same  material,  H^  =  Hz  and  el  =  ez  =  e,  and  we  have 


(8) 


If  the  bodies  are  perfectly  elastic,  we  have  el  =  e^  =  i  and  equations  (7)  reduce  to 
equations  (3),  page  345.  If  the  bodies  are  non-elastic,  el  =  o,  <?2  =  o,  vl  =  v2  =  vy  and 
equations  (7)  reduce  to  equation  (i),  page  342. 

Velocities  in  the  direction  of  u^  are  positive,  in  the  opposite  direction  negative. 

Examples. — (i)  If  an  iron  sledge  of  mass  m\  =  30  Ibs.,  h  =  6  inches  long  and  At.  =4  square  inches  area  of 
face,  strikes  an  immovable  lead  plate  li  =  f  inch  thick  and  Ai  =  2  square  inches  area,  with  a  velocity  u\  =  50 
ft.  per  sec.,  find  the  force  of  impact  and  the  compression  of  the  sledge  and  plate,  taking  E\=2Q  ooo  ooo  and 
Ei  =•  700  ooo  pounds  per  square  inch. 

ANS.  We  have  H\  = —       —  =  232  ooo  ooo  pounds  per  ft.  ( 

2 

Hi  =  2X700000  _  xg  goo  ooo  pounds  per  ft.     We  also  have  Ui  =  o  and  m*  =  oo ;  hence,  from  (6),  the 

72 

force  of  impact  is 

232000000x16800000 

.-^      2488ooooo > 

or,  taking^-  =  32  ft.-per-sec.  per  sec., 

Fc  =  247  350  pounds. 
From  (4)  we  have  for  the  compression  of  the  sledge  and  plate 

/r  7^ 

AI=  — -  =0.0016  ft.  =  0.0127  in.,  Aa  =  —£  =0.0147  ft.  =0.177  i°. 

Hi  fit 


350  KINETICS  OF  A  MATERIAL  SYSTEM-  IMPACT.  [CHAP.  V. 

(2)  In  the  preceding  example  consider  the  sledge  as  perfectly  elastic  and  tne  plate  as  inelastic. 
ANS.  We  have  «»  =  o,  m\  =  °o  ,  e\  =  i  and  et  =  o. 

Hence,  from  (7), 

=  s°  - 

That  is,  the  sledge  rebounds  with  this  velocity. 

(3)  Find  the  velocities  after  impact  of  two  steel  masses,  if  the  velocities  before  impact  are  *,  =  10  and 

*t  =  —  6ft.  per  sec.,  and  the  masses  mi  =  jo  Ids.,  m*  =  40  Ibs.,  taking  e  =  -. 


ANS.  w»  =  10  —  16  .  ^f  i  +  -J  =  -  4.22  ft.  per  sec., 
*/,  =  -  6  +  16  ^(i  +  |j  =  +  4.665  ftpersec. 


That  is,  m\  rebounds  and  mt  rebounds. 

Earth  Consolidation.  —  When  a  maul  strikes  a  mass  of  soft  earth  it  compresses  the  earth 
with  a  certain  force  F.  Let  s  be  the  depth  of  penetration,  m  the 
mass  of  the  maul,  and  h  the  height  from  which  it  is  let  fall.  Then 
the  energy  of  the  maul  before  it  is  dropped  is  mh.  Since  this 
energy  is  expended  in  compressing  the  soil,  we  have 


ff  ,  T?  mk 

Fs  =  mn,     or     F  =  — . 
s 

If  we  divide  this  force  F  by  the  cross-section  A  of  the  maul,  we  have  for  the  unit  force 
of  compression 

F  _mk 
~A~~As 

The  resistance  F  of  soils  to  the  penetration  of  the  maul  is  generally  variable  and 
increases  with  the  depth  s  of  penetration.  In  many  cases  we  may  assume  it  to  increase 
directly  with  the  penetration.  In  such  case  we  should  have 

—Fs  =•  mh,     or     F  = ,    and     -r  =  —3 — , 

or  twice  as  much  as  before. 

p 
If  A  is  taken  in  square  inches  and  s  and  h  in  feet  or  inches  and  m  in  Ibs.,—  is  the 

number  of  pounds  per  square  inch  resistance  of  the  soil.     Allowing  a  factor  of  safety  of  IO, 
we  could  then  safely  load  the  compacted  soil  up  to  —  --.-. 

IO  /i 

Example. — A  maul  whose  mass  is  120  Ibs.  falls  from  a  height  of  18  inches,  and  the  earth  is  compressed  one 
eighth  of  an  inch  by  the  last  blow.  The  cross-section  is  16  square  inches.  Taking  a  factor  of  safety  of  io,find 
the  safe  load. 

ANS.   —  =  —^  =  —      —  =  1080  pounds  per  sq.  inch.      Taking  a  factor  of  safety  of  10,  we  have  108 
i6Xg 

pounds  per  sq.  inch  safe  load. 


CHAP.  V.] 


OBLIQUE  CENTRAL  IMPACT. 


353 


making  the  angle  with  OtOt  given  by 


tan  /ffj  =  — 


sin 


-  -  tan 


(8) 


For  non-elastic  bodies  e  =  o,  and  for  perfectly  elastic  bodies  e  =  I.  For  non-elastic 
bodies,  then,  from  (6),  (7),  (8),  we  have  vl  =  o,  >wl  =  MI  sinar  l  , 
tan  Pi  =  oo  or  ft  =  90°.  That  is,  the  velocity  vl  along  the 
line  of  impact  is  annihilated,  that  at  right  angles  is  unchanged, 
and  the  body  moves  after  impact  in  the  direction  O^F  at  right 
angles  to  O^O^  with  the  speed  uv  sin  ar 

For  perfectly  elastic  bodies  vl  =  —  »x  cos  al  ,  w^  =  u^  , 
tan  /?j  =  —  tan  ar  That  is,  the  velocity  along  the  line  of 
impact  is  reversed,  the  angle  of  reflection  NO^B  is  equal  to 

the  angle  of  incidence  NO^A,  and  the  body  moves  after  impact  in  the  direction  O^B  with  the 
original  speed  ur 

For  imperfect  elasticity  we  have,  from  (8), 

tan  « 


tan  /V 

or  the  modulus  of  elasticity  is  equal  to  the  ratio  of  the  tangent  of  the  angle  of  incidence  to 
the  tangent  of  the  angle  of  reflection. 

Example.  —  Two  balls,  m\  =  30  Ibs.,  m*  =  30  Ibs.,  impinge  with  the  velocities  u\  =  20  and  u*  =  25ft.  per 
sec.,  making  the  angles  with  the  line  of  impact  cti  =  21"  sf  and  a,  =  65°  20'.  Find  the  velocities  after 
impact  if  the  bodies  are  non-elastic. 


Hence 


ANS.  ui  sin  a,  =    7.357  ft.  per  sec.,       u,  sin  or»  =  22.719  ft.  per  sec. 
Ui  cos  a,  =  18.598  "      ••    "  ut  cos  0.1  =  10.433  "    "      " 


vi  =  18.598  -  (18.598  -  10.433)  §3  =  1  3-495  ^  Per  sec.. 
vt  =  10.433  +  (18.598  -  10.433)  §3=  J3-495  .....  ' 


The  resulting  velocities  are  then 


=  V 1 3.495"  +    7.357s  =  15.37  ft.  per  sec., 


w*  =  |/I3-4952  +  22.719*  =  26.42 
making  the  angles  ft\  and  ft*  with  the  line  of  impact  given  by 


'   or  *  = 


=  59°  if. 


Friction  of  Oblique  Central  Impact.  —  The  pressure  between  the  colliding  bodies  gives 
rise  to  friction.  If  Pis  the  pressure  due  to  impact,  F  the  friction  and  //  the  coefficient  of 
friction,  we  have 

F  =  JiP. 


354 


KINETICS  OF  A  MATERIAL  SYSTEM-IMPACT. 


[CHAP.  V 


Let  the  mass  of  the  impinging  body  be  ;«1  ,  and  the  initial  and  final  velocity  along  the 
line  of  impact  be  ul  and  i»lt  and  /  be  the  very  short  time  of  impact.  Then  we  have  for  the 
impulse  (page  255) 


=  *,(.,-,,),     or     F  =  "-'<"' 


Hence  the  friction  is 


That  is,  the  impulse  of  the  friction  divided  by  the  mass  is  equal  to  //  times  the  change  of 
velocity  along  the  line  of  impact  ;  or  the  change  of  velocity  due  to  friction,  at  right  angles  to  the 
line  of  impact,  is  equal  to  /*  times  the  change  of  velocity  along  the  line  of  impact. 

This  change  of  velocity  is  always  a  retardation,  since  friction  is  a  retarding  force. 

Thus  if  a  mass  w,  falls  vertically  with  a  velocity  ;/,  upon  a  horizontal  sled  of  mass  mz 
moving  horizontally  with  the  velocity  u2,  and  if  the  velocity  ul  is  entirely  lost  by  the 
collision,  we  have  for  the  friction 


But  the  retarding  force  during  the  time  /  for  both  masses  in  contact,  if  v  is  the  common 
velocity  after  impact,  is  also 


Hence  we  have  for  the  change  of  velocity  of  the  sled 


If  a  body  of  mass  m^  strikes  an  immovable  mass  with  a  velocity  «x  at  an  angle  alt  we 
have  from  equations  (7),  page  349,  for  the  change  of  velocity 
along  the  line  of  impact,  since  u2  =  o  and  m.2  =  00,  /2  =  oo  and 


«,  cos  al  —  i\  =  ;/,  cos  ari       '2 
Hence  the  change  of  velocity  due  to  friction  is 

^ul  cosa,(l  +',), 
and  after  impact  the  component  w,  sin  «,  becomes 

«j  sin  «!  —  /^M,  cosor,(l  +  ^2)  =  [sin  ai  —  fit  cos  ^(i  +  ^2)]wi 
For  perfectly  elastic  bodies  e2  =  I  and  (3)  becomes 

(sin  al  —  2J*  cos  «,)^1, 


CHAP.  V.]  FR.ICT10K   OF  OBLIQUE  CENTRAL  IMPACT.  355 

and  for  non-elastic  bodies  e2  =  o  and  (3)  becomes 

(sin  a1  —  /*  cos  a^)ur 

The  friction  often  causes  bodies  to  turn  about  an  axis  through  the  centre'  of  mass,  or  if 
before  impact  rotation  exists,  that  motion  is  changed. 

Thus  let  r  be  the  radius  of  a  sphere,  O3l  its  initial  and  GO  its  final  angular  velocity 
during  the  time  t  of  impact. 

Then  the  moment  of  the  friction  is  /  .    --  -  (page  322).     But  the  moment  of  the 

.     .  m^u   cos  #,(1  -j-  e2}r 

friction  is  also  —  -  !  --    -=-  .      Hence 


Equation  (4)  gives  the  change  of  angular  velocity. 

Example.  —  A  billiard-ball  strikes  the  cushion  with  a  velocity  u\  =  15  ft.  per  sec.,  the  angle  of  incidence 
being  ai  =  45*  .     //  et  =  0.55  and  the  coefficient  of  friction  is  fj.  =  0.2,  find  the  motion  after  impact. 
ANS.  The  velocity  after  impact  along  the  line  of  impact  is 

v  .  =  —  e-iUi  cos  ori  =  —   0.55  X  15  cos  45°  =  —  5.833  ft.  per  sec. 
The  velocity  parallel  to  the  cushion  is 

«i  sin  ttt  —  jj.ui  cos  a^i  +'<?»)  =  7.319  ft.  per  sec. 
Hence  the  angle  of  reflection  /3t  is  given  by 


tan  /*,  =     |!9,     or    ft,  =  51"  2/, 

5-°33 


and  the  velocity  after  impact  is 


— 
cos  51°  27' 


per  sec> 


The  ball  also  acquires  the  angular  velocity  ooz  about  a  vertical  axis  through  the  centre  of  mass.     Since 
/  =  —  Wira,  where  tnt  is  the  mass  and  r  the  radius  of  the  ball  : 

o.2m\  x   15  cos  4;°  x  i.Z^r      8.22 

ooz  =  --  =  -  ^  --  32  -  22.  =  —  —  radians  per  sec. 
2  '       0  r 

—mir* 

The  ball  also  has  angular  velocity  aox  about  a  horizontal  axis  through  the  centre  of  mass  at  right  angles 
to  iv\  given  by 

wi       9-36 
oox  =   —  =  -  radians  per  sec. 

The  resultant  angular  velocity  is  then 


oo  =  4/Gox*  +  aoS  =  radians  per  sec. 

about  an  axis  through  the  centre  of  mass  in  a  vertical  plane  at  right  angles  to  wJt  making  an  angle  with 
the  vertical  whose  tangent  is  ~?  =  |^—  =  1.138,     or    48°  42'. 


'356 


KINETICS  OF  A  MATERIAL  SYSTEM-IMPACT. 


[CHAP.  V. 


Strength  and  Impact.  —  Let  the  mass  **t  ,  moving  with  the  velocity  «,  ,  impinge  on 
the  mass  mt  which  is  supported  by  the  rod  AB  of  uniform  cross- 
section  A  and  length  /.  Let  v  be  the  velocity  of  both  masses  during 
impact.  Then,  by  conservation  of  momentum, 


/ 

\u\  —  (m\ 


_ 

or     v  = 


1  and  the  work  in  foot-pounds  necessary  to  bring  the  combined  masses 


*oim« 


work  = 


where  — -  =  h  is  the  height  of  fall  of  mr 

This  work  is  equal  to  the  work  of  stretching  or  compressing  the  rod.  If  Fis  the  force 
of  impact  at  the  end  of  the  compression  or  stretch,  A,  —  is  the  average  force  and  —  .  A  is  the 
work.  But,  from  page  346,  within  the  limit  of  elasticity  we  have 


where  E  is  the  coefficient  of  elasticity.      Hence 


/I 


2  2/         »*,  +  **,' 

From  (3)  we  can  find  the  strain  A  of  the  rod  caused  by  the  impact.      Let  the  rod  be 
strained  up  to  the  elastic  limit  unit  stress  Se.     Then  we  have  from  (2),  by  putting  F=  SfA, 


w 


and  hence,  from  (3), 


But  ^4/  is  the  volume  of  the  rod  V.     The  velocity  of  impact 


which  is  necessary  to  strain  the  rod  up  to  the  limit  of  elasticity,  is  then  given  by 


The  quantity  —  £r  is  called  the  coefficient  of  resilience  (page  516).     We  see  from  (5)  that 

the  greater  the  volume  or  mass  of  the  rod,  the  greater  the  blow  it  can  bear.       Hence  the 
mass  of  bodies  subjected  to  impact  should  be  made  as  great  as  possible. 

Since  »/t  and  >«2  fall  during  impact  through  the  distance  A,  we  have  more  correctly 


CHAP    V-]  STRENGTH  AND  IMPACT.  357 

and  hence,  instead  of  (5),  we  have 


If,  finally,  we  wish  to  take  into  account  the  mass  m^  of  the  rod,  we  have,  since  its  centre 
of  mass  moves  through  the  distance  -A, 


and  hence,  instead  of  (5),  we  have 


,  .          . 

^      "  s^  •  ~~<~  •  ^- 

If  a  mass  ml  moving  with  a  velocity  u^  puts  in  motion  another  mass,  mt,  by  means  of  a 
chain  or  rope,  we  have  in  the  same  way  for  the 
velocity  of  both  bodies  during  impact 


and  the  work  in  foot-pounds  expended  in  stretching  the  chain  is 


ME.**,  ! 

work  =    -*-* ( 


m 


where   //  =  —  is  the  height  due  to  the  velocity  »r 

We  have  then,  if  the  chain  is  stretched  to  the  elastic  limit  unit  stress  Se, 

S?     A,_     mjn*     ' 

7-*  •   -ti  ^    •  j  *    /fr« 

2h  m^  -|-  m2 

where  A  is  the  cross-section  and  /the  length  of  the  chain.      Hence 

k^  =  ^±^S4.Al.    .  (8) 

2g          mvmz       2E 

Example. —  The  two  opposite  suspension  rods  of  a  suspension  bridge  support  a  constant  load  of  5000 pounds, 

S a 
which  is  increased  by  6000  pounds  by  a  passing  load.      The  coefficient  of  resilience  ^p- for  wroitght  iron  is  7 

poitnds  per  square  inch  (page  jv<5).    The  length  of  rod  is  200  inches,  and  cross-section  1.3  square  inches.   Find 
the  height  of  fall  to  stretch  the  rods  to  the  limit  of  elasticity. 

ANS.  h  =  (5QOO  +  6000)  x  7  x  200  x  1.5  x  2  ^  ^  inches 
6000" 

If  then  a  cart  of  6boo  pounds  should  pass  over  an  obstacle  of  1.3  inches  high,  and  drop,  the  rods  are  in 
danger  of  being  stretched  beyond  the  elastic  limit, 


'356 


KINETICS  OF  A  MATERIAL  SYSTEM— IMPACT. 


[CHAP.  V. 


Strength  and  Impact. — Let  the  mass  wt ,  moving  with  the  velocity  u^ ,  impinge  on 
the  mass  w2  which  is  supported  by  the  rod  AB  of  uniform  cross- 
section  A  and  length  /.  Let  v  be  the  velocity  of  both  masses  during 
impact.  Then,  by  conservation  of  momentum, 


m\u\  =  (w>  +  ™t)v,     or     v  —  m    \lm  » 

1  ~T          2 

'  and  the  work  in  foot-pounds  necessary  to  bring  the  combined  masses 
to  rest  is 


work  = 


*g    w,  +  w2         /«!  -f-  w2 ' 


(0 


#  * 
where  — L  =  h  is  the  height  of  fall  of  mr 

This  work  is  equal  to  the  work  of  stretching  or  compressing  the  rod.  If  Fis  the  force 
of  impact  at  the  end  of  the  compression  or  stretch,  A,  -  is  the  average  force  and  -  .  A  is  the 
work.  But,  from  page  346,  within  the  limit  of  elasticity  we  have 

EA\ 


(2) 


where  E  is  the  coefficient  of  elasticity.      Hence 
i  „         EAK          m*h 


2l 


or     A  = 


ml  +  mi '  EA  ' 


(3) 


From  (3)  we  can  find  the  strain  A  of  the  rod  caused  by  the  impact.      Let  the  rod  be 
strained  up  to  the  elastic  limit  unit  stress  Se.     Then  we  have  from  (2),  by  putting  F=  S^ , 


=  £- 


(4) 


and  hence,  from  (3), 


2E 


But  Al  is  the  volume  of  the  rod  V.     The  velocity  of  impact 


which  is  necessary  to  strain  the  rod  up  to  the  limit  of  elasticity,  is  then  given  by 


'  2E 


(5) 


The  quantity  — ^r  is  called  the  coefficient  of  resilience  (page  516).     We  see  from  (5)  that 

the  greater  the  volume  or  mass  of  the  rod,  the  greater  the  blow  it  can  bear.       Hence  the 
mass  of  bodies  subjected  to  impact  should  be  made  as  great  as  possible. 

Since  ml  and  ma  fall  during  impact  through  the  distance  A,  we  have  more  correctly 


work  = 


i 


CHAP    V-J  STRENGTH  AND  IMPACT.  357 

and  hence,  instead  of  (5),  we  have 

m,  +  mz   Sl_  (mv  -f  mtf     Sel 

If,  finally,  we  wish  to  take  into  account  the  mass  #/3  of  the  rod,  we  have,  since  its  centre 
of  mass  moves  through  the  distance— A, 


+  (*•  +  ">>+?»,)*, 


and  hence,  instead  of  (5),  we  have 


m  '  2E 


'  '  ~~~  '  ~' 


If  a  mass  wx  moving  with  a  velocity  «x  puts  in  motion  another  mass,  m2,  by  means  of  a 
chain  or  rope,  we  have  in  the  same  way  for  the 
velocity  of  both  bodies  during  impact 


and  the  work  in  foot-pounds  expended  in  stretching  the  chain  is 

m.u2          i  m.m9        u2  m  m 

work  =    •       ---  (m  -f  mXv2  =  -     \-*-  .  -i-    —   .  mim*        L 
2  *  «  ^ 


where   h  =  —  is  the  height  due  to  the  velocity  ur 

We  have  then,  if  the  chain  is  stretched  to  the  elastic  limit  unit  stress  Se, 


. 

2h  m^  -f-  mz 

where  A  is  the  cross-section  and  /the  length  of  the  chain.      Hence 


Example.—  The  two  opposite  suspension  rods  of  a  suspension  bridge  support  a  constant  load  of  3000  pounds, 
which  is  increased  by  6000  pounds  by  a  passing  load.      The  coefficient  of  resilience  ^g  for  wrought  iron  is  7 

pounds  per  square  inch  {page  j/d).     The  length  of  rod  is  200  inches,  and  cross-section  1.5  square  inches.   Find 
the  height  of  fall  to  stretch  the  rods  to  the  limit  of  elasticity. 

ANS.  h  =  (5QOO  +  6000)  x  7_x  200  x  1.5  x  2  =  i<2g  .nches< 
6000" 

If  then  a  cart  of  6000  pounds  should  pass  over  an  obstacle  of  1.3  inches  high,  and  drop,  the  rods  are  in 
danger  of  being  stretched  beyond  the,  elastic  limit, 


358  KINETICS  OF  A  MATERIAL  SYSTEM-IMPACT.  [CuAP.  V. 

Impact  of  Beams.  —  Let  a  mass  ml  fall  from  a  height  //  upon  a  beam  AB  of  uniform 

cross-section  A  and  span  /,  supported  at  the  ends. 

Let  6  be  the  density  of  the  beam,  and  Fits  volume. 
*  Then 

=5?  V=Alt 


and  the  mass  m,  of  the  beam  is 


Let  the  mass  ml  strike  the  beam  at  the  centre  C,  and  let  the  greatest  velocity  at  the 
centre  be  ve,  and  at  any  other  point  of  the  beam  be  v. 

Then  the  work  in  foot-pounds  necessary  to  bring  the  combined  masses  to  rest  is 


work  = 


If  y  is  the  deflection  at  any  point,  and  A  the  deflection  at  the  centre,  we  have 


We  have  then 


mtf   ,   *Av*    r^dx 
work  =  —  -  I    - 

*g  2g   J0     ^ 


Now  if  Pis  the  pressure  at  the  centre,  we  have,  from  page  543. 
Pts 


where  E  is  the  coefficient  of  elasticity,  /  the  moment  of  inertia  of  the  cross-section  of  the 
beam  relative  to  the  horizontal  axis  through  the  centre  of  mass  of  the  cross-section  at  right 
angles  to  the  length  of  the  beam,  and  x  is  the  distance  of  any  point  from  the  left  end.  We 
have  then 


and  substituting  in  the  expression  for  the  work  and  integrating,  we  have 

work  =  —  —  .  — . 

2ff  35        2f 

The  distance  through  which  the  point  C  moves  in  the  short  time  /is  -' .  /.  If  we  divide 
the  work  by  this  distance,  we  have  the  force  which  brings  the  bodies  to  rest.  This  force 
should  equal  -~ '  Hence 

,    176 Al  »>  » 

mfl~mf. +-—.„,  or  ,,= 


CHAP.  V  ]  IMPACT  OF  BEAMS.  .     359 

If  we  insert  this  value  of  vt,  \ve  have  for  the  work  of  bringing  the  combined  masses 
to  rest  ^ 


work  = 


m.-\ 

l 


35    /  35 


But  this  work  is  also  equal  to  —Pd.      If  the  elastic  limit  is  reached,  we  have,  from 


page  500, 


and     ^= 


vl  \2vE' 


where  Se  is  the  elastic  limit  unit  stress,  and  v  is  the  distance  of  the  most  remote  fibre  of  the 
cross-section  from  the  neutral  axis. 
Hence 

\JdAl 

I      -  s<n         m*h  i,-?L  !L  35 

*  *l~'" 


Since  dAl  is  the  mass  m2  of  the  beam, 


// 


If,  for  instance,  the  cross-section  of  the  beam  is  a  rectangle  of  breadth  b  and  depth  d, 


we  have  /  =  —  bd*  and  v  =  -d.      Hence  for  this  case 

12  2 


s.«  w     >         * 

~a2*  9  '        «,2 

Example,  —  Find  the  height  from  which  a  mass  of  200  Ibs.  must  fall  in  order  that,  striking  the  centre  of  a 
plate  of  cast  iron  j6  inches  long,  12  inches  wide  and  3  inches  thick,  supported  at  both  ends,  it  may  bend  it  to 
the  elastic  timit. 

ANS.  If  we  take  the  coefficient  of  resilience  (page  516), 

—  =  1.2  pounds  per  square  inch, 
iE 

we  have,  since  S  for  cast  iron  is  about  0.259  Ibs.  per  cubic  inch,  m*  =  12x3x36x0.259  =  335.7  Ibs.     Hence 
for  height  of  fall 

/ 

=  1.2x12x3x36    aoo  +  ^  335-7  = 
9  40000 


36° 


KINETICS  OF  A  MATERIAL  SYSTEM-IMPACT. 


[CHAP.  V. 


Impact  of  Rotating  Bodies.— Let  two  bodies  of  mass  ;«,  and  m2  rotate  about  fixed 
axes  at   Ol  and   Oa  and  impinge,   and   let  AB  be  the  line  of  impact.      Let  the  normals 
O^A  =  al  and  O2B  =  at. 

Then  (page  322)  we  can  reduce  the  masses  ml  and  ;«2  to  the 
equivalent  masses  at  A  and  B 

w,*-,'2  w2/c/ 

1  a!         and       -nr"! 

where  *•/  and  /c-2'  are  the  radii  of  gyration  relative  to  Ol  and  02. 
If  then  we  substitute  these  masses  in  the  place  of  ;«,  and  ;«2  in  the 
equations  for  central  impact  (page  349),  we  have  for  bodies  of  the  same  material 


' 


:,'V 


where  «,  and  «2  are  the  velocities  at   ^4   and  .#  before   and  ^  ,  ?'2  after  impact,  and  e  the 
modulus  of  elasticity. 

If  e,  and  e2  are  the  angular  velocities  before  and  00,  ,  &?2  after  impact,  we  have,  taking 
counter-clockwise  rotation  as  positive,  the  origins  at  (7j  and  O2  ;  and  nl  ,  ti3  as  coinciding  with 
the  axes  of  Y  for  each  origin, 


=  —  u2,          alcol  =  —  ?', 


Hence 


G7,    =    6,    — 


,£,'2 


Examples. —  The  moment  of  inertia  of  the  shaft  O\B  relative  to  its  axis  of  rotation  at  O\  is  m\K\'*  =  40000 
Ib.-ft*,  ami  that  of  the  trip-hammer  BO*  relative  to  its  axis  of  rotation  at  O,  zs  w,Ky»  =  130000  lb.-ft*.  The 
arm  O\B  of  the  shaft  is  a\  =  2  ft.,  and  that  of  the  hammer  BO*  is  tit  =  6  ft.  The  angular  velocity  of  the 
shaft  before  impact  is  f\  =  1.05  radians  per  sec.  Find  the  velocity  after  impact  and  the  loss  of  energy  at  each 
impact,  supposing  both  bodies  inelastic. 

ANS.  The  angular  velocity  of  the  shaft  after  impact  is,  since 


,.o5jcj5ooc     _  =  0.74,  radians  per  sec. 


40000  x  36  +  i  50000  x  4 
The  angular  velocity  of  the  hammer  after  impact  is 


O, 


6  x  2  x   1.05 
a»,  =  —  -  -=  =  0.247  radians  per  sec. 


The  loss  of  energy  at  each  impact  is,  in  foot-pounds, 


=  20..  63  foot-pounds. 


CHAP.  V.] 


IMPACT  OF  AN  OSCILLATING  BODY. 


36t 


Impact  of  an  Oscillating  Body. — If  a  body  of  mass  ml  impinges  upon  a  body  of  mass 
w/2  which  is  suspended  from  an  axis  at   O2,  the  equations  of  the  preceding  article  apply  if 

we  put  in   equations  (i)  ml  in  place  of  — ^  ,  an<^  —  a2ez  m  P^ace  °f  ^2  > 

*i  I     Og 

and  —  a.2&).2  in  place  of  v2.       We  have  then  for  the  velocity  of  the  mass 

;//j  after  impact,  taking  clockwise   rotation  as  positive,  origin  at  O2,  and 
O2'O2  as  axis  of  Y,  if  /c/  and  /c2'  are  the  radii  of  gyration    for   axes  at  Ol       m/ 
and  O2, 


and  for  the  angular  velocity  of  the  mass  m2  after  impact 


If  the  mass  m2  were  at  rest  before  impact,  we  have  e2  =  o  and 

072   =    —      «x(  I    -j-   ^)  —  — (4) 

If  ml  is  at  rest  and  the  oscillating  body  impinges  on  it,  we  have  ur  =  o,  and  hence 

(6) 

yo  c*>  I     •••  v*       i      *"  /    -^     I       .  #a    i  " 


The   angular  velocity   &>2  in  the  first  case,  equation  (4),  or  the  velocity  z/x  of  wx  in  the 
second  case,  equation  (5),  is  a  maximum  when 


is  a  maximum,  or  when 


is  a  minimum.      Putting  the  first  differential  coefficient  equal  to  zero,  we  have  for  the  value 
of  a2  when  the  maximum  velocity  is  imparted 


(7) 


362 


KINETICS  OF  A  MATERIAL   SYSTEM. 


[CHAI-.  V. 


Hence  the  maximum  velocity  imparted  to  the  oscillating  body  w2  when  at  rest  and  struck 
by  ;//,  is  given  by 


2/c,  V   »/. 


and  the  maximum  velocity  imparted  to  ml  when  at  rest  and  struck  by  #/2  is 


(8) 


(9) 


REACTION  OF  THE  Axis. — Let  the  force  of  impact  be  F,  and  the  reaction  of  the  axis  be 
R.     Then,  from  page  339, 


(10) 


If  we  give  to  a2  its  value  from  (7),  we  have  for  the  reaction  of  the  axis  when  the  maxi- 
mum velocity  is  imparted 


;n) 


The  centre  of  percussion  (page  339)  is  at  the  distance 


from  the  axis.  If  the  impact  takes  place  at  this  distance,  there  is  no  reaction  of  the  axis. 
Ballistic  Pendulum. — The  ballistic  pendulum  consists  of  a  large  mass  »/2  hung  from  a 
horizontal  axis  O' ' .  It  is  set  in  oscillation  by  a  cannon-ball  shot 
against  it,  and  is  used  to  determine  the  velocity  of  the  ball. 
In  order  to  render  the  impact  inelastic,  the  mass  ;«2  consists  of 
a  box  filled  with  sand  or  clay,  so  that  the  ball  enters  the  mass 
and  oscillates  with  it. 

In  order  to  determine  the  velocity  of  the  ball,  the  angle 
of  oscillation  is  measured  by  a  pointer  directly  below  the  centre 
of  mass  O,  which  moves  on  a  graduated  arc  AB. 

Let  ml  be  the  mass  of  the  ball.  Then,  from  equation 
(4)  of  the  preceding  article,  making  e  =  O,  we  have  for  the 
angular  velocity  after  impact 


(O 


where  /c2'  is  the  radius  of  gyration  of  the  pendulum  relative  to  the  axis  at   O',  and  a2  is  the 
distance  of  the  point  of  impact  below  this  axis. 

Let  /  be  the  length  of  the  equivalent  simple  pendulum  which  oscillates  in  the  same  time 
as  the  ballistic  pendulum,  and  let  the  angle  measured  on  the  arc  AB  be  0. 


CHAP.  V.]  ECCENTRIC  IMPACT.  363 

We  have  then  for  the  simple  pendulum  (page  336)  the  angular  acceleration 

g  sin  0 


—• 

If  O'  O  =  d  is  the  distance  of  the  centre  of  mass  O  of  the  ballistic  pendulum  from   O' 
we  have  (page  317) 

(ml  -f-  m^gd  sin  0  =  I'  a  =  (m^a*  +  m2K^}a, 

or 

\n  0 


Equating  these  two  values  of  a,  we  obtain  for  the  length  of  the  equivalent  simple  pen- 
dulum 

l  =  mfa  +  Zjd (2) 

The  height  of  displacement  is 

a 
h  =  I—  /cos  e  =  2l  sin2-. 

Hence  the  velocity  at  the  lowest  point  of  swing  is 

3 

v  =   V2gh  =  2  Vgl  sin  ~> 
and  the  corresponding  angular  velocity  is 

00  =  7  =  2  V  7  '  sin  I- 

Equating  this  to  (i)  and  inserting  the  value  of  /  from   (2),  we  have  for  the  velocity  of 
the  ball 

(m,-\-m<\d     ,—f            6  ,  \ 

u,=2\-      — 2-i/^/.sm- •  .      •      (3) 


If  the  pendulum  makes  «  vibrations  per  minute,  the  duration  of  a  vibration  is 


77       60 
=7f\/-—  —  , 
V  g        n 


t=7f\-—  —       and  hence 
g 


Hence 


Eccentric  Impact.—  Let  ^,  ,  m2  be  the  masses,  6>t  and  O2  their  centres  m, 
of  mass,  BN  the  line  of  impact,  and  BO2  =  p  the  distance  from  the  line  of  (<Vj 
impact  to  the  centre  of  mass. 

By  conservation  of  momentum  we  have,  so  far  as  translation  is  concerned, 


where  «x  and  «2  are  the  initial  and  vv  ,  vz  the  final  velocities  of  wt  and  ;«a. 


364  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  V. 

The  mass  mt  reduced  to  B  (page  322)  is 


where  K,  is  the  radius  of  gyration  of  m3  relative  to  axis  at  O2,  and  the  velocities  of  the  point 
B  before  and  after  impact  due  to  rotation  about  axis  at  O2  are  —  /e2  and  —  /«2,  or  opposite 
in  direction  to  u^  when  /,  e2,  o?2  are  positive. 

We  have  then,  so  far  as  rotation  is  concerned, 


(2) 


If  the  bodies  are  non-elastic,  ml  and  the  point  B  move  together  after  impact  and  we 
have 


»,  ...........     (3) 

Eliminating  z/2  and  <»2  from  (i)  and  (2)  by  means  of  (3)  we  have 


If  the  bodies  are  perfectly  elastic,  these  values  are  twice  as  great  (page  345).  If  the 
bodies  are  imperfectly  elastic  and  of  the  same  material,  these  values  are  (i  -J-  e)  times  as  great 
(page  349)  where  e  is  the  modulus  of  elasticity. 

We  have  then  in  general 


(5) 


These  equations  are  general.  If  the  impact  is  central,  p  =  o  and  &?2  in  (6)  is  unchanged 
and  equal  to  ea,  while  (4)  and  (5)  reduce  to  equations  (8),  page  349. 

If  the  bodies  are  perfectly  elastic,  e  =  i.  If  non-elastic,  e  —  o.  If  w2  moves  towards 
«*j,  "2  is  negative.  If  ml  is  initially  at  rest,  ul  =  o.  If  m2  is  initially  at  rest,  ;/2  =  o  and 
62  =  o.  If  7«t  is  fixed,  we  can  take  w,  =  oo  and  i/,  —  o. 

Examples.— ( i)  A  prismatic  bar  AR  falls  through  a  height  h,  retaining  its  horizontal  position  until  one  end 
p        o  strikes  a  fixed  obstacle  D.     Find  the  motion  after  impact,  considering  the  bodies 

1 B      non-elastic. 

ANS.  Let  m*  be  the  mass  of  the  bar,  /  its  length,  and  »,  =  —  |/  igh  the 
velocity  of  the  centre  of  mass  O  at  the  instant  of  impact.  We  have  e  =  o, 
ei  =  o,  mi  =  oo  ,  «i  =o.  and  hence  the  velocity  of  translation  after  impact  is 

*»*  u^  f,\ 

Vl=  U»  —  Ut   .     .  .    ,,,  =       .    . l*J 


CHAP.  V.]  IMPACT— EXAMPLES.  365 

and  the  angular  velocity  about  0  after  impact  is 


/*                       /                                                 / 
In  the  present  case  /<•,»  =  — ,  p  =  -\ —  if  A  strikes,  and^  = if  B  strikes.      Also,  ut  =  — 

12  2  2 

Hence  in  both  cases  of  A  or  B  striking 


If  A  strikes, 


If  B  strikes, 


=  +  2l 


The  plus  (+)  sign  denoting  counter-clockwise  and  the  minus  (—  )  sign  clockwise  rotation. 
We  can  obtain  (i)  and  (2)  directly  as  follows  :  We  have  (page  318) 

foot  =  moment  of  momentum  =  m*utp,     or     oaa  =  -  miutp  —  _  —  «a/>  — 

Also,  at  the  instant  of  impact 

v,  =p<a,. 
Hence 


v*  = 
The  momentum  after  impact  is  then 

The  impulse  is 

~~*S+J>*   "  4"" 
The  velocity  at  any  point  distant  x  from  O  after  impact  is 


where/  and  x  are  positive  towards  B  and  negative  towards  A.     Hence  for  A  striking 


and  for  B  striking 


After  impact  the  centre  moves  in  the  same  vertical  with  the  uniform  acceleration^,  while  the  angular 
velocity  &?.,  remains  unchanged. 


366 


KINETICS   OF  A   MATERIAL   SYSTEM. 


[CHAP.  V. 


(2)  An  inextensible  string  is  wound  around  a  cylinder  and  has  its  free  end  attached  to  a  fixed  point. 
The  cylinder  falls  through  a  height  h,  and  at  the  instant  of  impact  the  string  is  vertical 
and  tangent  to  the  cylinder.     Find  the  motion  after  impact. 


ANS.  v  = 


(3)  An  iron  ball  of  mass  m\  =  65  Ibs.  moving  with  a  velocity  of  jb  ft. 
Per  sec.  strikes  a  pine  beam  of  uniform  rectangular  cross-section  in  the 
centre  line  of  a  side  and  at  ri^ht  angles,  at  a  distance  p  =  t\ft.  above  the 
centre  of  mass.  The  mass  of  the  beam  is  842.4  Ibs.,  its  length  j  ft.  and 
breadth  2  ft.  If  the  beam  is  at  rest,  find  the  motion  after  impact,  con- 
sidering the  impact  as  non-elastic. 

ANS.  The  moment  of  inertia  (page  38)  is  the  same  as  for  m-  concentrated  at  a  corner 
We  have  then 


2.4i6;«a,    or 


2.416. 


Hence  the  velocity  of  the  ball  after  impact  is 


4 


The  velocity  of  the  centre  of  mass  of  the  beam  after  impact  is 

2.364  ft.  per  sec. 

The  angular  velocity  of  the  beam  after  impact  is 
mipui 


OJ,    =  — 


(mi 


S  +  rrti 


5  =  —  1.712  radians  per  sec. 


(4)  A  ballistic  pendulum  weighing  30,000  Ibs.  is  set  in  oscillation  by  a  6-lb.  ball,  and  the  angular  displacement 
is  75°.  If  the  distance  d  of  the  centre  of  mass  from  the  axis  is  5ft.,  and  the  distance  at  of  the  point  of  impact 
below  the  axis  is  5.5  ft.,  and  the  number  of  oscillations  per  minute  is  n  =  40,  find  the  velocity  of  the  ball. 

3006        1 20  x  32.2  x   5 

V  '  40  x  3.14-6  x  55  S1"  7i   =  l828  ft>  PCr  ^ 


ANS. 


CHAPTER   VI. 

ROTATION   ABOUT   A    TRANSLATING   AXIS. 

Effective  Forces — Rotation  about  a  Translating  Axis. — Equations  (5),  page  159,  give 
the  components  of  the  acceleration  for  any  particle  of  a  rotating  and  translating  body.  If 
the  axis  of  rotation  passes  through  the  centre  of  mass,  we  have  ^  —  o,  y  =  o,  J  —  o.  The 
co-ordinates  x,  y,  s  are  taken  from  the  centre  of  mass  O. 

If  then  we  make  these  changes  in  equations  (5),  page  159,  multiply  each  term  by  the 
mass  m  of  the  particle  and  sum  up  for  all  the  particles,  we  have,  since  x,  y,  z  are  taken  from 
the  centre  of  mass  O  and  hence  2mx  =  o,  2my  =  o,  2mz  =  o,  for  the  component  effective 
forces  for  a  body  of  mass  m  =  2m  rotating  about  any  translating  axis 


Hence  the  effective  force  in  any  direction  for  a  body  rotating  about  a  translating  axis  is 
the  same  as  for  a  particle  of  mass  equal  to  the  mass  of  the  body  having  the  acceleration  of 
the  centre  of  mass  in  that  direction. 

If  we  take  mass  in  Ibs.  and  acceleration  in  ft.-per-sec.  per  sec.,  we  have  force  in 
poundals.  For  force  in  pounds  divide  by  g  (page  171). 

Moments  of  the  Effective  Forces  —  Translating  Axis.  —  Equations  (10),  page  160,  give 
the  component  moments  of  the  acceleration  for  any  particle  of  a  rotating  and  translating 
body.  If  the  axis  of  rotation  passes  through  the  centre  of  mass,  we  have  ~x  =  o,  y  =  0,^  =  0, 
and  if  it  does  not  change  in  direction,  we  have  ooxK)y  —  O,  GOXOO,  =  o,  ooyoax  =  o,  <&,<*),  =  o, 
aogcox  =  O,  os.  cov  —  o.  If  we  make  these  changes  in  equations  (10),  page  160,  multiply  each 
term  by  the  mass  in  of  the  particle  and  sum  up  for  all  the  particles,  we  have,  since  x,  y, 
z  are  taken  from  the  centre  of  mass  O  and  hence 


=  o,  my  =  o,  mz  =  o, 

=  Ix  ,          SM(**  +  x^  =  Iy,          2m(x*  -\-  /)==/„ 

where  Ix,  Iy,  Ix  are  the  moments  of  inertia  of  the  body  for  the  axes  of  X,   K,  Z  through  the 
centre  of  mass  O, 


M'fx  =  ni/J'  —  ftt/jj  —  <*y2myx  —  a^mzx  +  (&?/  —  oo^'Sntzy  -f  Ixa 

M'fy  -    mfxz  -  mfzx  -  az^mzy  -  a^mxy  -f  (^  -  <a.9)2mxx  +  Iyay,     -  (2) 


M'fz  =  fo.fyX  —  m/xy  —  ax2mxz  —  ay2myz  -f  (G?/  —  < 

If  the  translating  axis  coincides  with  one  of  the  co-ordinate  axes,  as,  for  instance,  O'X', 
we  have  o?y  =  o,  GOZ  =  o,  ay  =  o,  at  =  o,  and 

mfyJ+  fxax, 


M'fy  =  mfx  ~z  —  mfzx  +  co^mxz  —  ax2mxy,       ......      (3) 


M'ft  —  mfyx  —  mfxy  —  o 

367 


368  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  VI. 

Since  we  can  take  any  co-ordinate  axes  we  please,  let  the  co-ordinate  axes  O'X',  (7V, 
O'Z'  be  principal  axes  at  O1  ',  the  intersection  with  the  axis  of  rotation  of  a  plane  through 
the  centre  of  mass  O  at  right  angles  to  the  axis.  (Figure,  page  153.)  Then  the  parallel  axes 
OX,  OY,  OZ  at  the  centre  of  mass  O  will  be  principal  axes,  and  we  have  2tnxy  =  o, 
2m  ys  =  o,  2msx  =  o,  and  equations  (2)  become 


Mfx  =  m/.  J  - 

^=m/J  -mfs  +  J,a,,  I  ........  (4) 

Mft  =  m/y.r  -  mfxy  +  IMetf.  j 

If  the  axis  of  rotation  is  a  principal  axis,  let  it  coincide  with  O'X',  and  we  have  in  (4) 
ay  =  o,  a,  =  O.  If  the  axis  passes  through  the  centre  of  mass,  we  have  x  =  o,  y  =  Q,  3=0 
and  Mfx  =  7>,  ,  Mfy  =  Iyay  ,  Mfx  =  Itat  . 

If  we  take  distance  in  feet  and  mass  in  Ibs.,  these  equations  (2),  (3),  (4)  give  moments  in 
poundal-feet.  For  pound-feet  divide  by  g  (page  171). 

Momentum  —  Translating  Axis.  —  Equations  (4),  page  154,  give  the  component  velocities 
for  any  particle  of  a  rotating  and  translating  body. 

If  we  multiply  each  term  by  the  mass  in  of  the  particle  and  sum  up  for  all  the  particles, 
we  have,  since  JT,  y,  z  are  taken  from  the  centre  of  mass  O  and  hence  2mx  =  o,  2  my  =  o, 
2mz  =  o 

2mvx  =  mF,  ,          2mvy  =  mz^  ,          2mvM  =  mvx  ......     (5) 

Hence  the  momentum  in  any  direction  for  a  body  rotating  about  a  translating  axis  is 
the  same  as  for  a  particle  of  mass  equal  to  the  mass  of  the  body  having  the  velocity  of  the 
centre  of  mass  in  that  direction. 

Moment  of  Momentum  —  Translating  Axis.  —  Equations  (10),  page  155,  give  the  com- 
ponent moments  of  velocity  for  any  particle  of  a  rotating  and  translating  body.  If  we 
multiply  each  term  by  the  mass  m  of  the  particle  and  sum  up  for  all  the  particles,  we  shall 
obtain  the  component  moments  of  momentum  for  any  co-ordinate  axes  we  please. 

Let  us  take  these  axes  as  principal  axes  at  the  point  O',  the  intersection  with  the  axis 
of  rotation  of  a  plane  through  the  centre  of  mass  O  at  right  angles  to  the  axis.  (Figure, 
page  153.)  Then  the  parallel  axes  OX,  OY,  OZ  at  the  centre  of  mass  O  are  principal 
axes,  and  we  have 

2mxy  =  o,          2myz  =  o,          2mzx  =  o,          2mx  =  o,          2nty  =  o,          2tnz  =  o, 


where  /,,  Iy,  /,  are  the  moments  of  inertia  of  the  body  for  the  parallel  axes  OX,  OY,  OZ. 
We  have  then,  from  equations  (10),  page  155, 


(6) 


If  the   axis   passes    through   the   centre   of  mass,   we  have  ^  =  o,  y  =  o,  z  =o  and 


CHAP.  VI.]  PRESSURES  ON   TRANSLATING  AXIS.  369 

Pressures  on  Translating  Axis  —  Permanent  Axis.—  We  have  just  as  on  page  319  for 

fixed  axis,  and  using  the  same  notation, 

x  =  mfx  ,         R,'  +  Ry"  +  2F,  =  m/7, 

.....     (7) 


We  have  also 


-  mf,x  + 


mfsy  -  mfyz  +  Ixax. 


.     (8) 


For  a  principal  axis    through   the   centre   of    mass   we    have   ~2mxy  =  o,   2myz  =  o, 
=  o  and  x  =  o,  y  =  o,  J  =  o,  and  these  equations  become 


xy  =  o,          -     .,  ,  j  -        ,x  =  o, 

2F.y-  2Fyz  =  Ixax. 
We  see  from  these  last  equations  that  if  we  have  always 

M,  =  ^Fyx  -  2Fxy  =  o,          M  =  2F^  -  2F,x  =  o, 


that  is,  if  the  moments  Mz  ,  My  about  the  axes  of  Z  and  Y  of  the  impressed  forces  F  are 
always  zero,  there  is  no  rotation  of  the  axis,  and  hence  even  if  this  axis  is  unconstrained  it 
will  not  change  its  direction.  The  axis  is  then  a  permanent  axis  of  rotation. 

Hence  if  a  body  rotates  about  a  principal  axis  through  the  centre  of  mass  and  Mg,  My 
are  always  zero,  the  axis  of  rotation  is  a  permanent  axis  and  will  not  change  in  direction 
even  if  the  pressures  on  the  axis  are  removed. 

Mx  and  My  are  always  zero  when  there  are  no  impressed  forces  ;  when  the  impressed 
forces  always  reduce  to  a  resultant  force  through  the  centre  of  mass,  or  to  a  resultant  force 
through  the  centre  of  mass  and  a  couple  whose  plane  is  at  right  angles  to  the  axis  ;  when 
the  impressed  forces  all  lie  in  a  plane  through  the  centre  of  mass  at  right  angles  to  the  axis, 
or  reduce  to  a  resultant  force,  or  force  and  couple,  in  this  plane. 

Conservation  of  Moment  of  Momentum  _  Translating  Axis.  —  We  have  from  equations 
(6)  for  the  component  moments  of  momentum  for  a  body  rotating  about  a  translating  axis, 
taking  the  co-ordinate  axes  as  principal  axes  at  O'  ', 


Mvy  =  mz^J  —  mv^x  H-  fy<x*y, 
MV1  =  mvyX  —  mvxy  -f-  /,<»,, 

and  from  equations  (4)  for  the  component  moments  of  the  effective  forces 

Mfx  =  m/J  -  m^J+/,a, 


If  in  equations  (b)  we  have  ax  =  o,    ay  =  O,  at  =  o,   and  also  fx  =  o,  fy  =  o,  /,  =  o,  we 
shall  evidently  have  Mfx  =  o,   Mfy  =  o,   Mft  =  o,  and  ooxt    aay,    «*,,    vx,    ~vy,  ~vt  constant  in 


37°  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAI>.    VI. 

equations  (a).  But  since  the  motion  of  the  centre  of  mass  is  the  same  as  if  all  the  impressed 
forces  were  applied  to  a  particle  of  mass  m  at  the  centre  of  mass  (page  299),  when  fx  =  o, 
f,  =  o,/t  =  owc  must  have  the  algebraic  sum  of  the  components  of  all  the  impressed  forces 
in  three  rectangular  directions  equal  to  zero,  or 

2FX  =  o,         2F,  =  o,         2FM  =  o. 

Also,  since  by  D'Alembert's  principle  the  moment  of  the  effective  forces  is  equal  to  the 
moment  of  the  impressed  forces,  when  M/,,  =  o,  Mfy  =  o,  M/t  =  o  we  have  the  moment  of 
the  impressed  forces  for  the  axis  of  rotation  zero  also.  If  we  have  at  the  same  time 

Mfm  —  o,     Mfy  =  o,     Mf,  =  o,     and     2F,  =  o,     2Fy  =  o,     2F,  =  o, 

the  impressed  forces  then  form  a  system  of  forces  in  static  equilibrium,  and  from  (b)  we  see 
that  ax  =  O,  a,  —  O,  or,  =  O. 

Hence  if  the  impressed  forces  acting  upon  a  body  rotating  about  a  translating  axis  form 
a  system  of  forces  in  static  equilibrium,  the  moment  of  momentum  about  that  axis  is  constant. 

Kinetic  Energy  —  Translating  Axis.  —  The  kinetic  energy  of  a  particle  of  mass  m  and 

velocity  v  is-tntf.  If  a  particle  has  the  component  velocities  vx,  vyt  vtt  we  have  v*  = 
vx*  +  v*  +  v*,  and 

Lffg^  =  -m(v*  -)-]  Vy*  _|_  vv)m 

Equations  (4),  page  154,  give  the  component  velocities  for  any  particle  of  a  translating 
and  rotating  body.  If  we  square  these  component  velocities,  multiply  each  term  by  —  m, 

sum  up  for  all  the  particles  and  add,  we  shall  have  the  kinetic  energy  for  a  rotating  and 
translating  body  for  any  co-ordinate  axes  we  please. 

Let  us  take  these  axes  as  principal  axes  at  the  point  O',  the  intersection  with  the  axis 
of  rotation  of  a  plane  through  the  centre  of  mass  O  at  right  angles  to  the  axis.  (Figure,  page 
153.)  Then  the  parallel  axes  OX,  OY,  OZ  at  the  centre  of  mass  O  are  principal  axes  and 
we  have 

2mxy  =  o,      2myz  =  o,      2mzx  =  o,      2mx  —  O,      2  my  —  o,      2mz  =  o, 


where  Ix,  fy,  f,  are  the  moments  of  inertia  of  the  body  for  the  parallel  axes  OX,  OY,  OZ. 
We  have  then  from  equations  (4),  page  154,  for  the  kinetic  energy 


(9) 


If  the  axis  of  rotation  coincides  with  one  of  the  principal  axes,  as,  for  instance,  O'Z', 
we  have  GOX  =  o,  coy  =  o,  and 


(10) 


The  kinetic  energy  is  then  the  sum  of  the  kinetic  energy  of  translation  and  rotation. 
If  we  take  mass  in  Ibs.,  these  equations  give  kinetic  energy  in  foot-poundals.  For  foot- 
pounds divide  by  g  (page  I71)- 


CHAP.  VI.] 


INSTANTANEOUS  AXIS  ROTATION—TRANSLATING  AXIS. 


371 


Instantaneous  Axis. — If  a  body  rotates  about  a  translating  axis,  the  instantaneous  axis 
is  parallel  to  the  translating  axis  and  passes  through  a  point  whose  co-ordinates  are  given 
by  equations  (15),  page  I57»  f°r  the  instantaneous  axis  of  rotation,  and  by  equations  (i5)» 
page  161,  for  the  instantaneous  axis  of  acceleration. 

If  we  take  the  translating  axis  as  the  axis  of  OX,  we  have  coy  =  o,  cot  =  o,  and  for  the 
instantaneous  axis  of  rotation  the  co-ordinates  from  the  centre  of  mass  are 


We  also  have  ay  =  o,  or,  =  o,  and  the  co-ordinates  of  the  instantaneous  axis  of  accel- 
eration are 


Examples.— (i)  A  circular  disc  of  mass  m  and  whose  plane  is  -vertical  has  a  force  of  P  pounds  applied  at 
the  centre,  and  rolls  upon  a  horizontal  plane.     Determine  its  motion. 

ANS.  Let  the  force  P  make  the  angle  9  with  the  horizontal  and  act  in  the  plane  of  the  disc.     The  force 
is  Pg  poundals,  and  the  horizontal  component  Pg  cos  0,  and  the  vertical  com- 
ponent Pg  sin  0.     Let  r  be  the  radius,  and  A  the  point  of  contact. 

The  moment  of  the  impressed  forces  about  A  is  —  Pg  cos  9  .  r.  Let  K  be 
the  radius  of  gyration  for  axis  through  the  centre  of  mass  O  at  right  angles 
to  the  disc,  and  /'  the  moment  of  inertia  for  parallax  axis  through  A.  Then 
/'  =  m  (tf2  +  r1*),  and  we  have  the  moment  of  the  effective  forces  fa.  By 
D'Alembert's  principle, 


—  /'a  —Pg  cos  9  .  r  =  o, 


_  _  Pgr  cos  0 

~ 


The  axis  at  A  is  the  instantaneous  axis.     The  acceleration  at  the  centre  O  is  then 


7--_^_ 
~ 


For  a  disc,  K*  =  —,  hence 


_  iPg  cos  0 


iPg  cos  9 
3m 


Both  angular  and  linear  accelerations  are  then  constant  if  P  is  constant. 

(2)  A  disc  of  mass  m  whose  plane  is  vertical  rolls  (without  sliding)  down  a  rigid  inclined  plane.    Determine 
its  motion. 

ANS.  Let  the  radius  be  r,  the  radius  of  gyration  about  an  axis  through  the  centre  of  mass  O  perpendicular 
to  the  plane  of  the  disc  be  K,  and  0  the  inclination  of  the  plane,  and  A  the  point  of  contact. 

Then  the  weight  is  mg-,  the  force  parallel  to  the  plane  is  mg  sin  0,  and  its 
moment  about  A,  —  rn.gr  sin  0.     We  have  then,  as  in  the  preceding  example, 


I'a  = 


+  ra)a  =  —  vagr  sin  0,     or     a  =  — 


gr  sin  0 
+   r* 


Also,  since  A  is  the  instantaneous  axis,  the  linear  acceleration  of  the  centre  is 


Since  K*  =  —,  we  have 


37*  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  VI. 

Both  linear,  tangential  and  angular  accelerations  are  constant  and  the  velocity  after  any  time  may  readily 
be  determined. 

(3)  Find  the  time  a  rigid  homogeneous  cylinder  will  take  to  roll  from  rest  down  a  plane  20  ft.  long  and 
inclined  30"  to  the  horizon,  the  axis  of  the  cylinder  being  horizontal. 

ANS.  f-9j  sec. 

(4)  A  rigid  homogeneous  circular  disk  of  mass  m  and  radius  r,  whose  plane  is  vertical,  moves  in  contact 
with  a  smooth  inclined  plane  whose  angle  is  0.     From  a  point  in  the  same  vertical  plane  as  the  disc,  and  at  a 
distance  from  the  inclined  plane  equal  to  the  diameter  of  the  disc,  a  string  is  carried  parallel  to  the  inclined 
plane  and  is  wrapped  round  the  edge  of  the  disc,  and  its  end  is  fixed  in  the  circumference.     Find  the  tension  T 
in  the  string,  the  linear  acceleration  f of  the  centre,  and  the  angular  acceleration  a  of  the  disc. 

m?K*  sin  8       me  sin  0  m  sin  6  ., 

ANS.   T  =    ^  =  -3E-; poundals  or  — - —  Ibs. ; 


radians-per-sec.  per  sec. 

(5)  A  perfectly  flexible  and  inextensible  ribbon  is  coiled  on  the  circumference  of  a  homogeneous  circular 
disc  of  radius  r  and  mass  m,  and  has  its  free  end  attached  at  a  fixed  point.     A  part  of  the  ribbon  is  unrolled 
and  vertical,  and  the  disc  is  allowed  to  fall  from  rest  by  its  own  weight.     Find  the"  acceleration  f  of  the  centre 
and  the  angular  acceleration  a.  before  the  ribbon  becomes  wholly  unrolled,  and  the  distance  s  which  the  centre 
"will  descend  in  one  second. 

~P  3*  ~~2        ~3* 

(6)  A   homogeneous   hemisphere  of  radius  r  performs  small  oscillations  on   a    rough  horizontal  plane. 
Find  the  periodic  time. 

ANS.   For  the  simple  pendulum  of  length  /and  mass  m  we  have 

mg  x  /  sin  0  = 

For  the  hemisphere  let  </be  the  distance  OC  from  the  centre  of 
A  the  hemisphere  O  to  the  centre  of  mass  C,  and  let  K  be  the  radius  of 

gyration  about  an  axis  through  the  centre  of  mass  C  parallel  to  the 
instantaneous  axis  at  A.     Then  the  moment  of  inertia  for  the  instantaneous  axis  at  A  is 

/'  =  «[*•  +  (rfsin  0)'  +  (r  -  rfcos  fl)1]. 
If  0  is  small,  we  may  put  sin*  0  =  o  and  cos  0  =  i,  and  we  have 

r  =  «[*•  +  (r  -  </)•]. 

We  have  then 


Equating  (i)  and  (2),  we  have  for  the  length  of  the  equivalent  simple  pendulum 

,  _  **  +  (r  -  //)* 
d 

The  periodic  time  is  then 


(7)  A  homogeneous  circular  hoop  moving  in  a  vertical  plane  in  contact  with  a  rough  horizontal  surface 
kits  at  a  given  instant  an  angular  velocity  opposite  in  direction  to  that  which  would  enable  it  to  roll  in  the 
direction  of  its  translation  at  that  instant.  Determine  its  motion. 


CHAP.  VI.]  ROTATION -TRANSLATING  AXIS-EXAMPLES.  373 

ANS.  Let  m  be  the  mass  of  the  hoop.     The  forces  acting  on  the  hoop  are  its  weight  mg  at  the  centre  C, 
the  upward  pressure  of  the  plane  R  at  A,  and  the  friction  F  opposite  to  the 
direction  of  translation  AX. 

The  acceleration  of  the  centre  is  then 


and  the  angular  acceleration  upon  the  axis  through  C  perpendicular  to  the   p 
plane  of  the  hoop  is,  if  K  is  the  radius  of  gyration  for  this  axis, 

— £ 

We  have  also 

R  —  mg  =  o,     or    R  =  ing. 

If  u  is  the  coefficient  of  sliding  friction,  we  have 

F  =  nmg. 
Hence 


If  vi  and  <»,  are  the  initial  values  of  the  linear  and  angular  velocities,  we  have  then  for  the  linear  and 
angular  velocities  after  any  time  / 

ugrt 
v  =  vi-  Hgtt         GO  =  cot  -  ^5-. 

If  at  any  instant  there  is  no  slipping,  we  have  at  that  instant  the  velocity  at  A  zero,  or 

v7  +  roo  =  o. 
If  we  eliminate  ~v  and  oa  by  means  of  this  equation,  we  have  then  for  the  time  after  which  slipping  ceases 


_ 

~ 


At  this  instant  there  is  no  tendency  to  slip  and  n  becomes  zero,  and  hence  at  this  instant  /=  o  and 
a.  —  o.     Hence  after  the  time  /  the  linear  and  angular  velocities  are  constant  and  given  by 


If  rvt  —  K'OJ  is  negative,  v  is  negative.    Hence  if  oot  is  positive  and  greater  than  ^r.the  translation  of  the 

hoop  after  the  lime  /  will  be  in  the  opposite  direction  to  the  initial  translation. 

(8)  A  homogeneous  beam  is  supported  horizontally  on  two  supports.  Find  where  one  of  them  must  be 
placed  in  order  that  when  the  other  is  removed  the  instantaneous  force  exerted  on  the  former  may  be  equal  to 
half  the  weight  of  the  beam. 

ANS.   Let  d  be  the  required  distance  of  the  permanent  support  from  the  centre  of  the  beam,  K  the  radius 
R      of  gyration  of  the  beam  about  a  normal  axis  through  the  centre,  m  the  mass  of 
the  beam,   a   its  angular   acceleration,   and    R  the  reaction  of  the  permanent 
'         d        1        support  immediately  after  the  removal  of  the  other.     Then 


R  —  r 
mg 

and  for  the  centre  of  the  beam 


Rd,     or     a  = 
For  the  end  of  the  beam 

Equating  these  two  values  of  a,  we  have  d  =  K. 


374  KINETICS  OF  A   MATERIAL  SYSTEM.  [CHAP.  VI. 

(9)  A  homogeneous  circular  cylinder  of  radius  r,  radius  of  gyration  about  the  axis  K,  rotating  about  its 
axis  with  angular  velocity   cut  ,  is  placed  with  its  a.tis  horizontal  on  a  rough  inclined  plane  {coefficient  of 
friction  u,  inclination  0,  so  that  n  =  tan  0),  the  direction  of  rotation  being  that  which  it  would  have  if  the 
cylinder  were  rolling  without  sliding  up  the  plane.     Show  ///<«/  the  axis  <>/  the  cylinder  will  be  stationary  for  a 

time  t  =  ^-j-  ,  at  the  end  of  which  the  angular  velocity  will  be  zero. 

(10)  A  uniform  square  is  supported  in  a  vertical  plane  with  one  diagonal  horizontal  by  two  supports,  one 
at  each  extremity  of  the  diagonal.      Show  that  the  initial  force  on  one  support  when  the  other  is  removed  is 
equal  to  one  fourth  of  the  weight  of  the  square. 

(11)  A  uniform  horizontal  bar,  suspended  from  any  two  points  in  its  length  by  two  parallel  cords,  is  at 
rtst.     If  one  of  the  cords  be  cut,  find  the  initial  tension  in  the  other. 

wr 

ANS.    T  —jr wi    where  /  is  the  length  of  the  bar,  d  the  distance  from  its  centre  of  mass  to  the 

point  of  attachment  of  the  uncut  cord,  and  W  is  the  weight  of  the  bar. 

(ip)  A  uniform  beam  of  weight  IV  rests  with  one  end  against  a  smooth  vertical  wall  and  the  other  on  a 
smooth  horizontal  plane,  the  inclination  to  the  horizon  being  0.  //  is  prevented  from  falling  by  a  string 
attached  to  its  lower  end  and  to  the  wall.  Find  the  force  between  the  upper  end  and  the  wall  when  the  string 
is  cut. 

ANS.  —  W  cot  9. 

(13)  A  sphere  is  laid  upon  a  rough  inclined  plane  of  inclination  8.      Show  that  it  will  not  slide  if  the 
coefficient  of  friction  is  equal  to  or  greater  than  -    tan  0. 

(14)  A  sphere  of  radius  r  whose  centre  of  mass  is  not  at  its  centre  of  figure  is  placed  on  a  rough  horizontal 
plane,  coefficient  of  friction  H.     Find  whether  tt  will  slide  or  roll. 

ANS.  Let  K"  be  the  radius  of  gyration  of  the  sphere  about  the  line  through  the  point  of  contact  at  right 
angles  to  the  plane  of  the  centres  of  figure  and  mass.  Then  if  the  initial  distance  of  the  centre  of  mass 

from  a  vertical  through  the  centre  of  figure  is  greater  than  - —  ,  it  will 'begin  to  slide;   if  less,  to  roll. 


CHAPTER   VII. 

ROTATION   ABOUT   A   FIXED    POINT. 

Effective  Forces—  Rotation  about  a  Fixed  Point.  —  Equations  (5),  page  159,  give  the 
components  of  the  acceleration  of  any  particle  of  a  rotating  and  translating  body. 

For  rotation  only  we  should  omit  all  terms  containing  fx  ,  fy  ;  fz.  If  we  make  these 
changes  in  equations  (5),  page  159,  multiply  each  term  by  the  mass  m  of  the  particle  and 
sum  up  for  all  the  particles,  we  have,  since  2mx  =  o,  2  my  =  o,  2mz  =  o,  for  the  compo- 
nent effective  forces  for  a  rotating  body  of  mass  m  =  2m  for  any  co-ordinate  axes  we  please 
through  the  fixed  point  O' 


—  1&}>Goxa)y  -f- 
=  mJ  coycoz  -\- 

—  mxGozoax  -\-  mj7  oozooy  —  rascal  —  TS^ZGO*  -\-  myax  — 

If  any  one  of  the  co-ordinate  axes  is  fixed,  as,  for  instance,  O'  Z'  we  have  oozos  =  o, 
GOZGOX  —  o,  oaxojy  =  o,  ooyoo,  =  o.  For  O'  Y'  fixed  we  have  o^yoox  =  o,  ooyco,  =  o,  cox&)z  =  o, 
aozGox  =  o.  For  O'X  '  fixed  we  have  Goxcoy  =  o,  ooxcoz  =  o,  GoyK>z  =  o,  (jozwy  =  o.  For  all  the 
co-ordinate  axes  fixed  we  have  equations  (3),  page  316,  for  a  fixed  axis. 

If  the  fixed  point  is  at  the  centre  of  mass,  we  have  ~x  =o,  ~y  =  o,  J  =  o,  and  hence 
~2mfx  =  o,  ^infy  =  o,  2mfz  =•  o.  That  is,  if  the  centre  of  mass  is  fixed,  the  components  of 
the  effective  forces  are  zero. 

If  we  take  distance  in  feet  and  mass  in  Ibs.,  equations  (i)  give  force  in  poundals.  For 
force  in  pounds  divide  by  g  (page  I/1)- 

Moment  of  Effective  Forces—  Rotation  about  a  Fixed  Point.  —  Equations  (10),  page 
160,  give  the  component  moments  of  the  acceleration  of  any  particle  of  a  rotating  and 
translating  body. 

For  rotation  only  we  should  omit  all  terms  containing  fx,  fy  ,  fx  ,  and  put  x'  ',  y'  ,  z'  in 
place  of  ~x  -\-  x,~y  -\-y,~z  -\-  z.  If  we  make  these  changes  in  equations  (10),  page  160, 
multiply  each  term  by  the  mass  m  of  the  particle  and  sum  up  for  all  the  particles,  we  obtain 
the  component  moments  of  the  effective  forces.  In  making  the  summation  we  have 


where  I'x  ,  /,','/,'  are  the  moments  of  inertia  of  the   body  for  the   co-ordinate   axes   O'X', 
O'Y',  O'Z'.      We  also  have 


2mx>z  =  I'yz  ,          2  my*  =  !'„  ,  2mz*  =  I'xy  , 

where  7xy  ,  fy't,  I'zx  are  the  moments  of  inertia  of  the  body  for  the   co-ordinate  planes  X'Y', 
Y'Z',  Z'X', 

375 


—  cotwy2mx'y'  —  ct^mz'y'—  ot^mx'y'  -f  (o^  — 


376  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  VII. 

We  have  then 


(2) 


Since  we  can  take  any  co-ordinate  axes  we  please,  let  us  take  them  principal  axes  at 
the  fixed  point  £X.  Then  we  have  "Stnx'y1  =  o,  ^nty'z'  =  o,  2mz'x'  =  o,  and  equations  (2) 
become 

M'x     =     l'tXOOtOOy       —      I'xyWyCOt      +      l'X   (X  X     , 


M'ft  = 


If  any  co-ordinate  axis  is  fixed,  as,  for  instance,  O'Z',  we  have  GO,C*)X  —  o,  oot<ay  =  o, 
coxcoy  =  o,  oaycax  =  o.  If  O'  Y'  is  fixed,  we  have  ooy(ax  =  o,  ooyoof  =  o,  ooxooI  =  o,  6L>zo?v  =  o. 
If  O'X'  is  fixed,  we  have  a?,oa>,  =  o,  ^G?,  =  o,  cayo?,  =  o,  &)zcoy  =  o.  If  all  are  fixed, 
equations  (2  and  (3)  become  equations  (4)  and  (6),  page  317,  for  fixed  axis. 

If  the  co-ordinate  axes  are  not  fixed,  we  have  ootoay  —  ooyoo^  ooxcof  =oaf  cox,  ooycox  —  ooxcoy', 
and  since  we  have 


equations  (3)  become 

Mf,  =  (f.  ~  fya>.<»,  +  7>,'  I 

Mf,  =  (I'x  ~  Ow,w.  +  />,.    |      .......      (4) 

M,.  =  (/;  -  /;)«,«,  +  /;«..  j 

If  we  take  distance  in  feet  and  mass  in  Ibs.  ,  all  these  equations  give  moments  in  poundal- 
feet.  For  pound-feet  divide  by^  (page  171). 

Momentum—  Rotation  about  Fixed  Point.  —  From  equations  (2),  page  154,  we  have  the 
component  velocities  for  any  particle  of  a  rotating  body, 

vx  =  (z  +  z)wy  —  (J 


If  we  multiply  by  the  mass  m  of  the  particle  and   sum  up,  we  have,  since  2mx  =  o, 
my  =  O,  2mz  =  o  for  the  components  of  the  momentum 


,  =  mxao.-m.3Go,,          .....     ...     (5) 


For    axis    through    the    centre    of    mass    we    have  *  =  o,  J  =  o,  ~z  =  o,    and     hence 
x  =  o,   2£fnvy  =  o,  2mvt  =  o. 
Hence  the  momentum  for  a  body  rotating  about  a  fixed  point  is  the  same  as  for  a 
particle  of  equal  mass  at  the  centre  of  mass. 


CHAP.  VII.]  MOMENT  OF  MOMENTUM  -ROTATION  ABOUT  FIXED  POINT.  377 

Moment  of  Momentum  —Rotation  about  Fixed  Point.  —  Equations  (10),  page  155,  give 
the  component  moments  of  velocity  for  any  particle  of  a  rotating  and  translating  body.  If 
the  origin  O'  is  a  fixed  point,  we  should  omit  terms  containing  vxt  vy,  vz,  and  x,  J7,  J,  and 
put  .*•',  y'  r  z'  in  place  of  ^r,  y,  2. 

If  we  make  these  changes  in  equations  (10),  page  155,  multiply  each  term  by  the  mass  m 
<>f  the  particle  and  sum  up  for  all  the  particles,  we  shall  obtain  the  component  moments  of 
momentum  for  any  co-ordinate  axes  we  please. 

Let  us  take  these  axes  as  principal  axes  at  O'  '  .  Then  we  have  ~2mx'y'  =  o,  ~2my'z'  =  o, 
=  o.  We  have  then  from  equations  (10),  page  155,  since  2m(y*  -f-  z'2)  =  Ix, 


JC  =  I'xoox  ,         M'Ty  =  fycoy  ,          M'vz  =  fgCo,  .......     (6) 

Pressure  on  Fixed  Point.  —  Let  the  component  pressures  at  the  fixed  point  be 
Rx,  Ry,  Rz.  Let  the  components  of  all  other  impressed  forces  be  2FX  ,  2Fy  ,  2FZ.  We 
have  then  from  equations  (i),  by  D'Alembert's  principle, 


Rx  -j-  2FX  —  myGoxGoy  -f-  lR-ZGaxGox  —  m.xc0*  —  IR.XGO?  -\-  Viz  ay  —  mj7a-z 

Ry  -j-  2Fy  —  T&zK}ycds  -\-  mx&)yojx  —  myoa?  —  mjo?^2  -{-mxax  —  mFax  ,   }-      .     (7) 

R,  -j-  2FZ  =  m.x(*)z  GJX  -\-  mJ/Ci?,  ojy  —  m^  ojx2  —  TCLZ  oof  -f-  VLyax  —  ?&xay 


Equations  (7)  give  Rx1   Ry,   R,.        If  the  centre   of  mass  is  the   fixed  point,  we  have 
~x  =  o,  y  =  o,   J  =  o,  and  hence 


that  is,  when  the  centre  of  mass  is  fixed,  the  pressure  on  the  fixed  point  is  the  same  as  if 
there  were  no  rotation,  and  is  found  by  the  conditions  of  static  equilibrium. 

Conservation  of  Moment  of  Momentum—  Rotation  about  Fixed  Point.  —  We  have  from 
equations  (6)  for  the  component  moments  of  momentum  for  a  body  rotating  about  a  fixed 
point  O't  taking  the  co-ordinate  axes  as  principal  axes  at  O', 

M^x=Ixcox,          M'vy=Iy'(ay,          -0C  =  /,'<»,,      ......      (a) 

and  from  equations  (4)  for  the  component  moments  of  the  effective  forces 


If  in  equations  (ff)  the  co-ordinate  axes  O'  X'  ,  O'  Y'  ,  O'  Z'  are  fixed  (figure,  page  153),  we 
have  GO,CO  =  o,  coxoo,  =  o,  GO  oox  =  O,  and  the  axis  of  rotation  is  fixed.  If  also  we  have 
ax  =  ot  a  =  o,  otz  =  o,  we  have  Mfx  =  o,  Mfy  =  o,  Mft  =  o,  and  cox,  COY,  ooz  in  equations  (a) 
are  constant.  But  since,  by  D'Alembert's  principle,  the  moment  of  the  effective  forces  is 
equal  to  the  moment  of  the  impressed  forces,  we  have  the  moment  of  the  impressed  forces 
zero. 


3?8  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAI-.  \  II. 

Hence  if  the  moment  of  the  impressed  forces  about  any  fixed  axis  through  the  fixed  point 
is  always  zero,  the  moment  of  momentum  of  the  body  about  that  axis  is  constant,  and  if  the 
moment  of  inertia  about  that  axis  does  not  change,  then  the  angular  velocity  about  that  axis  is 
also  constant. 

Invariable  Axis. — If  the  moment  of  the  impressed  forces  relative  to  the  fixed  point  O' 
is  always  zero,  the  resultant  must  either  be  zero  or  always  pass  through  O'.  In  such  case 
the  co-ordinate  axes  do  not  change  in  direction,  and  we  have  <»,<#,  =  O,  coxoat  =  O,  coycox  =  O, 
and  also  Jlff(  =  o,  Mfy  =  o,  Mfl  =  O.  Hence,  from  equations  (b),  ax  =  O,  a,  =  O,  a,  =  O,  and 
cox,  &?y,  «.  in  equations  (a)  are  constant. 

We  have  then  the  moment  of  momentum 


//  trT*      '      i      T1'*      '     i       /"'*      ' 

CO  =  V  /,  COX      +    I,   G0y     +    /,    GO,         ........ 

about  an  axis  through  the  fixed  point  whose  direction  cosines  are 

Ix'oox  a       I'cOy  7/cw, 


and  this  axis  is  then  invariable  in  direction. 

Hence  when  the  moment  of  the  impressed  forces  relative  to  the  fixed  point  is  always  zero, 
or  if  there  are  no  impressed  forces,  and  a  body  rotates  about  any  axis  through  the  fixed  point 
at  any  instant,  there  will  be  a  certain  axis  through  the  fixed  point  for  which  the  moment  of 
momentum  is  constant.  This  axis  is  invariable  in  direction,  and  its  direction  cosines  are  given 
by  equations  (//).  The  moment  of  momentum  about  this  axis  is  given  by  equations  (c). 

Kinetic  Energy—  Rotation  about  a  Fixed  Point.—  The  kinetic  energy  of  a  particle  of 

mass  m  and  velocity  v  is  —mvz.  If  a  particle  has  the  component  velocities  VM,  vv,  VM,  we 
have  v2  =  vx  -\-  vy*  -j-  v*  and 

I  i  .    I        ,    .    I 

-mir  —  ~mv*  +  -mv*  -\--mv?. 

From  page  154  we  have  for  the  component  velocities  for  any  particle  of  a  rotating 
body 

vx  =  z'coy  —  y'oo,  ,          vy  —  X'GO,  —  z'oox  ,          vt  =  y'oo,  —  x'ooy  , 

where  x  '  ,  y'  ,  z  are  to  be  taken  from  the  fixed  point  O'  as  origin.  If  we  square  these  com- 
ponent velocities,  multiply  each  term  by  —m,  sum  up  for  all  the  particles  and  add,  we  shall 

have  the  kinetic  energy  for  a  rotating  body  for  any  co-ordinate  axes  we  please.  Let  us 
take  these  axes  as  principal  axes  at  O'.  Then  we  have  2mxy  =  o,  2myz  =  o,  ~2ntzx  =  o. 
We  have  then,  since  2m(y*  +  z>*}  =  /,',  2m(zlZ  -f  x*)  =  //,  ^m(x»  +  y*}  =  //,  for  the 
kinetic  energy 

5C  =  ^/c,?,2+^/;aJ/  +  ^/>,2.    ..........     (7) 

If  we  take  mass  in  Ibs.,  this  equation  gives  kinetic  energy  in  foot-poundals.  For  foot- 
pound divide  by  g. 


CHAP.  VII.] 


ROTATION  ABOUT  A  FIXED  POINT— EXAMPLES. 


379 


9 


Examples.— (i)  A  circular  disc  supported  at  its  centre  of  mass  Oj'otates  about  the  principal  axis  OZ  at 
right  angles  to  the  plane  of  the  disc  with  an  angular  "velocity  ooz  =  ^/j  radians  per  second.  The  plane  of  the 
disc  makes  an  angle  9,  =  jo°  with  the  horizontal.  If  now  an  angular 
•velocity  O0y  =  i  radian  per  second  is  given  to  the  disc  about  a  principal 
axis  O  Y  in  the  plane  of,  the  disc,  find  the  motion. 

A\S.    From  equations  (4),  page  376,  we  have  for  the  component 
momt.its  of  the  effective  forces 

Mfx    •=.    (7Z    —    Iy)  GUfOOy    +  Ix&X  , 

Mfy  —  (Ix  —  7,)*»je»,  +  lyOCy , 
Mfz  =  (Iy-Ix)<»y«>x  +  Izaz. 

By  D'Alembert's  principle  the  moments  of  the  effective  forces 
are  equal  to  the  moments  of  the  impressed  forces.  In  the  present 
case  the  only  impressed  force  is  the  weight  mg  acting  at  O  and  the 
equnl  ;md  opposite  reaction  R  of  the  support  at  O.  The  component 
moments  of  the  impressed  forces  are  therefore  zerc,  and  hence 
Jlffx  —  o,  Mfy  =  o,  J\ffi  =  o.  We  have  also  a>x  =  o  and  ax  =  o,  ay  =  o,  nrz— o,  and  the  angular  velocities 
ooy,  &?z  are  therefore  constant. 

By  the  principle  of  page  378,  we  have  in  this  case  an  invariable  axis  of  rotation  OZi  fixed  in  space  for 
which  the  moment  of  momentum  is  constant. 

Let  7i  be  the  moment  of  inertia  for  this  invariable  axis,  and  -j-   be  the  angular  velocity  about  it,  so  that 
7i  -J-  is  the  moment  of  momentum.     Since  72  —  27,  for  the  disc,  we  have 

/t^=  I/AW +  //"/  = 
If  0  is  the  angle  Z<9Zt,  we  have 


(0 


OOy 


From  page  35, 
and  therefore,  from  (i)  and  (2), 


..  ^V>       y  4*»a2  +  &>y*  T  ^       V 

1  dt  dt 

7,  =  Iz  cos2  9  +  7^  sin1  9  =  Ty(\  -+  cos"  9). 


(2) 


(3) 


(4) 


dt  i  +  cos2  9 

Equations  (2)  give  the  angle  ZOZ\.  of  OZ  with  the  invariable  axis  OZ\,  and  the  motion  of  the  disc  is  the 
same  as  if  the  axis  OZ  fixed  to  the  disc  were  to  rotate  about  OZi,  always  making  the  angle  9  with  it,  with 

the  angular  velocity  -j~  given  by  (4). 

Inserting  numerical  values,  we  have 

cos  9  =  ?    _  .  =  0.95958,      or     9  =  1 6"  21', 
1/13 

j-  =  ^—  =1.87  radians  per  sec. 
dt  25 

2D  SOLUTION. — From  page  168  we  have  Euler's  geometric  equations, 

o  =  GOX  =  — —  sin  0  —  —  sin  9  cos  0, 
dt  dt 

dj,  =—  cos  <t>  -H  -77  sin  9  sin  <f>, 


dt 


KINETICS  OF  A  MATERIAL  SYSTEM. 


[CHAP.  VII. 


where  —   is  the  angular  velocity  about  OZ\,   -j-  the  angular  velocity  of  OZ  about  the  line  of  nodes  ON  (see 
figure,  page  167),   -%-  the  angular  velocity  of  OY  relative  to  ON. 

From  the  first  two  of  these  equations,  eliminating  -r-,  we  obtain 


,  sin 
sin  0 


=    <By  COS 


dt  -       sin  9    '  dt   ~  dt 

We  have  also  the  moment  of  momentum  for  the  invariable  axis, 


<«>  sin   <f>  cos  Q 
sin  6 


and,  as  before, 


sin  6  = 


We  have  then    sin  <f>  =         CQ^  Q,    and 


</9         ooy  cos  6 


«v ^> !!1  _    ">  wo  "    4/,    ,    ___»  „ 

dt       sin  9(i  +  cos'  0)  '  dt  ~  i  +  cos'9  V 

or,  putting  in  the  values  of  sin  9,  cos  9, 

>,*  4-  <a}?)\  dQ  2,<of<oy  ^4.03^ 


dt 


/,  =  7j,(i?+  cos1  9). 


d(f> 
~dt 


ooy  cos  9 


sin  6(1  +  cos3  &) 


dt  -  8oV  +   ^ 

Inserting  numerical  values,  we  have  9  =  16°  21',  —^  =  1.87  radians  per  sec.,  -j  =  3.08  radians  per  sec., 

dd> 

—   =  —  0.07  radians  per  sec. 
tit 

(2)  Let  a  vertical  disc  of  mass  m  =  40  Ibs.  and  radius  r  =  3  ft.  roll  on  a  horizontal  plane  with  angular 
velocity  toy  =  -f  2  radians  per  sec.,  while  its  centre  O  describes  a  horizontal  circle  of  radius  O'O  =  y  =  g  feet 
with  angular  velocity  ea»  about  a  fixed  vertical  axis  O1  Z' .  At  each  extremity  of  a  diameter  ab  let  a  mass 
m  =  10  Ibs.  be  attached.  Find  the  pressure  on  the  horizontal  plane  when  these  masses  are  in  a  vertical  and  in 
a  horizontal  line.  Take g  =  32  ft.-per-sec.  per  sec. 

ANS.  Let  O  be  the  origin,  OZ'  the  fixed  axis,  and  take  <oy  in 
the  direction  O1  Y'.     The  axes  OX'.  Cf  Y',  a  Z'  are  principal  axes. 
We  have  a>x  =  o  and  crx  =  o,  cty  =  o.    ft*  =  o.     Also,  since  O'Z1  is 
fixed,  oflyoa,  =  o,  03,09,  =  o,  oof(ox  =  o,  oa,ooy  =  o. 
From  equations  (3),  page  376,  we  have  then 


and  since  the  disc  rolls, 

rooy  =  —  J( 


or     o>,  = = 

y 


Now  I'xy ,  or  the  moment  of  inertia  relative  to  the  plane  X1  Y',  is  the  same  as  the  moment  of  inertia  for 
the  horizontal  axis  cd  through  the  centre  of  mass.     Hence  if  9  is  the  angle  of  Oa  with  the  vertical, 

mr1 
/«',  =   +  2tnr*  cos*  9. 

Let  R  be  the  vertical  pressure  of  the  horizontal  plane  at  b.     Then,  by  D'Alembert's  principle, 
Ry~  -(m  +  im)sy  =  Mfx  =  -f  —  +  2mr* cos" Q\<oy<o,. 


CHAP.  VII.]  ROTATION  ABOUT  A  FIXED  POINT— EXAMPLES.  381 

Substituting  the  value  of  <*>,,  we  have 

/?  =  (m  +  2m)g  +  ~>-Y^!  +  vmr*  cos8  0 

y   \  4 

If  we  take  distance  in  feet  and  mass  in  Ibs.,  this  equation  gives  R  in  poundals.      For  R  in  pounds  we 
divide  by g  and  obtain 


R  =  m  +  2m  +  -=%  ( +  2mr*  cos"  91. 


(0 


We  see  from  (i)  that  the  pressure  R  is  greater  than  the  total  weight  m  +  2m.  The  greatest  possible 
pressure  for  given  r  and  y  is  when  cos  9  =  i  or  0  =  o,  that  is,  when  the  masses  m  are  at  a  and  b  in  the  figure. 
The  least  possible  pressure  is  when  cos  9  =  o,  or  9  =  90°,  that  is,  when  the  masses  m  are  at  c  and  d  in  the 
figure. 

In  the  first  case,  when  the  masses  m  are  in  the  vertical  line  ab,  we  have  9  =  o  and 

R  =  60  H 2(90  +  180)  =  61.25  pounds. 

In  the  second  case,  when  the  masses  are  in  the  horizontal  line  cd,  we  have  9  =  90  and 

R  —  60  +  ^~-  =  60.4166  pounds. 
If  the  masses  m  are  removed,  we  have  in  all  positions 

R  =  60  +  ^-^  =  60.4166  pounds. 

We  see,  then,  that  for  a  disc  rolling  as  in  the  example  the  pressure  on  the  horizontal  plane  is  greater 
than  the  weight  of  the  disc. 

(3)  In  the  preceding  example  let  the  horizontal  plane  on  which  the  disc  rolls  be  above  instead  of  below  the 
disc.     Find  the  force  R  necessary  to  keep  the  disc  in  contact  with 
the  plant. 

ANS.  In  this  case  we  have 


rooy  =  y  <*>* ,    or    ooz  =  +  -=*, 


and,  as  before, 


/'„  = 


cos*  9, 


and 


Mf*  =    —  I'xyVy00*' 

We  have  then,  as  before, 


Ry  —  (m  +  -zm  *)gy  =  M/x  =  —  (—  +  2wr2  cos"  9  )ay», 

\  4  / 


Substituting  the  value  of  &>,,  we  have 
or  for  R  in  pounds 


R  =  (m  +  2m)g  -  ^fc-  +  2mr*  cos*  9J, 


_l^.pl   +  2;«r'cos'e 


We  see  in  this  case  that  the  force  R  is  less  than  the  total  weight  (m  +  2m). 
If  the  masses  m  are  vertical  at  a  and  b,  we  have  9  =  o  and 

R  =  60  -  ~  (90  +  180)  =  58.75  pounds. 


382  KINETICS  OF  A  MATERIAL   SYSTEM. 

If  the  masses  m  are  horizontal  at  c  and  d,  we  have  0  =  90  and 


[CHAP.  VII. 


If  in  (i)  we  put  /?  =  o,  we  have  for  the  value  of  oa,  for  *rhich  no  force  K  will  be  requirtd 


+  2wr*  cos1  6  , 


(2) 


If  the  masses  m  are  removed,  we 


19.6  radians  per  sec. 


If  the  disc  rolls  with  this  angular  velocity,  R  will  be  zero,  that  is,  the  disc  is  self-supporting  and  will  roll 
round  on  the  horizontal  plane  abov<^  i*  without  any  force  R  being  necessary  to  keep  it  in  contact. 

If  ooy  is  greater  than  this,  the  disc  will  press  against  the  horizontal  plane  above  it.     If  o^  is  less  than 
this,  a  force  R  will  be  necessary  to  keep  the  disc  in  position. 
(4)  Discuss  the  action  of  the  Gyroscope  and  Spinning  Top. 

DESCRIPTION.  —  The  gyroscope  consists  of  a  disc  aa  which  is  set  in  rotation  about  an  axis  in  the  direc- 
tion of  the  diameter  of  the  ring  A1.     This  ring  is  attached  to  the  rod  S,  which  passes  through  a  sleeve  at  h. 

This  sleeve  is  pivoted  in  the  fork^  so  that  S  can  rotate  in  a 
vertical  plane,  and  the  fork^  is  pivoted  at/  so  that  this  rod 
can  rotate  horizontally.  A  sliding  counterweight  G  can  be 
so  adjusted  that  the  centre  of  mass  of  the  apparatus  can  be 
made  to  lie  on  the  same  side  of  the  standard  as  the  disc,  on 
the  opposite  side,  or  directly  over  the  standard. 

SOLUTION.—  Let  O'Z'  be  the  axis  of  the  disc,  <a,  the 
angular  velocity  about  this  axis,  the  rotation  being  clockwise 
as  we  look  from  O'  to  C,  the  centre  of  the  disc.  Take  the 
co-ordinate  axes  O'X,  O'Y',  O'Z'  principal  axes  at  the  fixed 
point  O1. 

Let  the  counterweight  be  adjusted  so  that  the  centre  of 
mass  O  is  on  the  same  side  of  the  standard  as  the  disc.  Let 
m  be  the  mass  of  the  apparatus,  and  mg  its  weight  acting  at 
the  centre  of  mass  O.  The  impressed  forces  are  then  fag 
and  the  reaction  A*  of  the  standard  at  (7.  These  always  pass 

through  *he  axis  O'Z'.  Hence  the  moment  about  the  axis  O'Z'  of  tbe  impressed  forces  is  always  zero,  and 
by  D'Alembert's  principle  we  have  M/t  =  o,  or  the  moment  of  the  effective  forces  about  O'Z'  also  zero. 
From  equations  (4),  page  376,  we  have  then  V 

Mf*  =  o  =  (//  —  Ix')a>y<ox  +  7,'a,. 

In  the  case  of  a  circular  disc  //  is  equal  to  fx',  and 
hence  7»'a»  =  o,  or  a,  =  o.  Hence  the  angular  velocity  <», 
about  the  axis  O  '  Z'  is  constant  whatever  the  posjtion  of 
the  axis. 

Let  the  axis  O'Coi  the  disc  make  the  initial  angle  6, 
with  the  vertical  O  '  V,  and  the  disc  be  allowed  to  fall  under 
the  action  of  gravity  to  the  position  in  the  figure  where  the 
axis  O'C  makes  the  angle  0  with  the  vertical.  The  angu- 
lar acceleration  ay  will  be  always  at  right  angles  to  the  axis 
O'Coi  the  disc  in  the  direction  shown  in  the  figure  if  O  falls, 
in  the  opposite  direction  if  O  rises.  Hence  the  axis  O'C 
rotates  about  O"  V  clockwise  as  we  look  from  O"  to  V  \i  O 
falls  so  that  the  angular  velocity  <»„  about  O*  V  is  as  represented  in  the  figure. 

Just  as  in  the  preceding  example,  then,  there  is  a  certain  value  of  <»„  for  which  the  disc  is  self-supporting. 
and  when  that  value  is  obtained  there  is  no  farther  fall. 


CHAP.  VII.  ]  ROTATION  ABOUT  A  FIXED  POMT—  EXAMPLES.  383 

If  the  counterweight  is  so  adjusted  that  the  centre  of  mass  O  is  on  the  opposite  side  of  the  standard 
from  the  disc,  the  axis  O'C  o(  the  disc  will  move  upwards,  the  angular  acceleration  a  and  angular  velocity 
cav  will  be  opposite  in  direction  to  that  shown  in  the  figure,  and  the  axis  O'C  then  rotates  about  O1  V  in  a 
clockwise  direction  as  we  look  from  V  to  O1. 

If  the  counterweight  is  so  adjusted  that  the  centre  of  mass  is  over  the  standard,  the  axis  O'C  of  the  disc 
neither  rises  nor  falls,  there  is  no  angular  acceleration  a  and  the  axis  O'C  does  not  turn  about  O1  V. 

Let  the  centre  of  mass  O  be  as  in  the  figure,  and  suppose  the  axis  O'C  of  the  disc  has  fallen  from  the 
initial  angle  ffi  till  it  makes  the  angle  6  with  O'  V. 

Take  the  co-ordinate  axes  O  'X'  ',  <7  V,  O'Z'  fixed  in  the  body,  and  let  the  angular  velocities  about  these 
axes  be  oox,  K)y  and  oof. 

The  moment  of  the  impressed  forces  m^  and  R  is  always  zero  about  the  fixed  axis  O1  V,  and  by  conser- 
vation of  momentum  (page  390)  the  moment  of  momentum  about  this  axis  is  constant  for  all  positions  of 
the  body. 

In  the  initial  position  the  moment  of  momentum  about  O1  V  is  fz'oot  cos  6j  ,  since  oox,  <oy  for  this  position 
are  zero.  In  the  final  position  the  moment  of  momentum  about  O1  V  is 

I,'ms  cos  6  +  7/ca,  cos    VO1  Y'  +  Sx'oot  cos  VOX'. 
If  we  insert  the  values  of  the  cosines  from  equations  (5),  page  168,  we  have  then 

II'MI  cos  6,  =  I'tOOf  cos  6  +  fy'coy  sin  6  sin  0  —  /*'<»*  sin  6  cos  <f>  ........     (i) 

where  (page  167)  </>  is  the  angle  of  O'  Y'  with  the  line  of  nodes. 

The  initial  kinetic  energy  is  —  /,'&V  (page  378).  and  the  initial  potential  energy  relative  to  a  horizontal 

plane  at  a  distance  y  below  O',  if  y  is  the  distance  O'O  of  the  centre  of  mass,  is  mg(~y  +  y  cos  61)  when  O 
is  on  the  same  side  of  the  standard  as  the  disk,  and  wg(y  —  y  cos  61)  when  O  is  on  the  opposite  side  of 
the  standard  from  the  disc.  The  total  initial  energy  is  then 


,  =  -/,'<»,»  +  &g(y±y~  cos  61). 


The  final  kinetic  energy  is  (page  378) 


and  the  final  potential  energy  is  mg-(y~±  y  cos  6).     The  total  final  energy  is  then 

g  =  —  /*'  *»**  +  ~fy  <>y  +  -J*'00**  +  ™g(y~  ±  7COS  9  )' 

If  we  disregard  friction,  we  have,  by  the  principle  of  conservation  of  energy  (page  304),  §  =  Si,  or 

—/3t'aoi*  +  —I,'tDy*  =  ±  mg7(cos6i  —  cos  6),     .     .    ..     .     .     .     .     .     .     .     (2) 

where  the  (+)  sign  is  to  be  taken  for  centre  of  mass  O  on  same  side  of  standard  as  the  disc,  and  the  (-)  sign 

when  it  is  on  the  opposite  side. 

We  have  also,  from  Euler's  geometric  equations  (page  168), 


oox  =  -3-  .  sin  0  —  -j-  sin  6  cos  <p, 
fit  at 


L>V  =  —   cos  <f>  +  -r- 
y       dt  dt 

(ill)  ,         d<p 

,=--  cose  +  -_. 


.         . 

sin  o  sin  <f>, 


(3) 


384  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  VII. 


i#  'Sthe  anKular  velocity  oa,  about  W,  or  the  angular  velocity  of  precession;  -^  is  the  angular 

velocity  of  ffZ'  about  the  line  of  nodes  (see  figure,  page  167),  or  the  angular  velocity  of  nutation  ;  -j-  is  the 

angular  velocity  of  O1  Y  relative  to  the  line  of  nodes. 

Squaring  and  adding  the  first  two  of  equations  (3),  we  have 


Substituting  this  in  (2),  we  have,  since  for  the  disc  /,'  =  //, 

*'        +  /,'fLsin*9=±  imgy  (cose.-cose)  ..........     (4) 


Substituting  the  values  of  o>M,  ooy  from  (3)  in  (i),  we  have 

7*'^  sin'  G  =  ff'oo,  (cos  6,  -  cos  6)  ...........     (5) 

We  have  also,  from  the  last  of  equations  (3), 

§  +  ^cosO=«,  ...............     (6) 

Equations  (4),  (5),  (6)  are  the  differential  equations  for  the  gyroscope.    When  6  =  6,,  or  at  the  beginning 
of  motion,  we  have 


From  ,(5)  we  have  the  angular  velocity  «„  about  (7  V,  or  the  angular  velocity  of  precession, 

_  dij)  _  I*'oOf     cos  61  —  cos  0 

~  ~dt  ~    Ix'    '  sin*!  ............     (7) 

and  substituting  this  in  (4)  we  have  for  the  angular  velocity  of  CfZ'  about  the  line  of  nodes,  or  the  angular 
velocity  of  nutation, 


T'    "7"sin'6v -cos  6)1 (8) 

where  the  (  +  )  sign  is  taken  for  centre  of  mass  O  on  same  side  of  standard  as  disc,  and  the  (— )  sign  when  it 
is  on  the  opposite  side. 

From  (8),  taking  the(  +  )  sign,  we  see  that  for  the  centre  of  mass  O  on  same  side  of  standard  as  the  disc,  — 

dt 

is  imaginary  when  6  is  less  than  6|.     Taking  the  (— )  sign,  we  see  that  for  the  centre  of  mass  O  on  the  oppo- 
site side  of  standard  from  disc,  —  is  imaginary  when  6  is  greater  than  6,.     Hence  the  centre  of  mass  O 

always  falls  from  its  initial  position  and  can  never  rise  above  it. 

From  (7),  then,  if  o>,  is  positive,  that  is,  if  the  rotation  of  the  disc  when  looking  from  O1  to  Z'  is  clock- 

ji 
wise,  -j-  is  positive,  or  the  rotation  about  <7  V  looking  from  (7  to  V  is  clockwise  if  the  centre  of  mass  O  is 

on  the  same  side  of  standard  as  the  disc. 

If  O  is  on  the  opposite  side  and  oot  is   positive,   -^-  is  negative.     If  the  centre  of   mass  is  over  the 


standard,  J  —  o  and  the  angle  6,  remains  unchanged.      Hence  cos  6,  —  cos  6  =  0,     and 


—  =    ,     -3 

ill  at 


The  axis  O  '  Z'  then  remains  stationary. 

These  conclusions  can  be  verified  by  the  apparatus  (figure,  page  382)  by  shifting  the  counterweight. 

If  we  put       =  o  in  (8),  we  obtain  the  maximum  and  minimum  values   of  6.     We  have  '-  =  o  when 


CHAP.  VII.]  ROTATION  ABOUT  A  FIXED  POINT— EXAMPLES.  385 

0=0,,  and  this  is  the  minimum  value  of  9,  for  we  have  just  seen  that  O  always  falls,  and  hence  0  cannot  be 

less  than  0j.     We  also  have—-  =  o  and  0  a  maximum  when 
at 

cos  0i  —  cos  0  =  ±  zmgy  •  —^-^ — 5 — .     •••••••••••    (9) 

If  we  insert  this  value  in  (7),  we  have  for  the  maximum  value  of  on, 

max.  <»„  =  ±  ~Y^~.   . (io) 

The  minimum  value  of  oav  is  when  Q  =  0i ,  or  min.  oav  =  o.     We  see  from  (io)  that  tjie  maximum  value 
of  oov  is  independent  of  0j  and  0,  or  of  the  initial  and  final  positions  of  the  disc. 
From  (9)  we  obtain  for  the  maximum  value  of  0 

Let  us  put  fof  brevity  and  convenience  the  quantity 


Then  we  can  write  equation  (i  i) 


COS0max.  =    ±  ft*  +    -f/I  T  2/5*  COS  QI  +  ft*  ...........      (12) 

where  the  upper  signs  are  for  O  on  same  side  of  standard  as  disc,  and  the  lower  signs  for  O  on  the  opposite 
side  from  disc. 

We  see,  from  (12),  that  the  maximum  value  of  0  depends  upon  ft,  and  that  this  maximum  value  can  be 
o°  ot  1  80°,  that  is,  cos  0max.  can  be  +  i  or  —  I  only  when  ft  =  o.  But  ft  can  be  zero  only  when  oj»  =  o. 
Hence  any  velocity  of  rotation  GO,  of  the  disc,  however  small,  is  sufficient  to  prevent  the  axis  CfZ'  from 
reaching  the  vertical.  The  self-sustaining  power  of  the  gyroscope  is  thus  proved. 

From  (12)  we  have 


If  ft  or  oot  is  very  great,  cos  0i  —  cos  0max.  is  very  small.  Hence,  by  increasing  the  value  of  oot  ,  0max.  —  0, 
can  be  made  less  than  any  assignable  quantity.  This  proves  the  apparently  paradoxical  result  that  the  disc 
when  rotating  rapidly  and  set  at  any  angle  0t  with  the  vertical  does  not  visibly  rise  or  fall.  But  we  see, 

from  (7),  that  for  0  =  61  ,  -£  is  zero,  and  hence  the  disc  must  rise  or  fall  in  order  to  generate  rotation  about 
at 

Cf  V.     If  oof  is  great,  this  rise  or  fall  is,  however,  very  small  and  may  not  be  visible. 

Let  A  be  the  length  of  the  simple  pendulum  which  would  oscillate  about  QfX'  or  £7  Y'  in  the  same  time 
as  the  apparatus,  when  oox  is  zero.  Then  (page  337)  we  have 


and  equation  (8)  becomes 

sin5  0  —  =  j£f~±  sin5  0  -  2/y(cos  0,  -  cos  0)l(cos  0,  -  cos  6). (14) 

and  equation  (7)  becomes 

sina0-^  =  2/3A/£.(cosQi  —  cos  6) (15) 

and  equation  (io)  becomes 

/T 

(16) 


3&6  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  Vll. 

The  centre  of  mass  then  oscillates  up  and  down  between  the  minimum  and  maximum  values  of  0i  and 
max.  8  as  given  by  (12),  while  the  angular  velocity  of  the  centre  of  mass  about  O1  V  varies  from  -^  —  o.  when 

the  axis  is  in  its  initial  position,  to  the  maximum  value  given  by  (16). 

The  complete  solution  of  the  problem  requires  the  integration  of  the  differential  equations  (4),(  5)  and 
(6).  This  necessitates  the  use  of  elliptic  functions.  If,  however,  we  assume  that  the  velocity  of  rotation  of 
the  disc  <»,  is  very  great,  and  hence  cos  Oi  —  cos  0  max.  or  0  —  0,  very  minute,  we  may  obtain  integrals  of  (4) 
and  (5)  which  will  express  the  motion  with  all  requisite  accuracy. 

Let  us  then  assume  <o,  or  ft  very  large  and  0  —  0i  very  small,  the  centre  of  mass  O  being  on  the  same  side 
of  (7  as  the  disc. 

Let  0  —  Oi  =  u,  or 

0  =  0i  +  u,        dB  =  du, 
where  u  is  a  very  small  angle. 

Then  we  have 

sin  0  =  sin  0i  cos  u  +  cos  0,  sin  u, 

cos  0  =  cos  0,  cos  u  —  sin  0t  sin  u. 
Also  by  series,  since  u  is  a  very  small  angle,  neglecting  higher  powers  of  u  than  the  square, 

«* 
sm  u  =  u,        cos  u  =  i . 

2 

Substituting,  we  have 

/        «*\ 
sin  0  =  sin  Oil  i I  +  u  cos  0i, 

cos  0  =  cos  0i  ( i ]  —  u  sin  0i. 


Hence,  neglecting  higher  powers  of  u  than  the  square, 

sin2  6  =  sin8  6,  —  «*  sin"  61  +  u*  cos*  0,  +  2«  sin  0i  cos  0i, 


and  therefore 


cos  0i  —  cos  0  =  u  sin  0,  H — u9  cos  0U       ,    (17) 


u  sin  81  H — u*  cos  0i 
cos  0.  -  cos  0  = _     (lg) 

sin"  0  sin*  0,  +  2u  sin  0i  cos  0»  +  u*  cos*  0,  —  «*  sin*  0i 

(cos  Q!  —  cos  6)* tt*  sin*  Q! f 

sin*! ~~  sin*  0,  +  2«  sin  0,  cos  0i  +  w*  cos*  0,  —  w*  sin*  6,  ~  * ('9' 


Inserting  (17)  and  (19)  in  (14),  we  obtain 


0,  +  «*  (cosO,  —  4/J*)' 
Since  ft  or  ao,  has  been  assumed  very  great,  cos  Oi  may  be  neglected  in  comparison  with  4/8*,  and  we  have 

du 
.  dt  =  -  =  —~  •  —  >  —  ;  —  ........    (20) 

4/2*  sin  9,  -  4/JV 


integrating,  since  when  /  =  o,  u  =  0  —  0,  =  o, 


versin-'         _  .  „  cos 

2ft  sin  0,  2ft 


'f,  _  4/M 
\         sin  fi,/ 


Hence 


CHAP.  VII.]  ROTATION  ABOUT  A  FIXED  POINT—  EXAMPLES. 

or,  since  cos  2A  =  i  —  2  sin*  A, 

u  =       sin  9l  si 


387 


(23) 


We  have  from  (18),  neglecting  the  square  as  well  as  higher  powers  of  u  (which  may  be  done  without 
sensible  error  owing  to  the  minuteness  of  u,  though  it  could  not  be  done  in  the  foregoing  values  of  dt  and  /, 
since  /?*  is  great  when  u  is  small), 


COS  0]   —   COS 


u  sin  0i 


sin"  0  sin*  0,  +  2»  sin  0,  cos  0/ 

The  greatest  possible  value  of  sin  0i  cos  0i  is  for  0,  =  45°,  or 

sin  0i  cos  0j  =  — . 

2 

Since  u  is  very  small,  we  have  then,  neglecting  2«  sin  0,  cos  0t, 

cos  0i  —  cos  0          u 
and  substituting  in  (15),  we  obtain 


sin*  0  sin  0t ' 


dt  r    A     sin  0i 

Inserting  the  value  of  u  from  (23),  we  have 


Integrating,  since  dty  —  o  when  /  =  o,  we  obtain 


Equations  (23),  (24),  (25)  give  with  all  requisite  accuracy  the  vertical  angular  depression  of  OZ'  ,  w  =  0  —  0,, 
the  horizontal  angular  velocity  —  ,  and  the  horizontal  angle  $  at  the  end  of  any  time  /,  provided  oox  is  very 

great. 

Referring  to  (20),  we  see  that  //  is  the  differential  equation  of  a  cycloid  generated  by  a  circle  whose  angular 

.  sin  0! 
diameter  is  —  -^j-  (page  140). 

Mf 

When,  starting  from  /  =  o  (and  therefore  u  =  o,  -j-  and  ^  =  o),  u  has  its  greatest  value,  we  have,  from 
(21),  (22),  (24),  (25) 


After  the  expiration  of  the  time  /  =  ^y  —  we  have 


and  O ' Z'  has  regained  its  original  elevation  and  the  horizontal  velocity  is  zero. 

The  axis  O' Z'  then  moves  as  if  it  were  the  element  of  a  right  circular  cone 
AO'Z1 ',  the  angle  AO ' Z'  being  equal  to  «,  which  rolls  on  the  cone  ZO 'A,  the 
angle  ZO'  A  being  equal  to  fl,.  0 

(5)  A  layer  of  dust  of  uniform  depth  d,  d being  small  compared  to  the  radius  of  the  earth,  is  formed  on  the 
earth  bv  the  fall  of  meteors  reaching  the  earth  from  all  directions.  Consider  the  earth  as  a  homogeneous  sphere 
of  radius  r  and  density  ^,  and  let  8  be  the  density  of  the  layer.  Find  the  change  in  length  of  the  day. 

ANS.   Let  c»i  be  the  angular  velocity  before  and  00  after  the  layer  is  formed,  and  /i  the  moment  of  inertia 


388  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  VII. 

of  the  earth  and  /  that  of  the  layer.    Since  there  are  no  forces  in  the  system  except  the  mutual  action  of  the 
particles  by  the  principle  of  conservation  of  moment  of  momentum  (page  376),  we  have 

/,<»,  =(/,+; 


The  mass  of  the  earth  is  —A*r*.    The  moment  of  inertia  of  the  earth  is  then 
3 


5       3  '5 

The  moment  of  inertia  of  the  dust-layer  is 


Hence 

/  _  S(r  +  <*)•-  8r* 


Expanding,  and  neglecting  —  j-  and  all  higher  powers,  we  have 


Therefore 


If  the  density  of  the  earth  is  taken  at  5.5  as  compared  to  water,  and  the  density  of  the  dust  is  taken 


at  2,  and  d  =  —  r,  we  have 
20 


10 


5.5x20 

The  length  of  the  day  in  this  case  would  be  ii  of  24  hours,  or  only  22  hours. 

12 

(6)  If  the  earth  gradually  contracted  by  radiation  of  heat,  so  as  to  be  always  similar  to  itself  as  regards 
its  physical  constitution  and  form,  show  that  when  every  radius  vector  has  contracted  an  nth  part  of  its  length 

where  —  is  a  small  fraction,  the  angular  velocity  has  increased  a  2nth  part  of  its  value. 

ANS.  Let  m  be  the  mass  of  the  earth,  r\  its  initial  and  r  its  final  radius,  and  I\  its  initial  and  /  its  final 
moment  of  inertia. 

Then 

/  —  2        *         /  —  2    r* 

~~$mi          ~~smr 

and 

/i          r\* 

But  r  =  fi  —  nri  =  r\(i  —  n).     Hence 

»=r7(r^»>=(r±nr<01' 

Expanding,  and  neglecting  »'  and  higher  powers, 

(V)  = 09,    =  (I    +  2ft)<Ui. 

I    ~"  2/1 


CHAP.  VII.] 


ROTATION  ABOUT  A  FIXED  POINT—  EXAMPLES. 


389 


(7)  If  two  railway  trains  each  of  mass  m  were  to  move  in  opposite  directions  from  the  poles  along  a 
meridian,  and  arrive  at  the  equator  at  the  same  time,  show  that  the  angular  velocity  of  the  earth  would  be 

decreased  by  ^r  of  itself,  where  E  is  the  mass  of  the  earth. 

ANS.  Let  7  be  the  moment  of  inertia  of  the  earth,  and  o>i  and  oo  the  angular  velocities  before  and  after. 
Then  /=  —  Er*,  where  r  is  the  radius  of  the  earth,  and  we  have 

—  Er 


Hence,  neglecting  -^  and  higher  powers, 


=  _E^_f 
£+$>»      \ 


x_ 


(8)  Suppose  a  mass  of  ice  m  to  melt  from  the  polar  regions  for  20  degrees  round  each  pole  to  the  extent  of 
about  a  foot  thick,  or  enough  to  give  i-faft.  over  those  areas,  which  spread  over  the  whole  globe  would  raise  the 
sea-level  by  only  some  such  undiscoverable  difference  as  three  fourths  of  an  inch.  Show  that  this  would 
diminish  the  time  of  the  earth's  rotation  by  one  tenth  of  a  second  per  year. 

ANS.  Let  0  be  the  angle  from  the  pole,  and  8  the  density  of  the  ice.     Then  the  mass  m  is 

m  =  47t$r*(  i  —  cos0), 

where  r  is  the  radius  of  the  earth.  We  have  rdB  =  ds  and  the  mass  of  a  strip  is 
iitSxds.  But  x  =  r  sin  0,  hence  the  mass  of  a  strip  is  zxSr*  sin  fl  dQ.  The  moment 
of  inertia  is  then 


7=2 


cos  0(i  -  cos  0)(i  +cos  0), 


or,  substituting  the  value  of  m, 


mr* 

1=  -  COS  0(l  +COS  0). 

If  E  is  the  mass  of  the  earth,  the  moment  of  inertia  of  the  earth  is  /= 


lEr* 


If  o>i  is  the  angular 


velocity  before  and  GO  after  melting,  we  have,  by  the  principle  of  conservation  of  moment  of  momentum, 


COS  (XI  +COS  »>».  = 


The  final  angular  velocity  GO  is  then  greater  than  the  initial,  and  the  time  of  rotation  is  diminished. 
The  difference  of  time  for  one  day  is 

21C        21t 


Substituting  numerical  values,  the  difference  of  time  is  easily  found. 


CHAPTER   VIII. 

ROTATION   AND   TRANSLATION. 

Effective  Forces  —  Rotation  and  Translation.  —  Equations  (5),  page  159,  give  the 
components  of  the  acceleration  of  any  particle  of  a  rotating  and  translating  body.  For 
origin  at  the  centre  of  mass  we  have  Ic  =  o,  y  =  o,  T  =  o.  If  we  make  these  changes, 
multiply  each  term  by  the  mass  m  of  the  particle  and  sum  up  for  all  the  particles,  we  have, 
since  2mx  =  o,  2  my  =  o,  2mM  =  o,  for  the  component  effective  forces  for  a  rotating  and 
translating  body  of  mass  m  =  2m,  for  any  co-ordinate  axes  we  please, 


2m/,  = 


where,/^,  fyt  ft  are  the  component  accelerations  of  the  centre  of  mass. 

That  is,  the  effective  force  in  any  direction  is  the  same  as  for  a  particle  of  mass  equal 
to  the  mass  of  the  body  having  the  acceleration  in  that'  direction  of  the  centre  of  mass. 

Moments  of  Effective  Forces  —  Rotation  and  Translation.  —  Equations  (10),  page  160, 
give  the  component  moments  of  the  acceleration  of  any  particle  of  a  rotating  and  translating 
body.  If  we  multiply  each  term  by  the  mass  m  of  the  particle  and  sum  up  for  all  the 
particles,  we  shall  obtain  the  component  moments  of  the  effective  forces  for  any  co-ordinate 
axes  we  please.  Let  us  take  these  axes  as  principal  axes  at  the  origin  O'  .  Then  we  have 
2mx  =  o,  2  my  =  o,  2ms  =  o,  also  2mxy  =  o,  2mys  —  o,  2//isx  =  o. 

We  also  have 


2m(  y*  +  z>)  =  fx,          2m(z*  +  **)  =  Iy  , 

where  /,,  Iy  ,  f,  are  the  moments  of  inertia  of  the  body  for  the  co-ordinate  axes  OX,  O  Y,  OZ. 
We  also  have 

2mx*  =  Iys  ,          2  my2  =  /„  ,          2,,ix3  =  Ixy  , 

where  IXJ  ,   Iyt  ,   Itx  are   the    moments    of   inertia   of   the    body  for  the   co-ordinate   planes 
XY,   YZ,  ZX. 

We  have  then  from  equations  (10),  page  160,  for  the  component  moments  of  the  effective 
forces  for  principal  co-ordinate  axes  through  the  origin  O' 


M'fx  —  mfry  —  mf^  -f-  I^GO.GO,  —  Ix,<**y<*>t  +  I,  a,  , 
X1vx<#t  —  Iyf<*>,™x  +  lyi*,- 

*™,™*  ~  ^.w,  +  f,<*.  >  J 


M'fy  = 


390 


CHAP.  VIII.]  MOMENTUM-  ROTATION  AND   TRANSLATION.  391 

where  J',  J,  ~~  are  the  co-ordinates  of  the  centre  of  mass  and  /.,.,  fy,  fz  are  the  component 
accelerations  of  the  centre  of  mass. 

If  any  co-ordinate  axis,  as  O'X'  ',  is  fixed,  we  have  coxooy  =  O,  K)XOSZ  =  O  ;  if  O'  Y'  is  fixed, 
we  have  coyoj^  =  O,  &?yaj,  =  O  ;  if  O'  '  Z'  is  fixed,  we  have  ootoox  =  O,  cozooy  =  O. 

If  the  co-ordinate  axes  are  not  fixed,  we  have  coxooy  —  07,07,,  coyGof  —  cozcjy,  Got(ax  =  coxooz, 
and  since 

/„  -  /*  =  ^*(/  -  ^  =  I.  -  Jy  •  I*y  ~  4.  -  ^^  -  *2)  =  I-  I., 

4  -  /„  =  ^™(*?  -/)  =  //-  /,, 
equations  (2)  reduce  to 

Y¥A  -  m/J  -  m£F  +  (/.  -  />.«,  +  /A  ,  ] 

M'xy  =  mfjs  -  mf?  +  (Ix  -  Sz)GoxGot  +  /,«*,  [    ......     (3) 

J/;,  =  m/^  -  mfxy  +  (/,  -  /Ja,^,  +  1  A.  \ 

If  we  take  distance  in  feet  and  mass  in  Ibs.,  these  equations  give  moments  in  poundal- 
feet.  For  pound-feet  divide  by  g. 

Momentum  —  Rotation  and  Translation.  —  Equations  (4),  page  154,  give  the  component 
velocities  for  any  particle  of  a  rotating  and  translating  body.  If  we  multiply  each  term  by 
the  mass  m  of  the  particle,  and  sum  up  for  all  the  particles,  we  have,  since  ^mx  =  o, 
2my  =  O,  2mz  —  o,  for  the  components  of  the  momentum  for  a  rotating  and  translating 
body  of  mass  m  =  2m,  for  any  co-ordinate  axes  we  please, 

2mvx=  mvx,         2mvy  =  mvy,         2mvt  =  m^r,     .....     (4) 

where  vx,  vy,  vz  are  the  component  velocities  of  the  centre  of  mass. 

That  is,  the  momentum  in  any  direction  is  the  same  as  for  a  particle  of  mass  equal  to 
the  mass  of  the  body  having  the  velocity  in  that  direction  of  the  centre  of  mass. 

Moment  of  Momentum  —  Rotation  and  Translation.  —  Equations  (10),  page  155,  give 
the  component  moments  of  velocity  for  any  particle  of  a  rotating  and  translating  body.  If 
we  multiply  each  term  by  the  mass  m  of  the  particle  and  sum  up  for  all  the  particles,  we  shall 
obtain  the  component  moments  of  momentum  for  any  co-ordinate  axes  we  please.  We  have 
then  from  equations  (10),  page  155,  since  ~2nix  =  o,  2my  =  o,  2ms  =  o,  and  2m(y*  -j-  £} 
=  4,  2m(z*  -\-  x*)  =  Iy,  2m(x*  +./)  =  /2,  for  the  component  moments  of  momentum  for  a 
rotating  and  translating  body  of  mass  m  —  2m 


Mvx  = 

Mvy  = 


—  mz^  ~y  -f  7,67,  . 


(5) 


Conservation  of  Moment  of  Momentum  —  Rotation  and  Translation.  —  We  have  from 
equations  (4)  for  the  component  moments  of  momentum  for  a  body  rotating  about  a  trans- 
lating axis  through  the  centre  of  mass,  since  J  =  o,  jr  =  o,  ^  =  o, 

Mvx=Sxcox,          Mv  =  !,&,,          Mw  =  lr,<ot,        .     .-   V    V:V    .     (a) 
and  from  equations  (3)  for  the  component  moments  of  the  effective  forces 


Mfy  =  (Ix- 


39*  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  VIII. 

If  in  equations  (£)  the  co-ordinate  axes  OX,  OY,  OZ  (figure,  page  153)  are  fixed,  we 
have  07,07,,  =  o,  07,04  =  o,  07,07,  =  O,  and  the  axis  of  rotation  is  fixed.  If  also  we  have 
an  =  o,  ^  =  o,  a,  =  O,  we  have  Mff  =  o,  Mfy  =  O,  Mft  =  o,  and  oox,  07,,  07,  in  equations  (a] 
are  constant.  But  since,  by  D'Alembert's  principle,  the  moment  of  the  effective  forces  is 
equal  to  the  moment  of  the  impressed  forces,  we  have  the  moment  of  the  impressed  forces 
zero. 

Hence  if  the  moment  of  the  impressed  forces  about  any  fixed  axis  through  the  centre  of 
mass  is  always  zero,  the  moment  of  momentum  of  the  body  about  that  axis  does  not  change. 

Invariable  Axis. — If  the  moment  of  the  impressed  forces  relative  to  the  centre  of  mass  O 
is  always  zero,  the  resultant  must  either  be  zero  or  always  pass  through  O.  In  such  case  the 
co-ordinate  axes  do  not  change  in  direction  and  we  have  07,0?,,  =  o,  07,07,  =  o,  07,07,  =  o, 
and  also  Mfx  —  o,  Mfy  —  o,  Mfl  —  o.  Hence,  from  equations  (b),  otx  =  o,  ay  =  o,  a,  =  o  and 
a>M,  <a,,  o?,  in  equations  (a)  are  constant. 

We  have  then  the  moment  of  momentum 


about  an  axis  through  the  centre  of  mass  whose  direction  cosines  are 

7_07,  /,G7y  /.  07. 


and  this  axis  is  then  invariable  in  direction. 

Hence  if  the  moment  of  the  impressed  forces  relative  to  the  centre  of  mass  is  always  zero, 
or  if  there  are  no  impressed  forces,  there  will  be  a  certain  axis  through  the  centre  of  mass  for 
which  the  moment  of  momentum  is  constant.  This  axis  is  invariable  in  direction,  and  its 
direction  cosines  are  given  by  equations  (d).  The  moment  of  momentum  about  this  axis  is  given 
by  equations  (c). 

Kinetic  Energy  —  Rotation  and  Translation.  —  The  kinetic  energy  of  a  particle  of  mass 

m  and  velocity  v  is  -mv2.      If  a  particle  has  the  component  velocities  vx,    vy,  vt,  we  have 
^  =  v*  4-  v*  +  v»,  and 


mv   =  -mo 

Equations  (4),  page  154,  give  the  component  velocities  for  any  particle  of  a  rotating 
and  translating  body.      If  we  square  these  component  velocities,  multiply  each  term  by  -m, 

sum  up  for  all  the  particles  and  add,  we  shall  have  the  kinetic  energy  for  a  rotating  and 
translating  body  for  any  co-ordinate  axes  we  please.  Let  us  take  these  axes  as  principal 
axes  at  the  centre  of  mass.  Then  we1  have  2mx  —  o,  2my  =  o,  ^mz  =  o,  and  also 
2mxy  =  o,  2myz  =  O,  2mzx  =  o.  We  have  also  2m(f  +  z*)  —  /„,  2tn(i^  +  -v2)  =  Iy, 


We  have  then  for  the  kinetic  energy  of  a  rotating  and  translating  body 

(6) 


CHAP.  VIII.] 


SPONTANEOUS  AXIS   OF  ROTATION, 


393 


If  we  take  mass  in  Ibs.,  this  equation  gives  kinetic  energy  in  foot-poundals.  For  foot- 
pounds divide  by^. 

Spontaneous  Axis  of  Rotation.  —  The  axis  through  the  centre  of  mass  about  which  a 
body  rotates  at  any  instant  is  called  the  spontaneous  axis  of  rotation.  If  K)x,K)y,  a?,  are  the 
component  angular  velocities  for  any  co-ordinate  axes  through  the  centre  of  mass,  the 
resultant  angular  velocity  is 


and  the  direction  cosines  of  the  axis  are 


cos  a  —  — 
GO 


COS  y  =  — 
F  G3 


Instantaneous  Axis  of  Rotation.  —  The  instantaneous  axis  is  parallel  at  any  instant  to 
the  spontaneous  axis  and  passes  through  a  point  whose  co-ordinates  relative  to  the  centre 
of  mass  are  given  by  equations  (15),  page  156. 

Examples.  —  (i)  Find  the  motion  of  a  sphere  rolling  on  a  rough  plane, 

ANS.  Let  the  plane  be  the  plane  of  X'  Y',  and  let  the  components  of  friction  be  Fx,  Fy.  All  the 
other  impressed  forces  can  be  reduced  to  a  single  resultant  R  at  the  centre  of  mass  O  and  a  couple  which 
causes  angular  acceleration  a  about  an  axis  through  O. 

Let  Rx,  Ry,  Rz  be  the  components  of  the  resultant  R,  and 
ax,  ay,  az  the  components  of  the  angular  acceleration  a. 

Take  the  axes  OX,  OY,  OZ  through  the  centre  of  mass  O 
parallel  to  O'X',  O'  Y',  O'  Z'. 

We  have  from  equations  (i),  page  390,  the  component  effective 
forces,  m/lr,  mfy,  rafz,  and  from  equations  (3),  page  391,  since  Ix  = 
Iy  =  Iz,  for  the  moment  of  the  effective  forces  about  OX,  OY,  OZ, 
M/x  =  Ixax,  M/y  —  Iyay,  Mfz  =  Izaz. 

By  D'Alembert's  principle  we  have  then 


But 


-— -  •*  x  ~\~   RXI          ~^Jy  —      y  ~^~  l^yi         ni/z 
r  =  Fyr,          ly&y  —  ~  Fxr,          fzaz  =  o. 

Jx  =  ^ay-Jaz,         fy  =  : 


R., 


(0 


Y' 


and  taking  the  origin  at  the  point  of  contact  P, 


x  =  o,        y  =  o,         z  =  +  r. 


We  have  then    fx  =  ray,       fy=    —  r&x       /«  =  o,     or 
Substituting  in  equations  (i),  we  obtain 


•'— 


a    =  •— 


Or,  since  /*  =  Jy  =  mK*,  where  K  is  the  radius  of  gyration, 


(2) 


394  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  VIII. 

Now,  *•»  =  — r1    and    r*  +  *"*  =  -r1.     Hence  these  are  the  same  equations  as  for  a  particle  of  equal 

r»  e 

mass  acted  upon  by  -; r  =  -  of  the  acting  forces. 

Hence  the  motion  of  the  centre  of  mass  of  a  homogeneous  sphere  rolling  on  a  rough  plane  under  the  action 
of  any  forces  is  the  same  as  for  a  particle  of  the  same  mass  if  all  the  impressed  forces  except  friction  are 

reduced  to  —  of  their  former  value. 

Now    ^/r*  +  fy*  is  the  resultant  horizontal  acceleration  Jk  in  the  plane  X'Y',  and   I/AV  +  Ry*  is  the 
resultant  horizontal  impressed  force  H. 
Hence,  from  (2), 


m  4'A1  +  /,'  =  ftffi?  +  *S.      or     m/A  =  ^/f. 
If  /?,  is  the  normal  force  and  u  the  coefficient  of  friction,  then  uRx  is  the  friction,  and  hence 

m/A  =  H  —  uA't . 
Henc* 


If  then  the  coeffirJ-o.*  of  friction  is  equal  to  or  greater  than  ~-— ,  the  sphere  will  roll  without  sliding. 

7  At 

(2)  A  sphere  is  plcued  on  an  inclined  plane  sufficiently  rough  to  prevent  sliding,  and  a  velocity  in  any 
direction  1.1  conff-i-nicated  to  it.     Show  that  the  path  of  the  centre  is  a  parabola.      If  v  is  the  initial  horizontal 

velocity  of  the  centre,  and  a  the  inclination  of  the  plane,  show  that  the  latus  rectum  will  be  — -   : — . 

5  g  stn  a 

ANS.  The  acceleration  down  the  plane,  from  the  principle  of  the  preceding  example,  is  —g  sin  a.     If 

the  initial  velocity  v  makes  an  angle  6  with  the  line  of  slope,  the  velocity  down  the  plane  is  v  cos  8,  and  at 
right  angles  v  sin  6.  There  is  no  acceleration  at  right  angles.  The  distance/  passed  over  at  right  angles 
in  the  time  /  is  then 

y  =  -vt  sin  0, 
and  the  distance  x  down  the  plane  is 

x  =  vt  cos  6  +  --  p-/a  sin  a, 
1 4* 

Eliminating  /,  we  have  for  the  equation  of  the  curve 


-  _£_.  +   5  ^  sin  a 
tan  6       uv*  sin"  6     ' 


14  v' 

This  is  the  equation  of  a  parabola.     If  the  initial  velocity  is  horizontal,  8  =  90°,  sin  6  =  i,  tan  6  =  oo, 
and  we  have 


5  g  sin  a 

(3)  Show,  as  in  example  ('/),  that  the  motion  of  the  centre  of  mass  of  a  homogeneous  disc  rolling  on  a  rough 
plane  under  the  action  of  any  forces  is  the  same  as  for  a  particle  of  the  same  mass,  if  all  the  forces  excefit 
friction  are  reduced  to  two  thirds  of  their  former  value  ;  also  that  the  disc  will  roll  without  sliding  ij  the 

coefficient  of  friction  is  equal  to  or  greater  than  -  •  —  . 

(4)  Show  in  the  same  way  that  for  a  circular  hoop  the  motion  of  the  centre  of  mass  is  the  same  as  for  a 
particle  of  the  same  mass,  if  all  the  forces  except  friction   are  reduced  to  one  half  their  former  value;   also 

that  the  hoot)  will  roll  without  sliding  if  the  coefficient  of  friction  is  equal  to  or  greater  than  —  —  . 

2  At 


CHAP.  VIII.] 


ROTATION  AND   TRANSLATION-EXAMPLES. 


395 


(5)  Also  show  in  the  same  way  for  a  spherical  shell  that  the  forces  are  reduced  to  three  fifths  their  former 
value,  and  that  the  shell  will  roll  without  sliding  if  the  coefficient  of  friction  is  equal  to  or  greater  than  ~  —  ~. 

(6)  A  hoop  moving  in  a  vertical  plane  in  contact  with  a  rough  horizontal  surface  has  at  a  given  instant 
an  angular  velocity  opposite  in  direction  to  that  which  would  enable  it  to  roll  in  the  direction  of  its  translation 
at  that  instant.     Find  the  motion. 

AKS.   We  have  the  effective  force  mf  and  the  moment  of  the  effective  force  la,  where  /  is  the  moment 
of  inertia  for  axis  through  centre  of  mass  O  at  right  angles  to  the  plane  of  the  hoop. 
The  impressed  forces  are  the  weight  mg,  the  equal  and  opposite  pressure  R  and 
the  friction  F  =  umg  acting  opposite  to  the  direction  of  motion.     We  have  then, 
from  D'Alembert's  principle, 

m/  =  -  umg,     or    /  =  -  fig ; 


Ja  =  —  umgr,     or     a.  —  — 
where  K  is  the  radius  of  c:vration. 


ugr 


Hence  the  linear  and  angular  accelerations  are  constant. 

Let  v\  and  <y,  be  the  initial  values  of  the  linear  and  angular  velocities  and  ~v,  ca  the  final  values  at  any 
time  /,  and  we  have 


ugr, 

-2— 


K1 


At  the  instant  when  slipping  ceases  we  have  v  +  rco=  o. 
Eliminating  v  and  &?,  we  have  for  the  time  after  which  slipping  ceases 


where  u  is  the  coefficient  of  friction  for  slipping.     For  the  linear  and  angular  velocity  at  the  instant  when 
slipping  ceases  we  have  then 

r(rv>   -  K-2<»,}  K-'OJ,  -  rvi 


For  any  interval  of  time  less  than  /  as  given  by  (2)  equations  (i)  give  v  and  K>,  when  /<  is  the  coefficient 
for  slipping.  For  any  interval  of  time  greater  than  this  equations  (i)  also  give  v  and  ca,  but  /i  is  the  coeffi- 
cient for  rolling  friction. 

In  equations  (3),  if  rv,  —  K-OJ-.  is  negative,  v  is  negative.     Hence  if  oji  is  greater  than    ~.  the  hoop  at 

the  instant  sliding  ceases  will  have  translation  opposite  in  direction  to  the  initial  translation. 

Equations  (i),  (2),  (3)  hold  for  a  sphere  or  cylinder  also.  We  have  only  to  substitute  the  value  of  x-a  in 
each  case. 

(7)  A  disc  whose  plane  is  vertical  rolls  without  sliding  down  an  inclined  plane.     Find  its  motion. 

ANS.  The  effective  force  is  m/  parallel  to  the  plane  and  the  moment  of  the  effective  force  la,  where  / 
is  the  moment  of  inertia  for  axis  through  the  centre  of  mass  O  at  right  angles 
to  the  plane  of  the  disc. 

The  impressed  forces  are  the  weight  mg,  the  normal  pressure  R  =  mgcosi, 
where  /  is  the  angle  of  inclination,  and  the  friction  F  =  M K  =  urn?  cos  /acting 
opposite  to  the  motion. 

By  D'Alembert's  principle  we  have  then 

m/=  m£-  sin  /  —  umg  cos  /,     or    J  —  g  sin  /  —  ug  cos  /,      .     .     (i) 


la 


u'nrg  cos  /             urg  cos  /       .  . 
=  —  Fr  =  —  umrg  cos  /,     or     a  =  — j —        = ^ ,     (-) 


mg 


where  K  is  the  radius  of  gyration. 

Now  the  condition  for  rolling  without  sliding  \sf-\-  ra  =  o.      From  (2),  then,  we  have 

/V 


" gr1  cos  /' 


396  KINETICS  OF  A  MATERIAL  SYSTEM.  [CHAP.  VIII. 

Substituting  this  value  of  //  in  (i),  we  have 

7  =  *!l«i!L£,    or       a  =  -2^4.     .'  ..........     (3) 

J        K-*  +  r*  K1    +r* 

Equations  (3)  give  the  linear  and  angular  accelerations.     We  see  that  they  are  both  constant.  We  have 
then 


The  equations  hold  for  a  sphere  or  cylinder  also.  We  have  only  to  substitute  the  value  of  K*  in  each 
case. 

(8)  Find  the  time  a  cylinder  will  take  to  roll  from  rest  down  a  plane  20  ft.  long,  inclined  jo"  ,  the  axis  of 
the  cylinder  being  horizontal. 

r*  I 

ANS.  We  have  z/i  =  oand  *«  =  —  .   Hence,  since  sin  *  =  —  and  taking^  =  32, 


STATICS   OF   RIGID   BODIES. 


CHAPTER  I. 

FRAMED  STRUCTURES. 

Stress. — We  have  seen  (page  1 74)  that  the  exertion  of  force  upon  a  body  or  particle  is 
only  one  side  of  the  complete  phenomenon  which  consists  of  the  simultaneous  action  of 
equal  and  opposite  forces  between  two  bodies  or  particles.  We  call  these  internal  forces 
STRESSES.  Stress,  then,  is  a  force  internal  to  the  body  or  system  considered,  while  the  term 
force  is  reserved  for  external  action. 

We  speak,  then,  of  the  force  on  a  body  or  particle,  and  the  stress  in  a  body  or  between 
two  bodies  or  particles. 

Tensile  Stress  and  Force. — Let  a  body  AB  be  acted  upon  by  two   equal  and  opposite 

forces  -j-  Fl  —  F  in  the  same  straight  line.     These     __       r-r — ^         ^ 

forces   act   to  stretch  the   body,   and  are  therefore 
called  tensile  forces. 

If  there  is  equilibrium,  we  must  have  at  any  two  sections  A  and  .5  two  equal  and  oppo- 
site stresses,  4-  S,  —  S,  which  resist  the  extension.  These  are  called  TENSILE  STRESSES. 
The  tensile  forces  act  away  from  each  other,  tending  to  pull  A  and  B  apart.  The  tensile 
stresses  act  towards  each  other,  tending  to  draw  A  and  B  together.  We  have  thus  a  force  F 
and  an  equal  and  opposite  stress  in  equilibrium  at  A,  and  a  force  F  and  an  equal  and  oppo- 
site stress  in  equilibrium  at  B. 

We  see  also  that  if  the  stresses  are  tensile,  they  act  away  from  the  ends  A  and  B. 

Compressive  Stress  and  Force — In  the  same  way,  if  the  forces  are  reversed  in  direction, 
we  have  COMPRESSIVE  FORCE  and  COMPRESSIVE  STRESS.  The  compressive  forces  act  to 

make  A  and  B  move  towards  each  other,  and  the 
compressive  stresses  to  move  them  apart. 
^F We  see  also  that  if  the  stresses  are  compress- 
ive, they  act  toivards  the  ends  A  and  B. 

Shearing  Force  and  Stress. — The  algebraic  sum  of  the  components  parallel  to  a  section 
of  all  the  external  forces  on  the  right  of  that  section  tends  to  make  that  section  slide  upon 
the  consecutive  section  on  the  left.  The  algebraic  sum  of  the  components  parallel  to  a 
section  of  all  the  external  forces  on  the  left  of  a  section  tends  to  make  that  section  slide 
upon  the  consecutive  section  on  the  right. 

For  reasons  to  be  given  later  (page  402)  we  always  take  the  algebraic  sum  of  the 
components  parallel  to  a  section  of  all  the  external  forces  on  the  left  of  that  section,  and 
call  this  algebraic  sum  the  shearing  force  for  that  .section. 

We  define,  then,  the  shearing  force  for  any  section  as  the  algebraic  sum  of  the  components 
parallel  to  that  section  of  all  the  external  forces  on  the  left. 

It  is  resisted  by  the  shearing  stress  or  resistance  of  the  section  to  sliding  on  the 
consecutive  section  on  the  right.  This  shearing  stress  is  equal  and  opposite  to  the  shearing 
force.  Since  for  equilibrium  the  algebraic  sum  on  one  side  must  be  equal  and  opposite  to 
that  on  the  other,  we  can  define  the  shearing  stress  for  any  section  as  the  algebraic  sum  of 

397 


-S 


398 


ST /tTICS  OF  RIGID  BODIES. 


[CHAP.  1. 


the  components  parallel  to  that  section  of  all  the  external  forces  on  the  right.     Thus  in  the 

case  of  a  horizontal  beam  AB  acted  upon  by 
vertical  loads  Plt  P2,  P3  and  the  vertical 
upward  reactions  /?,  and  7?2,  the  shearing  force 
for  any  section  at  a  between  the  left  end  and 
Pl  is  by  definition  -\-  Rr  For  any  section  at 
b  between  Pl  and  PI  the  shearing  force  is 
-j-  /?,  —Pr  For  any  section  at  c  between  P2 
and  P3  the  shearing  force  is  -|-  /?,  —  Pl  —  P2. 
For  any  section  at  d  between  P3  and  the  right 
end  the  shearing  force  is  +  j?j  —  Pl  — />2  —  Py 
The  ordinates  of  the  shaded  area  thus  give  the 

shearing  force  to  scale.     The  shearing  stress  is  equal  and  opposite. 

If  we   have   a  load   of  w  per  unit   of  length,  uniformly  distributed,  we  have  for  the 

shearing  force  for  any  section  distant  x  from  the 

left  end 

shearing  force  = ivx, 

which  is  the  equation  to  the  straight  line  A"  B" 
passing  through  the  centre  of  the  span.  The 
ordinate  to  this  line  at  any  section  a  gives  the 
shearing  force  to  scale.  The  shearing  stress  is 
equal  and  opposite.  The  shearing  force  at  the 
centre  is  zero. 

Equilibrium  of  a  System  of  Bodies. — If  a  number  of  rigid  bodies  are  connected  by 
strings,  rods,  joints,  etc.,  and  the  system  is  in  equilibrium,  each  body  of  the  system  must  be 
in  equilibrium,  and  the  conditions  of  equilibrium  apply  to  each  body  as  well  as  to  the  whole 
system. 

Examples. — (i)  In  the  system  of  pulleys  sh<nvnfind  the  relation  bet-ween  P  and  Q  for  equilibrium,  disre- 
garding friction  and  rigidity  of  ropes. 

ANS.  The  tensile  stress  in  a  continuous  string  with  two  equal  and  opposite  forces  at  the  ends  must  be 
the  same  at  every  point  of  the  string. 

Let  T\,  Tt,  Tt,  etc.,  be  the  tensile  stresses  as  shown,  and  m  the  mass  of  each 
pulley. 

Then  we  have  for  equilibrium  in  gravitation  measure 


Tt  =  P, 
-  Tt  —  m  =  o. 

—  Tt  -  m  =  o, 

—  Tt  —  m  =  o, 


T,  =  2P-  m, 

Ti  =  2*P  -  (2*  - 
Tt  =  21P  -  (2*  .— 


or  in  general,  if  n  is  the  number  of  movable  pulleys, 

But  for  the  last  pulley  we  have 

2  Tn  —  Q  —  m  =  o. 
Hence  we  obtain  in  general 


Q  +  (2«  - 
Q  =  2"P  —  (2"  -  i)iw.     or     P  =  £-   — ^~- 


[Student  should  compare  solution  by  virtual  work,  example  (6),  page  218.] 


CHAP.  I.] 


FRAMED  S TRUC  TURES.— EXAMPLES. 


399 


(2)  In  ike  system  of  pulleys  shown  find  the  retation  between  P  and  Qfor  equilibrium,  disregarding  friction 

and  rigidity  of  ropes. 

[Student  should  compare  solution  by  virtual  work,  example  (7),  page  218.] 
ANS.  The  tensile  stress  in  the  rope  is  T  =  P  at  every  point.     If,  then,  n  is  the 

number  of  times  the  rope  would  be  cut  by  a  plane  AB  between  the  two  blocks,  and 

m  is  the  mass  of  a  block,  we  have 


or         = 


In  the  figure  n  =  6. 

(3)  In  the  system  of  pulleys  shown  find  the  relation  between  P  and  Q  for  equilib- 
rium, disregarding  friction  and  rigidity  of  ropes. 
ANS.  We  have,  if  m  is  the  mass  of  each  pulley, 


r,  =  P, 

T*  -  2  Ti  —  m  =  o, 
T3  -  2  Ti  -  m  =  o, 
etc. 


Ti  =  2P  +  m, 
r,  =  4/>  +  $m, 
T4  =  SP  +  7m, 


Tn  =  2*-lP  +  (2*-'  —  i)m, 
where  n  is  the  number  of  pulleys.     Also  we  have 

T,  +  T,  +   T3  +  .  .  .  Tn  =  Q. 
Substituting  the  values  of  7\,  7"»,  etc., 
P[i  +  2  +  4  +  .  .  .  2"—]  +  w[i  +  3  +  7  +  .  .  .  (2*-1  —  i)] 


CO) 


Hence 


,    or    P  = 


(2«-l) 

[Student  should  solve  by  virtual  work.] 

(4)  In  the  differential  pulley  shown  in  the  figure,  an  endless  chain  passes  over  a  fixed  pulley  A,  then  under 
a  movable  pulley  to  which  the  mass  Q  is  hung,  and  then  over  another  fixed  pulley  B  a  little  smaller  than  A. 
The  two  pulleys,   A  and  B,  are  in  one  piece  and  obliged  to  turn  together.      The  two 
ends  of  the  chain  are  joined.     The  force  P  is  applied  as  shown.     To  prevent  the  chain 
from  slipping,  there  are  cavities  in  the  circumferences  of  the  pulleys  into  which  the          / 
links  of  the  chain  fit.     Find  the  relation  between  P  and  Qfor  equilibrium.  I 

Q  A 

ANS.  We  have  27"—  Q  =  o,  or  T  =  — .   Let  a  be  the  radius  of  A,  and  b  the  radius 

of  B.     Then,  taking  moments  about  C, 

Ta  —  Tb  —  Pa  —  o, 


or,  inserting 


* 


LDP 


By  taking  a  and  b  nearly  equal  we  can  have  P  very  small. 

(5)  The  requisites  of  a  good  balance  are  as  follows :  ist.  It  should  be  "true,"  that  is,  when  loaded  unth 
equal  masses  the  beam  should  be  horizontal.  2d.  It  should  be  "  sensitive,"  that  is,  when  the  massed  differ  by  a 
small  amount  the  angle  of  /,">  beam  with  the  horizontal  should  be  large,  jd.  It  should  be  "  stable,"  that  is, 
wJien  moved  from  equilibrium  it  should  return  quickly.  Skow  the  conditions  necessary  for  these  requisites. 


400 


STATICS  OF  RIGID  BODIES. 


fCHAP.   I. 


i 


ANS.  Let  the  masse*  be  /'and  Q,  and  W  the  weight  in  pounds  of  the  balance  acting  at  its  centre  of  mass 

C.     Let  the  fulcrum  be.at  F,  above  C,  and  draw  FD  perpendicular 
to  the  beam  AB  at  Z>. 

Let  AD  -  a,  BD  =  b,  FC  =  d,  FD  =  h,  and  fl  the  angle  of  the 

C  _       .»      beam  with  the  horizontal. 

Thru  for  equilibrium,  taking  moments  about  F,  we  have 

Q(b  cos  8  —  h  sin  8)  -  P(a  cos  0  +  h  sin  6)  -  Wd  sin  6  =  0. 
Hence 


tane= 


(Q  +  P)h  +  Wd' 

I.  If  the  balance  is  "true,"  we  must  have  6  =  0  when  P  =  Q.     We  see,  from  (i),  that  to  satisfy  this 
requisite  the  arms  must  be  equal.     We  have  then  for  a  true  balance  a  =  b  and 


tanfl 


2.  If  the  balance  is  to  be  "sensitive,"  9  must  be  large  when  Q  —  P  is  small.     We  see  from  (2)  that  to 
satisfy  this  requisite  h  and  //must  be  small  relative  to  the  length  of  arm  a.     The  sensitiveness  is  increased, 
therefore,  by  either  increasing  the  length  of  arm  or  by  bringing  the  centre  of  mass  and  point  of  suspension 
nearer  the  beam. 

3.  If  the  balance  is  to  be  "  stable,"  or  to  return  quickly  when  disturbed,  the  moment  Wd  sin  6  of  W  about 
the  fulcrum  must  be  large.     Hence  stability  is  increased  by  moving  the  point  of  suspension  away  from  the 
beam. 

The  conditions,  then,  for  stability  and  sensitiveness  are  at  variance.  Stability  is  gained  at  the  expense 
of  sensitiveness,  and  vice  versa. 

In  scientific  measurements,  where  great  accuracy  is  required,  stability  is  sacrificed  to  obtain  great  sensi- 
tiveness. The  balance  recovers  slowly  from  a  disturbance,  and  time  is  required  for  it  to  come  to  rest.  For 
ordinary  commercial  purposes,  where  it  is  desirable  to  save  time,  sensitiveness  is  sacrificed  to  stability. 

Framed  Structures. — A  framed  structure  is  a  collection  of  straight  MEMBERS  joined 
together  at  the  ends  so  as  to  make  a  rigid  frame. 

The  simplest  rigid  frame  is  obviously  a  triangle,  because  that  is  the  only  figure  whose 
shape  cannot  be  altered  without  changing  the  lengths  of  its  sides.  All  rigid  frames  must 
consist,  therefore,  of  a  combination  of  triangles. 

Any  point  where  two  or  more  members  meet  is  called  an  APEX  of  the  frame. 

Determination  of  the  Stresses  in  a  Framed  Structure. — We  have  external  forces  acting 
upon  the  frame,  and  tensile  or  compressive  stresses  in  the  members.  If  >he  frame  is  in 
equilibrium,  the  stresses  and  external  forces  form  a  system  in  equilibrium. 

Also,  since  equilibrium  must  exist  at  every  apex  of  the  frame,  all  the  forces  and  stresses 
at  any  apex  form  a  system  of  concurring  forces  in  equilibrium.  Tensile  stress  acts  away 
from  an  apex,  compressive  stress  towards  an  apex  (page  397). 

We  have  then  two  methods  of  solution : 

I.    METHOD  BY  RESOLUTION  OF  FORCES. — In  the  figure  we  have  a  frame  acted  upon 
by  known  external  forces  /*,,  P2,  P3,  and  the  upward  pres- 
sures of  the  supports  Rl  and  R2  at  A  and  B. 

Take  any  apex,  as  a.  At  this  apex  we  have  Pl  and 
the  stresses  in  aA,  ae,  0/and  ab,  forming  a  system  of  con- 
curring forces  in  equilibrium. 

If  we  denote  these  stresses  by  Slt  S3,  S3,  S4,  the 
angles  of  the  members  with  the  horizontal  by  «,,  o-2,  as, 
*r4,  and  with  the  vertical  by  /?,,  /32,  /33,  /3t,  we  have  then 
for  equilibrium,  if  «5  is  the  angle  of  Pl  with  the  horizontal, 
and  /35  with  the  vertical, 

5,  cos  al  +  S,  cos  a2  -\-  S3  cos  «s  +  S4  cos  a4  -f-  PI  cos  <xs  =  o, 
St  cos  /3l  +  S,  cos  fa  -j-  S3  cos  /?3  +  S4  cos  fa  -f-  Pl  cos  /3S  =  o, 


CHAP.  I  ]  DETERMINATION  OF  THE  STRESSES  IN  A  FRAMED  STRUCTURE.  401 

or  in  general,  for  any  apex, 

2S  cos  a  -f  2P  cos  a  =  O,  1 
3S  cos  ft  +  2P  cos  >S  =  o.  J 


^         COS 

Since  we  have  thus  two  equations  of  condition,  this  method  can  be  applied  at  any  apex 
where  there  are  not  -more  than  tivo  unknoivn  stresses. 

Components  to  the  right  and  upward  are  positive,  to  the  left  and  downward  negative. 
If  then  a  stress  comes  out  with  a  minus  sign,  it  denotes  that  the  stress  acts  towards 
the  apex  and  is  compressive.  If  it  comes  out  with  a  plus  sign,  it  denotes  that  the  stress 
acts  away  from  the  apex  and  is  tensile  (page  397). 

2.  METHOD  BY  MOMENTS. — Suppose  the  frame  divided  into  two  parts.  Then  the 
stresses  belonging  to  the  cut  members  must  evidently  hold  in  equilibrium  the  external 
forces  on  either  side  of  the  section. 

Thus,  in  the  figure,  if  we  suppose  ab,  af,  ef  cut,  the  stresses  of 
these  members  must  hold  in  equilibrium  R  and  /\.  The  algebraic 
sum  of  the  moments  relative  to  any  point  must  then  be  zero.  ^  ^ 

If  then  we  wish  to  find  the  stress  in  ab,  we  can  take  the  centre    ^g^ I         ^  v>v>k. 

of  moments  at/".      The  moments  of  the  stresses  in  af  and  ef  will    1 
then  be  zero,  and  if  /  is  the  lever-arm  for  ab,  we  have  R 

—  ab  .  l-\-  2  moments  of  external  forces  on  left  of  section  =  o. 

Observe  that  the  moment  for  the  stress  in  ab  is  to  be  taken  with  the  sign  indicated  by 
the  arrow. 

So  if  we  wish  to  find  ef,  we  should  take  moments  about  a,  thus  eliminating  the 
moments  of  ab  and  af. 

Or  if  we  wish  to  find  af,  we  should  take  moments  about  A,  thus  eliminating  the 
moments  of  ab  and  ef. 

The  method,  then,  is  in  general  as  follows:  Divide  the  frame  and  consider  only  one 
portion,  as,  for  instance,  the  left-hand  portion. 

Place  arrows  on  each  cut  member  pointing  towards  the  section. 

To  find  the  stress  in  any  one  of  the  cut  members  take  the  intersection  of  the  other 
members  cut  as  a  point  of  moments.  This  will  eliminate  these  members. 

Place  the  algebraic  sum  of  the  moments  of  all  forces  and  stresses  for  this  point  of 
moments  equal  to  zero. 

This  method  is  general  and  can  always  be  applied  to  find  the  stress  in  any  member 
when  all  the  cut  members  whose  stresses  are  unknown,  except  the  one  whose  stress  is 
desired,  meet  in  a  point. 

If  two  of  the  cut  members  are  parallel,  their  intersection  is  at  an  infinite  distance,  but 

the  method  still  applies. 

Thus  if  we  wish  to  find  the  stress  in  cb, 
we  take  a  section  cutting  ab,  be  and  cd. 
The  intersection  of  ab  and  cd'\s  at  an  infinite 
distance.  We  therefore  have  the  lever-arm 
for  cb,  oo  cos  /?,  where  0  is  the  angle  of  cb 
with  the  vertical.  Hence 

Jjoo  —  Pjoo  —  P2oo  -f-  cb  X  oo  cos  /3  =  o,     or     cb  cos  ft  +  (Rv  —  /\  —  PJ  =  o, 
cb  —  —  (R^  —  Pl  —  Pa)  sec  /S. 


402 


STATICS  OF  RIGID  BODIES. 


[ClIAI'.     I. 


The  algebraic  sum  of  the  external  forces  (Rl  —  Pl  —  Pa)  is  the  SHKAKINC  PORCH 
(page  397).  For  horizontal  chords  and  vertical  forces  we  have  then  the  vertical  component 
of  the  stress  in  a  brace  in  equilibrium  with  the  shearing  force. 

It  is  evident  that  if  the  shearing  force  is  upwards  or  positive,  the  stress  in  cb  will  be 
compression  and  negative.  If  then  we  take  the  compression  as  negative  and  tension  as 
positive,  the  sign  will  denote  the  character  of  the  stress.  It  is  for  this  reason  that  we  have 
taken  shearing  force  as  the  algebraic  sum  of  all  the  forces  on  the  left  (page  397). 

Superfluous  Members. — We  see  from  equations  (i),  page  401,  that  we  have  two 
equations  of  condition  for  equilibrium  at  any  apex.  If  then  at  any  apex  there  are  more 
than  two  members  whose  stresses  are  necessarily  unknown,  the  frame  has  superfluous 
members. 

Criterion  for  Superfluous  Members. — The  simplest  rigid  frame  is  a  triangle,  and  all 
rigid  frames  must  consist  of  a  combination  of  triangles. 

Any  one  member  fixes  the  position  of  two  apices.  Every  other  apex  after  the  first  two 
requires  two  members  to  fix  its  position.  If  then  n  is  the  number  of  apices,  2(n  —  2)  will 
be  the  number  of  members,  lacking  one.  Let  in  be  the  number  of  members.  Then  if  there 
are  no  superfluous  members,  we  must  have 


M  —  I  =  2(n  —  2),      or     ;«  =  2n 

If  m  is  less  than  2/1  —  3,  there  are  not  members  enough, 
there  are  superfluous  members. 

Examples. — (i)  ///  the  cases  of  the  three  frames  represented,  show  that  in  the  first  case  there  are  not  enough 
members  and  the  frame  is  not  rigid;  in  the  second  case  the  frame,  is  rigid  and  there  are  no  superfluous 
members  ;  in  the  third  case  there  is  one  superfluous  member. 

IP  IP 


,v 


If  nf  is  greater  than  2n  —  3, 


ANS.    From  our  criterion  we  should 


have 

m  =  2«  —  3. 

But  we  have  in  the  first  case  n  =  6. 
Hence  m  should  be  9.  But  m  is  only  8, 
or  one  less  than  the  necessary  number. 

In  the  second  case  «  =  6  and  m  =  9, 
as  should  be.     In  the  third  case  n  =  6  and 
m  =  10,  or  one  more  than  necessary. 

(2)  A  roof -truss  has  a  span  of  48  feet  and  a  centre  height  of  18  feet.     Each  rafter  is  divided  into  two 
equal  parts,  and  the  lower  tie  into  two  equal  parts,  and  the  bracing 
is  as  shown  in  the  figure.     Find  the  stresses  in  the  members  for  a 
load  of  800  pounds  at  each  upper  apex. 

ANS.  The  reaction  at  each  end  is  R  =  1200  pounds.     We  find 
it  as  follows  : 

Take  B  as  a  point  of  moments.     Then  we  have 


R  x  48  -  800  x  36  —  800  x  24  -  800  x 
R=  +  1200. 


12  =  o,     or 


ist   METHOD.    For   the   different   members  we   have  for  the 


cosines  of   the   angles  with   the   horizontal   (cos  a)  and   vertical 

(COS  ft) 


Aa 

cos  a  =  0.8 
cos  ft  =  0.6 


0.8 
0.6 


Ad 

i 
o 


0.8 
0.6 


At  apex  A  we  have  then 


+  1200  +  Aa-x.  0.6  =o,     .' 

a  A   x  08  -f  Ad  =  o     .' 


Aa  —  —  2000  pounds. 

Ad  =    -f    1600 


CHAP.  L]  EXAMPLES.  403 

The  minus  sign  denotes  stress  towards  the  apex,  or  compression  ;  the  plus  sign  stress  away  from  the 
apex,  or  tension. 

At  apex  a  we  have 

—  Aa  x  0.6  —  800  +  ab  x  0.6  —  ad  x  0.6  =  o, 

—  Aa  x  0.8  +  ab  x  0.8  +  ad  x  0.8  =  o. 

Inserting  the  value  of  Aa  already  found  and  solving  for  #£and  ad,  we  have 

ab  =  —  13334,         ad  =  —  666f. 
At  apex  b,  since  be  is  evidently  the  same  as  ab,  we  can  put 

—  2ab  x  0.6  —  800  —  bd  =  o,     or     bd  =  +  800. 

2d   METHOD.    The  centre  of   moments  for  Aa  and  ab  should  be  taken  at  d,  for  ad  and  bd  at  A,  for 

We  have  then  for  the  lever-arms 

Aa          ab        Ad        ad         bd 
Lever-arm     1.44         14.4        9         14.4         24 

Hence  we  have 

— Aa  x  1 4.4  —  1 200  x  24  =  o,  Aa  =  —  2000  pounds  ; 

Adx.   9     —1200x12  =  0,  Ad  =  +  1600        " 

—  #3x14.4+    800  x  12'—  I2oox  24  =  o,          ab  =  —  1333^      " 

-  ad*.  14.4  —    800x12  =  0,  ad——    666f      " 

For  bd  we  have 

bdy.  24  —  800  x  12  +  cdy.  14.4  =  o, 

or,  since  cd  =  ad  =  —  666f,  bd=  +  800  pounds. 

(3)  A  bridge  truss  100  ft.  long  is  divided  into  five  equal  panels  in  the  lower  chord  and  four  equal  panels  in 
the  upper  chord.      The  depth  is  constant  and  equal  to  a  b  Q 

10  ft.      The  bracing  is  isosceles.     Find  the  stresses  for  a  /V  /\  f\  f\  7V 

load  of  800  pounds  at  each  lower  apex.  /        \     /         \     /        \      /       \/\ 


ANS.     ^^=  +  1600, 


jC 
CD  =  +  4800,          ab—  —  3200, 


=—  4800,         Aa  =  —  2262.4,        a£  =  +  2262.4,        2?<5  =  —  1131.2,        bC=  +  1131.2, 


CHAPTER   II. 


GRAPHICAL  STATICS.      CONCURRING   FORCES. 


FIG.  i. 


FIG.  2. 


Graphical  Statics. — While  the  solution  of  statical  problems  by  computation  and  ana- 
lytical methods  is  sometimes  tedious  and  involved,  they  may  often  be  solved  with  compara- 
tive ease  and  sufficient  accuracy  by  graphic  construction. 

The  solution  of  statical  problems  by  graphic  methods  gives  rise  to  GRAPHICAL  STATICS. 
We  shall  consider  only  co-planar  forces. 

Concurring  Co-planar  Forces. — Let  any  number  of  co-planar  forces  Flf  F2,  F3,  F4,  etc., 

given  in  magnitude  and  direction,  act  at  a  point 
A,  Fig.  i. 

In  Fig.  2,  from  any  point  o,  lay  off  to  scale 
the  line  representative  of  Fl  from  o  to  i,  then 
the  line  representative  of  F2  from  i  to  2,  then 
the  line  representative  of  F3  from  2  to  3,  then 
the  line  representative  of  F4  from  3  to  4,  and  so 
on.  The  polygon  01234  thus  obtained  we  call 
the  FORCE  POLYGON. 

If  all  these  forces  are  in  equilibrium,  the 
algebraic  sum  of  their  horizontal  and  vertical 
components  must  be  zero.  But  when  this  is  the  case,  evidently  4  and  o,  in  Fig.  2,  must 
coincide,  or  the  force  polygon  must  close.  We  have  then  the  following  principle: 

If  any  number  of  concurring  forces  are  in  equilibrium,  the  force  polygon  is  closed.  If 
the  force  polygon  is  not  closed,  the  line  04.  necessary  to  make  it  close  gives  the  magnitude  and 
direction  of  the  resultant  R.  If  we  consider  this  resultant  acting  at  the  point  of  application  t 1 
in  the  direction  from  4.  to  o,  obtained  by  following  round  the  polygon  in  the  direction  of  the 
forces,  it  will  hold  the  forces  at  A  in  equilibrium.  If  taken  as  acting  in  the  opposite  direction 
at  A ,  it  will  replace  the  forces. 

COR.  I.  The  order  in  which  the  forces  are  laid  off  in  the  force  polygon  is  immaterial. 
Thus,  in  Fig.  2,  if  we  had  laid  off  o  i,  then  the  line  representative  of  F3  from  i  to  3',  and 
then  the  line  representative  of  Ft,  we  should  arrive  at  3  just  as  before.  By  a  similar  change 
of  two  and  two  we  can  have  any  order  we  please. 

COR.  2.  Any  line  in  the  force  polygon,  as  o  2,  o  3,  or  I  3,  is  the  resultant  of  the  forces 
on  either  side.  Thus  o  2  is  the  resultant  of  Fl  and  Ft ,  and,  acting  in  the  direction  from  2  to 
o,  holds  FI  and  F2  in  equilibrium  and  replaces  F3 ,  F4  and  ./?. 

404 


CHAP.  II.  ] 


STATICS  OF  RIGID  BODIES. 


405 


FlG 


COR.    3.   If  the  forces  are  all  parallel,  the  force  polygon  becomes  a  straight  line.     Thus 
in  Fig.  i,  if  the  parallel  forces  Flt  F2,  F3,  F4,  etc.,  act  at  the  point  A,  we        FlG 
have  the  force  polygon   Fig.   2,   01234,   and  the  closing  line  40  is,  as 
before,  the  resultant  R  and  equal  to  the  algebraic  sum  of  the  forces. 

If  taken  as  acting  from  4  to  o,  it  will  hold  the  forces  at  A  in 
equilibrium.  In  the  opposite  direction  it  will  replace  the  forces. 

Notation  for  Framed  "Structures. — Let  the  figure  represent  a  roof-truss 
composed  of  two  rafters,  a  horizontal  tie-rod  and  intermediate  braces 
consisting  of  struts  and  ties. 

The  notation  which  we  adopt  in  order  to  designate  any  member  of 

a  framed  structure,  or  any  force  acting  upon  the  structure,  is  as  follows: 

We  place  a  letter   in  each   of  the  triangular  spaces   into  which   the 

frame  is  divided  by  the  members,  and  also  a  letter  between  any  two  forces. 

Any  member  or  force  is  then  denoted  by  the  letters  on  both  sides  of  it.     Thus  in  the  figure 

AB  denotes  the  force  F{,  BC  denotes  the  force  F2,  CD 
denotes  the  force  F3,  DE  denotes  the  upward  pressure 
of  the  right-hand  support  R2,  EA  denotes  the  upward 
pressure  of  the  left-hand  support  Rr  Also,  Aa,  Bb,  Cd, 
De  denote  the  portions  of  the  rafters  which  have  these 
letters  on  each  side.  The  portions  into  which  the  lower 
tie  is  divided  are  in  the  same  way  Ea,  EC,  Ee.  The 
braces  are  ab,  be,  cd,  de. 
The  student  should  carefully  adhere  to  this  notation  for  the  frame  whenever  using  the 
grapJiic  method. 

Character  of  the  Stresses. — The  determination  of  the  kind  of  stress  in  a  member 
of  a  frame,  whether  tension  or  compression,  is  as  important  as  the  determination  of  the 
magnitude  of  the  stress. 

In  the  preceding  figure,  suppose  we  know  the  upward  pressure  at  the  left  support •  Rl 
or  EA,  and  we  wish  to  find  the  stresses  in  the  members  Ea  and  Aa,  Fig.  I,  which  meet 
at  the  lower  left-hand  apex.  If  these  FlG.  x>  FlG.  2. 

stresses  and  Rl  are  in  equilibrium,  they  will 
make  a  closed  polygon.  If  then  we  lay  off 
EA  in  Fig.  2,  upwards,  equal  to  Rv,  and 
then  from  A  and  E  draw  lines  parallel  to 
Aa  and  Ea  in  Fig.  I,  and  produce  them 
till  they  intersect  at  a,  Fig.  2,  evidently 

the  lines  Aa  and  Ea  in  Fig.  2,  taken  to  the    [R*  'R» 

same  scale  as  EA,  will  give  the  magnitude  of  the  stresses  in  Ea  and  Aa  in  Fig.  I. 

Thus  lines  in  the  force  polygon  which  have  letters  at  each  end  give  the  stresses  in  those 
members  of  the  frame  denoted  by  the  same  letters  at  the  sides. 

Now  as  to  the  character  of  these  stresses,  the  directions  Aa  and  aE  in  Fig.  2,  obtained 
by  following  round  in  the  known  direction  of  Rlt  are  the  directions  for  equilibrium  (page 
404). 

Since  we  are  considering  the  concurring  forces  acting  at  the  left-hand  apex,  transfer 
these  directions  to  Fig.  I,  and  we  see  that  Aa  acts  towards  the  apex  we  are  considering  and 
thus  resists  compression,  and  aE  acts  away  from  it  and  therefore  resists  tension.  The  st' ess 
in  Aa  is  therefore  compressive  (— ),  and  in  aE  tensile  (-f-). 


GRAPHICAL  STATICS.-CONCURRING  FORCES. 


[CHAP.  II. 


In  general,  then,  if  we  take  any  apex  of  the  frame  in  Fig.  I,  and  consider  the  concurring 
forces  acting  at  that  apex  as  a  system  of  concurring  forces  in  equilibrium,  we  have  the 
following  rule: 

Follow  round  the  force  polygon  in  Fig.  2  in  the  direction  indicated  by  any  one  of  these 
forces  already  known,  and  transfer  the  directions  thus  obtained  for  the  stresses  to  the  apex  in 
Fig.  i  under  consideration.  If  the  stress  in  any  member  is  thus  found  acting  away  from  the 
apex,  it  is  tension  (-(-);  if  towards  the  apex,  it  is  compression  (—). 

Application  of  Preceding  Principles  to  a  Frame. — Let  Fig.  i  be.  a  frame  consisting  of 
two  rafters,  a  horizontal  tie-rod  and  bracing  as  shown,  carefully  drawn  to  a  scale  of  a  certain 
number  of  feet  to  an  inch.  This  we  call  the  FRAME  DIAGRAM. 

Let  the  forces  7^,  F2,  F3  act  at 
the  upper  apices,  and  let  the  reactions 
or  upward  pressures  of  the  supports 
be  ^  and  Rr  Notate  the  frame 
and  these  forces  as  directed,  so  that 
F,  =  AB,  F2  =  BC,  F3  =  CD, 
R2  =  -DE,  R^  =  EA,  while  the 
members  are  Aa,  Bb,  Cd,  De,  Ee, 
EC,  Ea,  ab,  be,  cd,  de. 
stresses  in  the  members.  These  outer 


FIG.  i. 


FIG.  2. 

The  outer  forces  acting  upon   the   frame  cause 
forces  must  first  be  all  known,  or  if  any  are  unknown,  they  must  first  be  found. 

Lay  off  these  outer  forces  AB,  BC,  CD,  DE,  EA  in  Fig.  2  to  a  scale  of  a  certain 
number  of  pounds  to  an  inch.  Each  force  in  Fig.  2,  having  letters  at  its  ends,  is  equal  and 
parallel  to  those  forces  in  Fig.  I  which  have  the  same  letters  at  the  sides. 

The  polygon  formed  by  AB,  BC,  CD,  DE,  EA  (in  this  case  a  straight  line,  Cor.  3, 
page  405)  we  have  called  the  FORCE  POLYGON. 

If  the  frame  is  in  equilibrium,  this  polygon  must  always  close,  that  is,  the  outer  forces 
acting  upon  the  frame  must  be  in  equilibrium.  If  it  does  not  close,  these  outer  forces  are 
not  in  equilibrium  and  the  frame  will  move.  That  is,  the  frame  itself,  so  far  as  its  motion 
as  a  whole  is  considered,  may  be  treated  as  a  point. 

Having  thus  drawn  and  notated  the  frame  Fig.  i  and  constructed  the  force  polygon 
Fig.  2,  we  can  find  the  stresses  in  the  members.  The  forces  and  stresses  at  each  apex  must 
be  in  equilibrium,  and  therefore  form  a  closed  polygon. 

Thus  consider  first  the  left-hand  apex,  Fig.  I.  At  this  point  we  have  the  reaction  EA 
and  the  stresses  in  Aa  and  Ea,  constituting  a  system  of  concurring  forces  in  equilibrium.  But 
we  already  have  EA  laid  off  in  Fig.  2.  If  then  we  draw  Aa  and  Ea  in  Fig.  2  parallel  to  Aa 
and  Ea  in  Fig.  I,  and  produce  to  intersection' a,  the  polygon  is  closed  and  we  have  in  Fig» 
2  the  stresses  in  Aa  and  Ea,  to  the  same  scale  employed  in  laying  off  EA.  Since  EA  acts 
upwards,  if  we  follow  round  from  E  to  A,  and  A  to  a,  and  a  to  E,  in  Fig.  2,  and  transfer 
the  directions  thus  obtained  for  Aa  and  aE  to  the  left-hand  apex  in  Fig.  i,  we  have  the 
stress  in  Aa  towards  this  apex  or  compression  (  — ),  and  the  stress  in  aE  away  from  the  apex 
and  therefore  tension  (-{-)• 

[The  student  should  follow  with  his  own  sketch  and  mark  each  stress  with  its  proper 
sign  as  he  finds  it.] 

Let  us  now  pass  to  the  next  upper  apex,  at  Flt  Fig.  i.  Here  we  have  Fl  or  AB  and 
the  stresses  in  Aa,  ab  and  Bb  in  equilibrium.  But  we  already  have  the  stress  in  Aa  and 
AB  laid  off  in  Fig.  2. 

If  then  we  draw  from  a  and  B  in  Fig.  2  lines  parallel  to  ab  and  Bb  in  Fig.  i,  and  pro- 


CHAP.  II.]  STATICS  OF  RIGID  BODIES.  407 

duce  to  intersection  b,  the  polygon  is  closed  and  we  have  in  Fig.  2  the  stresses  in  ab  and 
Bb.  Since  AB  is  known  to  act  downward,  we  follow  round  in  Fig.  2,  from  A  to  B,  B  to  d, 
d  to  a,  and  a  to  y4,  and  transfer  the  directions  thus  obtained  to  the  apex  at  Flt  Fig.  i, 
under  consideration.  We  thus  obtain  the  stress  in  Bb  towards  the  apex  or  compression,  the 
stress  in  ba  towards  the  apex  or  compression,  and  the  stress  in  aA  towards  the  apex  or  com- 
pression, just  as  already  found. 

Note  that  in  the  first  case,  when  we  were  considering  the  apex  at  R^  ,  we  found  the 
stress  in  aA  acting  towards  that  apex.  Now  when  we  consider  the  apex  at  F^  we  find  the 
stress  in  aA  acting  towards  that  apex — in  both  cases,  then,  compression  (page  397). 

Let  us  now  consider  the  second  lower  apex,  Fig.  i.  We  have  here  no  outer  force,  but 
the  stresses  in  Ea,  ab,  be  and  cE  must  be  in  equilibrium  and  therefore  form  a  closed  polygon. 
But  in  Fig.  2  we  have  already  found  the  stresses  in  Ea  and  ab.  If  then  we  draw  from  b  a 
line  parallel  to  be  in  Fig.  i,  and  produce  it  to  intersection  c  withEa,  the  polygon  closes,  and 
we  have  in  Fig.  2  the  stresses  in  be  and  cE.  We  have  already  found  aE  to  be  tension.  It 
must  therefore  act  away  from  the  apex  we  are  considering.  We  therefore  follow  round  in 
Fig.  2,  from  E  to  a,  a  to  b,  b  to^r,  and  c  to  E,  and  transfer  the  directions  thus  found  to 
the  corresponding  members  in  Fig.  i.  We  thus  obtain  the  stress  in  Ea  tension  and  the  stress 
in  ab  compression  as  already  found,  and  the  stress  in  be  tension  and  in  cE  tension. 

Let  us  now  consider  the  top  apex.  We  have  here  the  force  F2  =  BC,  and  the  stresses 
in  Bb,  be,  cd,  and  dC,  in  equilibrium.  But  in  Fig.  2  we  have  already  laid  off  BC,  and  we 
have  found  the  stresses  in  Bb  and  be.  If  then  we  draw  from  c  and  C  lines  parallel  to 
cd  and  Cd'm  Fig.  i,  and  produce  to  intersection  d,  the  polygon  closes  and  we  have  in  Fig.  2 
the  stresses  in  cd  and  Cd.  Since  BC  acts  downwards,  we  follow  round  from  £  to  C,  C  to  d, 
d  to  c,  c  to  b,  and  b  to  B.  Transferring  these  directions  to  the  corresponding  members  in 
Fig.  i,  we  obtain  the  stress  in  Cd  compression  and  in  dc  tension,  while  the  stress  in  cb  is 
tension  and  in  bB  compression  as  already  found. 

We  can  thus  go  to  each  apex  and  find  the  stresses  in  every  member. 

The  lines  in  Fig.  2  which  thus  give  the  stresses  in  the  members  constitute  the  STRESS 
DIAGRAM.  Each  stress  having  letters  at  its  ends  in  Fig.  2  is  parallel  to  that  member  in  Fig.  i 
which  has  the  same  letters  at  its  sides. 

Apparent  Indetermination  of  Stresses. — It  sometimes  happens  that  a  frame  has  no 
superfluous  members,  and  yet  in  applying  the  graphic  method  we  are  unable  to  find  any  apex 
at  which  all  the  forces  but  two  are  known.  In  such  case  the  difficulty  may  be  overcome  by 
taking  out  one  or  more  of  the  members  and  replacing  them  by  another  member,  and  then 
applying  the  method  until  we  find  the  stress  in  some  member  which  is  not  affected  by  the 
change.  Or  we  may  find  the  stress  in  this  member  by  the  method  of  sections  (page  401). 
Having  found  this  stress,  we  can  replace  the  members  taken  out  and  find  the  actual  stresses. 

Thus  let  Fig.  i  (page  408)  be  a  frame*  acted  upon  by  the  forces  Fl  ,  F2  ,  Fz  ,  F4, 
etc.,  and  the  reactions  or  upward  pressures  of  the  supports  J?lt  Rz. 

Notate  the  frame  and  the  forces  by  letters  on  each  side  as  directed  (page  405). 

Then  lay  off  to  scale  the  outer  forces  in  Fig.  2,  thus  forming  the  force  polygon 
ABCD  .  .  .  HIA.  This  polygon  is  a  straight  line  in  this  case,  because  all  the  forces  are 
parallel,  and  it  must  close,  that  is,  the  outer  forces  are  in  equilibrium. 

We  can  now  proceed  to  find  the  stresses  as  follows : 

Consider  first  the  left-hand  apex,  Fig.  i.  At  this  point  we  have  the  reaction  I  A  and  the 
stresses  in  Aa  and  la  constituting  a  system  of  concurring  forces  in  equilibrium.  But  we  already 


*  Disregard  for  the  present  the  dotted  member  in  Fig.  I. 


4o8 


GRAPHICAL  STATICS-CONCURRING  FORCES. 


[CHAP.  II. 


have  IA  laid  off  in  Fig.  2.  If  then  we  draw  Aa  and  la  in  Fig.  2  parallel  to  la  and  Aa  in 
Fig.  I,  and  produce  to  intersection  a,  the  polygon  is  closed  and  we  have  in  Fig.  2  the  stresses 
in  Aa  and  la  to  the  same  scale  employed  in  laying  off  the  forces.  Since  I  A  acts  upwards,  we 


FIG.  i. 


FIG.  2. 


follow  round   from  7  to  A,  A  to  #,  and  a  to  /,  in  Fig.  2,  and  transfer  the  directions  thus 
obtained  for  Aa  and  al  to  the  corresponding  members  in  Fig.  I. 

We  have  then  the  stress  in  Aa  towards  the  apex  we  are  considering  or  compression  (  — ), 
and  the  stress  in  rz/away  from  that  apex  or  tension  (+). 

Considering  now  the  next  upper  apex,  we  have  here  the  force  AB  known,  the  stress  in 
Aa  already  found,  and  the  stresses  in  ab  and  Bb  unknown.  If  then  in  Fig.  2  we  draw  ab  and 
Bb,  thus  closing  the  polygon,  we  obtain  the  stresses  in  ab  and  Bb. 

Since  AB  acts  down,  we  follow  round  in  Fig.  2  from  A  to  Bt  B  to  b,  b  to  a,  and  a  back 
to  A,  and  transfer  the  directions  thus  obtained  to  the  corresponding  members  in  Fig.  I.  We 
have  then  the  stress  in  Bb  towards  the  apex  we  are  considering  or  compression  ( — ),  the 
stress  in  ba  towards  that  apex  or  compression  (  — ),  and  the  stress  in  aA  also  towards  that 
apex  or  compression  (— ),  just  as  we  have  already  found  it. 

Note  that  when  we  were  considering  the  apex  at  R^  ,  we  found  the  stress  in  aA  acting 
towards  that  apex.  Now  when  we  consider  the  apex  at  F^  we  find  the  stress  in  aA  acting 
towards  that  apex.  In  both  cases,  then,  compression  (page  397). 

We  can  now  consider  the  next  lower  apex,  where  we  have  the  stresses  in  fa,  ab,  fcand 
cl  in  equilibrium.  We  already  know  la  and  ab,  and  if  we  draw  in  Fig.  2  be  and  cl,  we 
obtain  the  stresses  in  be  tension  (+)  and  in  cl  tension  (+). 

Thus  far  there  has  been  no  difficulty  in  the  application  of  the  graphic  method.  But 
now  we  cannot  consider  the  next  upper  or  lower  apex,  because  at  each  we  have  more  than 
two  unknown  forces.  If  we  should  start  at  the  right  end,  we  should  soon  come  to  the  same 
difficulty  on  the  right  side.  Apparently  we  can  go  no  farther. 

The  number  of  members  is  27  (we  disregard  the  dotted  member  in  Fig.  i).  The 
number  of  apices  is  15.  We  have  then,  applying  the  criterion  for  superfluous  members 
(page  402),  m  =  2n  —  3.  There  are  then  no  superfluous  members. 

If  now  we  remove  the  two  members  de  and  r/and  replace  them  by  the  dotted  member 
c'f,  where  e  takes  the  place  in  the  new  notation  of  the  two  letters  e  and  d,  we  have  still  a 
rigid  frame  with  no  superfluous  mcmbcrs.  For  the  number  of  members  is  now  in  =  25,  and 
the  number  of  apices  is  n  =  14.  We  have  then  ;;/  =  2n  —  3. 

But  this  change  has  evidently  not  affected  tlie  stress  in  the  member  Ig.  We  can  therefore 
now  carry  on  the  diagram  until  we  find  the  stress  in  Ig,  or  we  may  compute  the  stress  in  Ig 
directly  by  the  method  of  sections  (page  401). 


CHAP.  II.]  STATICS  OF  RIGID  BODIES.  4°9 

Thus  if  we  now  consider  the  apex  at  F2 ,  Fig.  i,  we  have  at  this  point  the  stresses  in  the 
members  Bb,  be,  ce'  and  e  C,  and  the  force  BC,  all  in  equilibrium.  We  know  BC,  Bb  and 
be,  and  if  we  draw  in  Fig.  2  ce'  and  e'C,  we  obtain  the  stresses  in  e'C  compression  and  in  ce 
compression. 

We  can  then  pass  to  the  apex  at  F3 ,  Fig.  I,  where  we  know  all  the  forces  except  the 
stresses  in  D/and/e'.  We  draw  then  Df  and  /<?'  in  Fig.  2,  and  obtain  the  stresses  in  Df 
Compression  and  in  fe'  tension. 

We  can  now  pass  to  the  next  lower  apex,  where  we  have  the  stresses  in  Ic,  ce'  and  //, 
and  can  therefore  find  fg  and  Ig.  We  draw  then  fg  and  Ig  in  Fig.  2,  and  obtain  the  stresses 
in  fg  and  Ig  tension. 

We  have  thus  found  the  stress  in  the  member  Ig,  and  since  this  is  unchanged  by  the 
removal  of  the  members  de  and  ef,  we  can  now  replace  those  members  and  remove  e'f. 

We  can  now  consider  the  second  lower  apex  and  find  the  stresses  in  cd  and  dg,  and  can 
then  pass  to  the  apex  at  Fs  and  find  the  stresses  in  ef  and  Df,  and  so  on.  We  can  thus  find 
the  stress  in  every  member  of  the  frame,  and  there  is  no  real  indeterminateness. 

Remarks  upon  the  Method. — The  method  just  illustrated  we  may  call  the  "graphic 
method  by  resolution  of  forces  "  The  student  will  note  that  he  must  always  know  all  but  two 
of  the  forces  concurring  at  any  apex  before  he  can  consider  that  apex. 

It  is  evident  that  if  the  frame  is  completely  divided  into  two  portions  by  cutting  the 
members,  the  stresses  which  existed  in  the  cut  members  before  the  section  was  made  must 
hold  in  equilibrium  the  outer  forces  acting  upon  each  portion  of  the  frame  (page  401). 

This  is  at  once  made  evident  by  Fig.  2,  page  405. 

Thus  suppose  a  section  cutting  the  members  Bb,  be  and  cE,  Fig.  I,  and  thus  dividing 
the  frame  into  two  portions.  We  see  from  Fig.  2,  page  406,  that  the  stresses  in  the  cut 
pieces  make  a  closed  polygon  with  EA  and  AB,  the  outer  forces  on  the  left-hand  portion,  or 
with  BC,  CD  and  DE,  the  outer  forces  on  the  right-hand  portion. 

If  we  solve  the  triangles  in  Fig.  2,  page  406,  we  obtain  algebraic  expressions  for  the 
stresses  identical  with  those  obtained  by  the  "algebraic  method  by  resolution  of  forces" 
(page  400). 

Thus,  since  the  algebraic  sum  of  the  horizontal  and  vertical  components  of  the  forces 

jr> 

acting  at  each  apex  must   be  zero,  we  have  -{-  R  -\-  Aa  cos  a  —  o,  or  Aa  =  -         —  ,  where 

COS  IX 

a  is  the  angle  of  the  rafter  with  the  vertical.  We  get  the  same  result  at  once  from  Fig.  2  by 
solving  the  triangle  AaE.  In  the  same  way  we  have  at  once,  from  Fig..  2,  ab  =  —  Fl  cos  ft, 
where  ft  is  the  angle  of  ab  with  the  vertical. 

We  see  also  from  Fig.  2,  page  406,  other  relations.  Thus  we  see  that  the  stress  in  ab 
will  be  the  least  possible  when  ab  is  perpendicular  to  the  rafter.  We  also  see  at  a  glance 
how  the  stress  in  any  member  is  affected  by  a  change  of  inclination  of  the  member. 

Finally,  the  application  of  the  method  is  equally  simple  no  matter  how  irregular  the 
frame  may  be. 

If  the  frame  is  symmetrical  with  respect  to  the  centre,  and  the  forces  Flt  F3  in  Fig.  2 
(page  406)  are  equal,  it  is  evident  that  the  stresses  in  each  half  will  be  the  same.  We  have 
then  Cd  =  Bb,  cd  —  cb,  and  so  on. 

Choice  of  Scales,  etc. — In  general  the  larger  the  frame  is  drawn  in  Fig.  I,  the  better, 
as  it  then  gives  more  accurately  the  direction  of  the  members  composing  it. 

The  force  polygon  Fig.  2,  on  the  other  hand,  should  be  taken  to  no  larger  scale  than 
consistent  with  scaling  off  the  forces  to  the  degree  of  accuracy  required,  so  as  to  avoid  the 
intersection  of  very  long  lines,  where  a  slight  deviation  from  true  direction  multiplies  the 


4io 


GRAPHICAL   STATICS.— CONCURRING   FORCES. 


[CHAP.  II. 


error.  If  an  error  of  one  twenty-fifth  of  an  inch  is  considered  the  allowable  limit,  the  scale 
should  be  so  chosen  that  one  twenty-fifth  of  an  inch  shall  represent  a  small  number  of 
pounds,  within  the  degree  of  accuracy  required. 

The  stress  polygon  Fig.  2  should  be  completely  finished  and  the  signs  for  tension  (-{-) 
and  compression  (  — )  placed  on  the  frame  for  each  member  as  its  stress  is  found,  to  avoid 
confusion,  before  the  stresses  are  taken  off  to  scale.  A  good  scale,  dividers,  straight-edge, 
triangle,  and  hard  fine-pointed  pencil  are  all  the  tools  required.  The  work  should  be  done 
with  care,  all  lines  drawn  light,  points  of  intersection  accurately  located  and  the  frame 
properly  notated  to  correspond  with  the  force  polygon.  Care  should  be  exercised  to  secure 
perfect  parallelism  in  the  lines  of  the  frame  and  stress  polygon.  Some  practice  is  necessary 
in  order  to  obtain  close  results.  It  should  be  remembered  that  careful  habits  of 
manipulation,  while  they  tend  to  give  constantly  increased  skill  and  more  accurate  results, 
affect  very  slightly  the  rapidity  and  ease  with  which  these  results  are  obtained. 


FIG.  i. 


Examples.— (i)  A  roof-truss  has  a  span  of  50  feet  and  rise  of  12.5  feet.  Each  rafter  is  divided  into  four 

equal  panels,  and  the  Imver  horizontal  tie  into  six  equal  panels.  The 
bracing  ts  as  shown  in  the  figure.  A  weight  of  800  Ibs.  is  sustained  at 
each  upper  apex.  Find  the  stresses. 

ANS.  Draw  the  frame  in  Fig.  i  to  a  scale  of,  say,  12  feet  to  an 
inch,  and  notate  it.  Then  construct  the  force  polygon  ABC. ..HI A, 
Fig.  2. 

Note  that  7v's  or  HI  and  K\  or  I  A  are  equal  and  eacli  2800  Ibs. 
The  force  polygon  then  closes  as  it  should.  We  can  take  the  scale 
of  Fig.  2  as  3200  Ihs.  to  an  inch.  Then  an  error  of  2'y  of  an  inch  will 
be  about  128  Ibs. 

We  can  then  find  the  stresses  as  shown  in  Fig.  2. 


Aa  Bb  Cd  Df  la 

—  6280     —  5816     —  4700     —  3580     —  5624 


Ic 
4832 


le  ab 

+  4024     -  720 


be 
+  720 


cd 
—  1060 


de  ef 

928     -  1452 


fg 
+  2400  Ibs. 


The  accurate  results  as  found  by  computation  (page  402)  are 
—  6260     -  5813     —  4696     —  3577      +  5600     +  4802     +  4003     —  720 
+  720     —  1081      +  920     —  1443     +  2401  Ibs. 

It  will  be  seen  that  the  greatest  error  is  only  30  Ibs.     The  above 
results  were  actually  obtained    from   the  diagram,  using  the  scales 
FIG.  2.  given. 

(2)  Sketch  the  stress  diagram  for  a  roof-truss  as  shown  in  the  following  Fig.  i,  equal  forces  acting  at 
every  upper  and  lower  apex. 

ANS.  The  student  should 
note  that  the  reactions  DE  and 
GA  are  each  equal  to  half  the 
sum  of  the  downward  forces,  or 
2$  forces. 

We  lay  off  then  in  Fig.  2 
AB,  BC.  CD  downwards.  Then 
DE  upwards  equal  to  2J  forces. 
Then  EF,  FG  downwards. 
Then  GA  upwards  equal  to  2| 
forces,  and  closing  the  force 


polygon. 

The  stresses  can  now  be  found  as  always. 


FIG.  i. 


FIG.  2. 


CHAP.  II.] 


STATICS   OF  RIGID  BODIES. 


411 


(3)    We  give  in  the  following  fi^ttres  a  number  of  frames  with  their  stress  diagrams*     For  the  sake  of 
generality  the  outer  forces  and  reactions  are  often  taken  inclined  as  well  as  vertical. 


rig.  i. 


Fig.  2. 


Fig.  3. 


Fig.  4, 


Fig.  5. 


Fig.  6. 


Fig.  9. 


*The  student  should  sketch  the   stress  diagrams   for  himself  in  each  case,  putting  down,  as  he  goes  alcng, 
the  sign  (— )  and  (-f-)  for  compression  and  tension  upon  each  member  of  the  frame  as  soon  as  he  finds  it. 


CHAPTER   III. 


GRAPHICAL  STATICS.     NON-CONCURRING   FORCES. 

Non-concurring  Forces. — Let  the  co-planar  forces  Flt  F2,  F^,  F4,  etc.,  act  at  the  points 
Al}  At,  A^  A4  of  any  rigid  body,  Fig.  i. 

If  we  lay  off  the  forces  to  scale  in  Fig.  2,  we  have  as  before  the  force  polygon  01234, 

and  the  closing  line  o  4  gives  as  before  the  resultant.      If  this  resultant  acts  in  the  direction 

FIG.  i. 

X 


B 


\ 


4  o  upon  the  rigid  body,  it  will  hold  the  given  forces  in  equilibrium.  If  it  acts  in  the  direc- 
tion o  4,  it  will  replace  the  given  forces. 

We  thus  know  the  magnitude  and  direction  of  the  resultant.  But  'imposition  in  the 
plane  of  the  forces  in  Fig.  I  is  as  yet  unknown. 

In  order  to  determine  this,  choose  any  point  O  in  Fig.  2,  and  draw  the  lines  Oo  and  (^4. 
This  point  O  we  call  the  POLE  of  the  force  polygon.  Now  since  every  line  in  the  force 
polygon  represents  a  force,  by  thus  choosing  a  pole  O  and  drawing  lines  Oo,  04  to  the 
extremities  of  the  resultant  O  4,  we  have  resolved  the  resultant  into  the  two  forces  represented 
by  Oo  and  $4.  This  is  evident  from  the  fact  that  these  two  lines  make  a  closed  polygon 
with  O  4,  and  hence  taken  as  acting  from  4  to  O  and  O  to  O,  as  shown  by  the  arrows,  hold 
the  forces  /*",,  F2,  F3,  Ft  in  equilibrium,  or  replace  the  resultant  4  o  (page  404).  As  the  pole 
O  is  taken  anywhere  we  please,  we  can  thus  resolve  the  resultant  4  o  for  equilibrium  into 
forces  in  any  two  directions  we  wish. 

Let  us  then  consider  the  resultant  40  for  equilibrium,  replaced  by  the  two  forces  4<? 
and  Oo.  Anywhere  in  the  plane  of  the  forces  in  Fig.  I  draw  a  line  s0  parallel  to  Oo  and 
produce  it  till  it  meets  /*",,  produced  if  necessary,  at  a. 

If  then  we  take  s0  and  Fl,  Fig.  I,  as  acting  at  a,  their  resultant  will  pass  through  a  and 
be  parallel  to  j,  in  the  force  polygon  Fig.  2,  because  sl  in  the  force  polygon  is  the 
resultant  of  T7,  and  J0,  since  it  closes  the  polygon  for  those  forces.  Through  a  in  Fig.  i, 
then,  draw  a  line  parallel  to  j,  and  produce  it  to  intersection  b  with  F2,  produced  if  necessary. 
The  line  s2  in  the  force  polygon  is  the  resultant  of  s}  and  F2.  Parallel  to  this  line  then  draw 
s2  through  b,  Fig.  i,  and  produce  to  intersection  c  with  F3,  produced  if  necessary.  The 
line  s3  in  the  force  polygon  is  the  resultant  of  s2  and  F3.  Parallel  to  this  line  then  draw 
ss  through  c,  Fig.  i,  and  produce  to  intersection  d  with  Ft,  produced  if  necessary.  Finally 
through  d  in  Fig.  I  draw  a  line  st  parallel  to  st  in  the  force  polygon. 


CHAP.  III.]  GRAPHICAL   STATICS— NON-CONCURRING  FORCES.  413 

We  thus  find  for  any  assumed  position  of  SQ  in  the  plane  of  the  forces  in  Fig.  I  the 
proper  corresponding  position  of  sr  Since  now  s0  and  J4  are  components  of  the  resultant  in 
proper  position  and  each  may  be  considered  as  acting  at  any  point  in  its  line  of  direction, 
we  have  only  to  prolong  them,  and  their  intersection  gives  a  point  e  on  the  line  of  direction  of 
the  resultant. 

We  prolong  s0  and  J4,  then,  in  Fig.  I  to  intersection  e.  The  line  of  direction  of  the 
resultant  passes  through  e.  Acting  in  the  direction  from  4  to  o,  it  will  hold  the  forces  in 
equilibrium.  We  thus  know  the  magnitude,  direction  and  position  of  the  resultant  for 
equilibrium. 

Position  of  Pole  and  Of  SQ  Indifferent. — The  method  is  evidently  general  no  matter 
where  in  the  plane  of  the  forces  in  Fig.  I  we  take  SQ  as  acting,  and  no  matter  where  we  take 
the  pole  in  Fig.  2. 

Pole,  Equilibrium  Polygon,  Rays,  Closing  Line. — The  point  O  we  call  the  POLE  in  the 
force  polygon.  It  may  be  taken  where  we  please.  The  polygon  abed  in  Fig.  i  we  call  the 
EQUILIBRIUM  POLYGON,  and  ab,  be,  cd,  etc.,  are  its  segments.  In  the  present  case  it  is 
FIG.  i.  FIG.  2. 


evidently  the  shape  a  string  would  take  if  suspended  at  any  two  points  as  A  and  B ',  in  Fig.  i, 
on  s0  and  s^  The  stresses  in  the  segments  would  be  tensile.  These  stresses  are  given  by 
the  lines  Oo,  O\,  O2,  in  the  force  polygon,  and  we  call  these  lines  RAYS.  In  general  forces 
may  act  up  as  well  as  down,  in  which  case  some  of  the  segments  would  sustain  compressive 
stresses  and  our  equilibrium  polygon  would  contain  struts  as  well  as  ties. 

Let  us  take  any  two  points,  as  A  and  B,  upon  the  end  segments  s0  and  s4,  Fig.  I,  and 
suppose  them  fixed.  The  force  SQ  acting  at  A  we  shall  then  have  to  replace  by  two  forces, 
one  parallel  to  the  resultant  and  one  in  the  direction  AB.  So  also  for  s4  at  B.  The  sum  of 
the  two  components  parallel  to  the  resultant  must  be  equal  and  opposite  to  the  resultant, 
and  the  component  in  the  direction  AB  must  be  resisted  by  a  strut  or  compression  member 
AB.  This  resolution  we  make  at  once  by  drawing  through  O  in  the  force  polygon  a  line 
OL  parallel  to  AB.  The  line  AB  we  call  the  CLOSING  LINE.  Thus  we  see  from  Fig.  2 
that  the  sum  of  the  components  ^L  and  Lo  equals  the  resultant. 

In  any  case,  then,  we  can  fix  any  two  points  of  the  equilibrium  polygon,  as  A,  B,  by 
drawing  the  closing  line  AB.  A  line  OL  through  O  parallel  to  AB,  in  the  force  polygon, 
gives  the  components  into  which  sa  and  st  are  resolved. 

We  can  then  consider  the  entire  polygon  AabcdB,  with  its  closing  line  AB,  as  a  frame 
in  equilibrium  with  the  given  forces,  and  can  apply  to  it  the  principles  of  page  406. 

Thus  take  the  apex  A.  Here  we  have  the  reaction  ~R^  —  Lo  in  equilibrium  with  the 
stresses  in  AB  and  Aa.  Following  round  in  the  force  polygon  from  L  to  o,  o  to  O,  and'  O 
to  L,  and  transferring  these  directions  to  the  apex  A,  we  find  50  away  from  A  or  tension, 
and  OL  towards  A  or  compression,  just  as  on  page  406. 


414 


ST/IT1CS  OF  RIGID  BODIES. 


[CHAP.  III. 


So  also  at  the  other  apex  B  we  have  R2  =  ^L  in  equilibrium  with  the  stresses  in  AB 
and  Bd.  Following  round  in  the  force  polygon  from  4  to  L,  L  to  O,  and  O  to  4,  we  find  $i 
away  from  B  or  tension,  and  LO  towards  B  or  compression,  as  before.  The  components 
J?l  and  R2  act  opposite  to  the  resultant  04  which  replaces  the  forces,  and  are  equal  to  it  in 
magnitude.  The  forces  at  A  and  B  parallel  to  OL  are  equal  and  opposite.  Hence  the 
frame  is  in  equilibrium. 

Recapitulation. — Our  method,  then,  is  as  follows: 

1st.  Draw  the  force  polygon  by  laying  off  the  forces  to  scale  one  after  the  other,  in 
any  order.  The  line  which  closes  this  polygon  gives  the  resultant  in  magnitude  and 
direction.  When  it  is  taken  as  acting  in  the  direction  obtained  by  following  round  the 
force  polygon  in  the  direction  of  the  forces,  it  will  cause  equilibrium.  In  the  opposite 
direction  it  replaces  the  forces. 

2d.   Choose  a  pole  (9,  and  draw  the  rays  s0,  slt  s2,  etc. 

3d.   Draw  the  equilibrium  polygon. 

4th.  Fix  any  two  points  in  the  end  segments  of  the  equilibrium  polygon  by  drawing 
the  closing  line  of  the  equilibrium  polygon  between  those  two  points. 

5th.  A  line  drawn  in  the  force  polygon  parallel  to  the  closing  line  of  the  equilibrium 
polygon  will  divide  the  resultant  into  the  two  reactions  at  the  ends.  We  thus  have  a  frame 
the  stresses  in  which  can  be  found  as  on  page  406. 

Graphic  Construction  for  Centre  of  Parallel  Co-planar  Forces. — Let  Fv  F2,  F3, 
etc.,  be  parallel  co-planar  forces  acting  at  the  points  Alt  A2,  A3,  etc.,  of  a  rigid  body. 

We  construct  the  force  polygon  Fig.  2  by  laying  off  the  forces  Flt  F2,  F3,  etc.  The 
resultant  is  then  the  algebraic  sum  of  the  forces  and  parallel  to  them. 

Then  choose  a  pole  O,  and  draw  the  rays  J0,  sl,  s2,  s3,  etc. 

Anywhere  in  the  plane  of  the  forces,  Fig.  i,  we  draw  a  line  parallel  to  s0  to 
intersection  a  with  Fl ;  then  ab  parallel  to  sl  to  intersection  b  with  F2;  then  be  parallel  to 
J2  to  intersection  c  with  F3;  then  s3  through  c  parallel  to  s3  in  Fig.  2. 

The  intersection  d  of  s0  and  s3  is  a  point  on  the  resultant  which  therefore  has  the 
direction  and  position  dC- 

Now  suppose  the  forces /^ ,  F2,  F3,  etc.,  all  turned  in  the  same  direction  through  a 
right  angle. 

FIG. 


FIG.  2. 

Oh 


Draw  the  new  equilibrium 
polygon  s0'a'&'c's3,  whose  sides 
are  respectively  perpendicular 
to  those  of  the  first. 

The  intersection  d'  of  s0' 
and  s3  is  a  point  on  the  result- 
ant, which  therefore  has  the 
direction  and  position  d'C. 

The  intersection  C  of  the 
two  resultants  gives  the  centre 
of  force  for  the  system  (page   189). 

COR.  The  same  construction  evidently  determines  the  centre  of  mass  (page  21),  if  we 
divide  a  body  into  a  convenient  number  of  portions,  and  take  the  weight  of  each  portion, 
Flt  /%,  F3,  etc.,  acting  at  the  centre  of  mass  of  that  portion. 

Properties  of  the  Equilibrium  Polygon. — The  equilibrium  polygon  has  many  interest- 
ing properties.  We  shall  call  attention  to  only  two. 

1st.   As  we  have  seen,  the  intersection  of  any  two  segments  is  a  point  in  the  resultant 


CHAP.  III.] 


STATICS  OF  RIGID  BODIES. 


415 


FIG.  i. 


FIG.  2. 


lA 


FIG.  i. 


of  the  forces  included  between  those  segments.      Thus  in  the  preceding  Fig.  i  the  intersec- 
tion d  of  SQ  and  s3  is  a  point  on  the  resultant  of  Fl ,  F2  and  Fy 

2d.  Let  s^ab,  Fig.  I,  be  a  portion  of  the  equilibrium  polygon,  and  Fig.  2  its  corre- 
sponding force  polygon. 

Take  any  line  fe  in  Fig.  I  parallel  to  Fl , 
and  draw  the  perpendicular  cd  =  x. 

Let  de  =  y  be  the  ordinate  between  s0 
and  sr 

In  the  force  polygon  Fig.  2,  draw  the 
perpendicular  OH  =  H  from  the  pole  to  01. 
This  is  called  the  POLE  DISTANCE  of  Fr  NS*\~ 

Then  by  similar  triangles  we  have 

y  :  x  : :  Fl  :  H,     or     F^x  =  Hy. 

But  F^x  is  the  moment  of  Fl  with  reference  to  any  point  on  the  line  fe. 
Hence  the  moment  of  any  force,  as  Fl,  with  reference  to  any  point  is  equal  to  the  ordinate 
through  this  point  parallel  to  Fl ,  included  between  the  segments  of  the  equilibrium  polygon 
which  meet  at  Fv,  multiplied  by  the  pole  distance  of  Fl  in  the  force  polygon. 

Application  to  Parallel  Forces. — The  outer  forces  acting  upon  framed  structures  are 
generally  weights  and  reactions  of  supports  due  to  these  weights.  We  have  then  in  general 
to  investigate  a  system  of  parallel  forces. 

Let  Flt  F2,  F3,  Fig.   I,  be  vertical  forces  acting  upon  a  rigid  body  or  frame. 

Lay  off  the  force  poly- 
gon 0123,  Fig.  2.  Choose 
a  pole  O,  and  draw  the  rays 

Then  in  the  plane  of  the 
forces  Fig.  i  draw  J0  to  meet 
Fl  at  a\  then  sl  through  a  to 
meet  F2  at  b\  then  s2  through 
b  to  meet  F3  at  c ;  and  finally 
Sy  We  thus  have  the  equilib- 
rium polygon  s^abcSy  We  see 
that  the  horizontal  component 
of  the  stress  in  any  segment  is  constant  and  equal  to  OH. 

Drop  verticals  through  A  and  B  which  meet  the  end  segments  s0  and  s3  in  A'  and  B'. 
If  we  fix  the  points  A' ' ,  B'  by  drawing  the  closing  line  A ' B' ' ,  the  reactions  at  A ' ,  B'  will  be 
the  reactions  at  A  and  B  of  the  frame. 

Therefore  in  Fig.  2  draw  OL  parallel  to  A' B'  and  we  have  Lo  =  Rlt  and  $L  =  R^. 
Draw  the  pole   distance  OH.      Through   the  apex  K  of  the   frame   drop  the  vertical 
Kkmn.     Then,  as   just   proved,  OH  (to  scale   of  force)  X  kn   (to   scale   of   distance)  =  the 
moment    of  Rv      Again,  OffX^n  =  the   moment   of  Fr      The  resultant   moment    is  then 
given  by  OH  X  (kn  —  ntn)  or  OH  X  km. 

That  is,  for  parallel  forces  the  pole  distance  multiplied  by  the  ordinate  of  the  equilibrium 
polygon  at  any  point,  parallel  to  the  forces  included  between  the  closing  'line  and  the  polygon, 
giires  the  resultant  moment  of  all  the  forces  on  either  side  of  the  ordinate  with  reference  to  any 
point  in  tJiat  ordinate. 

If-then  we  make  a  section  cutting  EK,  CK  and  CD,  and  take  the  centre  of  moments  at 


FIG.  2. 


sb 


4i6 


GRAPHICAL  ST /tTICS- NON-CONCURRING  FORCES. 


[CHAP.  III. 


K,  we  have  (page  401)  stress  in  CD  X  lever-arm  for  CD  =  algebraic  sum  of  moments  of  Rl 
and  Fl  with  reference  to  K.     But  this  algebraic  sum  we  have  just  seen  is  given  by  H  X  km. 

Hence  stress  in  CD  is  equal  to  ,—  ^~  . 

lever-arm  for  CD 

We  can  therefore  find  the  moment  graphically  at  any  point  by  multiplying  the  ordinate 
to  the  equilibrium  polygon  at  that  point  by  the  pole  distance. 

A  few  examples  will  make  the  application  of  the  preceding  principles  clear. 

Ex.  i.  Let  AB,  Fig.  I,  be  a  beam  or  rigid  body  or  framed  structure  subjected  to  two 
unequal  weights  T7,  and  F2  applied 
at  any  two  given  points.  Re- 
quired the  reactions  at  the  sup- 
ports A  and  B,  also  the  moment 
at  any  point  of  all  the  forces  right 
or  left  of  that  point  when  equilib- 
rium exists. 

Draw  the  force  polygon  Fig. 
2,  choose  a  pole  O,  and  draw  J0, 
slt  J2,  and  the  pole  distance  H. 

Construct  the  equilibrium  polygon,  Fig. 


FIG. 


with  FI\    through  a  a  parallel  to  j,   to   intersection  b  with  F2;   through 


by  drawing  a  parallel  to  s0  to  intersection  a 
b  a  parallel  to  sy 

Drop  verticals  from  A  and  B,  and  draw  the  closing  line  A' B' .  Parallel  to  A' B'  draw  OL 
in  Fig.  2. 

Then  Lo  and  2L  are  the  reactions  at  A  and  B\  and  since  they  act  upwards,  the  supports 
must  be  below  A  and  B. 

The  moment  at  any  point  k  is  equal  to  the  ordinate  kn  multiplied  by  the  pole 
distance  H. 

Ex.  2.  It  is  well  to  observe  that  the  order  in  which  the  forces  are  taken  makes  no 
difference  in  the  results,  although  the  figure  obtained  may  be  very  different. 


FIG.  i. 


FIG.  2. 


MI 

1 
i 

1  |K 

IF,  :* 

IF-  1 

Thus  take  the  same  example 
as  before,  but  number  the  forces 
in  inverse  order,  Fig.  i. 

We  form  the  force  polygon 
as  before,  choose  a  pole  and  draw 
s0 ,  j,  ,  s2.  Now  parallel  to  s0  we 
draw  a  line  till  it  meets  Fv  at  a 
[note  that  s0  must  always  be  pro- 
duced to  meet  F^\ ;  then  from  a  a 
parallel  to  5,  till  it  meets  F2  at  b;  then  from  b  a  parallel  to  s2.  Draw  the  closing  line 
A'B'.  A  parallel  to  it  in  Fig.  2  gives  the  reactions  Lo  and  2L  as  before.  At  apex  b  of  the 
equilibrium  polygon  we  find  s2  tension,  since  F2  acts  downwards.  At  apex  a  we  find  s0 
tension,  since  Fl  is  downward.  Hence  at  A',  SQ  acts  away  from  A',  and  following  round  in 
the  force  polygon  we  obtain  Lo  acting  upwards.  At  B1 ',  s2  acts  away,  and  hence  2L  acts 
upwards  also.  The  supports  at  A  and  B  must  then  be  below. 

As  to  the  moments,  the  moment  of  the  reaction  at  A  with  reference  to  any  point  k  is 
H  X  kn.  The  moment  of  F2  is  —  H  X  np.  The  resultant  moment  is  H  X  (km  —  np\  The 
lower  ordinates  subtracted  from  the  upper  will  give  us  the  same  figure  as  before. 

Whenever,  then,  we  obtain  a  double  figure  as  in  the  present  case,  it  shows  that  we  have 


CHAP.  III.] 


STATICS   OF  RIGID  BODIES. 


417 


FIG.  i. 


taken  the  forces  in  inconvenient  order.      We   have  only  to   change  the  order  to  obtain  the 
moments  directly  from  the  equilibrium  polygon. 

Closing  Line  at  Right  Angles  to  the  Forces — Choice  of  Pole  Distance. — It  makes  no 
difference  what  inclination  the  closing  line  may  have,  because,  as  we  have  seen,  the  ordinate 
in  the  equilibrium  polygon  parallel  to  the  resultant,  multiplied  by  the  pole  distance,  gives 
the  resultant  moment,  with  reference  to  any  point  on  that  ordinate,  of  all  the  forces  right  or 
left. 

We  can,  however,  if  we  wish,  always  cause  the  closing  line  to  be  at  right  angles  to  the 
parallel  forces.  We  have  only  to  find  first  by  preliminary  construction  the  reactions  or  the 
point  L.  If  then  we  take  a  new  pole  anywhere  in  a  line  through  this  point  at  right  angles 
to  the  forces,  the  closing  line  will  be  at  right  angles  to  the  forces. 

As  to  choice  of  pole  distance,  we  have  only  so  to  choose  the  position  of  the  pole  as  to 
give  good  intersections  for  the  polygon.  The  multiplication  may 'be  directly  performed  by 
properly  changing  the  scale  in  the  equilibrium  polygon.  The  ordinate  to  this  new  scale  will 
then  give  the  moment  at  once.  Thus  if  our  scale  of  length  in  Fig.  I,  preceding,  is 
five  feet  to  an  inch,  and  the  pole  distance  in  the  force  polygon  Fig.  2,  measured  to  the 
scale  of  force  adopted,  is  ten  pounds,  we  have  only  to  take  fifty  moment  units  to  an  inch  as 
the  scale  for  the  ordinates  and  they  will  give  the  moments  directly. 

Ex.  3.  Let  the  single  weight 
Fl  act  at  any  point  of  the  rigid  body 
AB.  Then  the  equilibrium  poly- 
gon is  A'aB'.  The  vertical  reac- 
tions at  A  and  B  are  Lo  and  iL, 
both  acting  up,  and  hence  the  sup- 
ports are  below  A  and  B. 

We  see  at  once  that  the 
moment  is  greatest  at  the  weight  and  decreases  to  zero  at  each  support. 

Ex.  4.  Let  Fl  act  outside  of  the  supports  A  and  B.  Observe  in  constructing  the  equi- 
librium polygon  that  SQ  is  always  produced  till  it  meets  F^\  also  that  the  closing  line  A' B' 

FIG.  2.  always  unites  the  two  points  ver- 

tically  under  the  supports,    upon 
the  two  end  segments. 

The  reactions  require  special 
notice.  Thus  the  reaction  RI  at  B 
is  the  resultant  of  the  stresses  in 
aB'  and  B' A' ,  or  iL  in  the  force 
polygon.  The  reaction  ./?,  at  A 
is  the  resultant  of  the  stresses  in 
A' a  and  A'B',  or  Lo  in  the  force 
polygon. 

Since  Fl  acts  downwards  at  apex  a,  we  have  ^  compression  and  s0  tension.  Therefore 
at  apex  A'  we  take  s0  acting  away,  and  hence  obtain  Lo  acting  down,  or  the  support  is 
above  A. 

At  apex  B'  we  take  ^  acting  towards  Bv  and  hence  obtain  iL  acting  up,  or  the  support 
is  below  B. 

Ex.  5.  ONE  DOWNWARD  AND  ONE  UPWARD  FORCE  BETWEEN  THE  SUPPORTS. — Here 
we  need  only  call  special  attention  to  the  fact  that  as  F2  acts  up  and  is  less  than  Flt  s2  in  the 
force  polygon  Fig.  2  lies  between  SQ  and  sl . 


GRAPHICAL  STATICS.— NON-CONCURRING  FORCES. 


[CHAP.  III. 


The  reaction  at  A  is  the  resultant  of  s0  and  L  or  Lo. 


of  sa  and  L  or  L2.     Since 
\*» 


FIG.  i. 


The  reaction  at  B  is  the  resultant 

is  down  at  a,  we  have  s0  tension,  and  since  F2  is  up  at  b,  we 
have  s2  tension.  At  apex  A',  then, 
s0  acts  away,  and  hence  L  is  com- 
pression and  Lo  acts  upwards  and 
support  at  A  is  below.  At  apex  B', 
s2  acts  away,  and  L  is  compression  as 
before  and  "2.L  acts  downwards,  or 
support  at  B  is  above. 

We  see  also  that  if  F2  were  less, 
FIG.  2.  so  that  2  falls  below  L  in  the  force 

polygon,  the  reaction  at  B  would  be  upward  also,  and  the  support  would  then  have  to  be 
belou'.     The  student  should  sketch  the  case  for  F2  greater  than  Fr 

At  the  point  A"  we  see  that  the  moment  is  zero.  If  AB  is  a  beam,  the  point  K  is  the 
"point  of  inflection,"  or  the  point  at  which  the  curve  of  deflection  of  the  beam  changes 
from  concave  to  convex.  The  beam  would  be  concave  upwards  as  far  as  K,  and  from  there 
on  convex  upwards. 

Ex.  6.  In  the  preceding  case,  let  the  forces  be  equal.  Laying  off  the  force  polygon 
Fig.  2,  the  first  force  extends  from  o  to  I,  and  the  second  from  I  back  to  o.  Choosing  a 
pole  O,  and  drawing  s0,  sl  ,  s2,  we 
find  that  s0  and  s2  coincide. 

Constructing  the  equilibrium 
polygon  and  drawing  the  closing 
line  A' B'  and  its  parallel  L  in  the 
force  polygon,  we  see  that  the 
reaction  at  A  or  the  resultant  of  s0 
and  L  is  Lo,  and  the  reaction  at  B 
or  the  resultant  of  s2  and  L  is  also 
Lo.  The  reactions  are  therefore 


FIG. 


FIG.  2. 


equal.     Since  SQ  and  s2  are  both  tension,  we  have  reaction  at  A  upward,  or  support  below  A, 
and  reaction  at  B  downward,  or  support  above  B. 

This  is   in   accord  with   the  principle   (page  186)  that  a   couple   can   only  be  held   in 
equilibrium  by  another  couple.      Moreover,  the  resultant  of  SQ  and  s2  in  Fig.  2  is  zero,  and 
the  point  of  application  is  at  the  intersection  of  s0  and  s2  in  Fig.  I,  or  at  an  infinite  distance. 
That  is,  the  resultant  of  a  couple  is  zero  at  an  infinite  distance  (page  186). 
At  K  the  moment  is  zero  as  before,  and  we  have  a  point  of  inflection. 
Ex.  7.    Two   EQUAL   WEIGHTS    BEYOND    THE    SUPPORTS.  —  The   figure  needs   no 
explanation,  except  to  call  attention  to  the  reactions. 

Thus  the  reaction  at  A  is  Lo 
acting  down.  At  B  it  is  2L  acting 
up. 

The  moment  at  any  point,  in 
all  cases,  is  the  ordinate  multiplied 
by  the  pole  distance  H.  The  shaded 
areas  then  show  how  the  moments 
vary. 

We  repeat  here  that  the  order 


B' 
Fie.  i. 

in  which  the  forces  are  taken,  .n 


FIG.  2. 


all   cases,  as  also  the  position   of  the  pole,  is  indifferent. 


CHAP.  III.] 


GRAPHICAL  STATICS— NON-CONCURRING  FORCES. 


419 


The    student    will    do    well    to    work   out   cases   to  scale   and   satisfy  himself   that   this   is 
true. 

Ex.  8.  Two  EQUAL  AND  OPPOSITE  FORCES  BEYOND  THE  SUPPORTS. — Observe  that 


FIG. 


FIG.  2. 


S0  is  produced  till  it  intersects  Fl  at 
a  in  Fig.  I  ;  then  sl  from  a  to  b ; 
then  s2  parallel  to  s2  or  s0  in  Fig.  2. 
The  closing  line  A' B'  is  then  drawn. 
A  parallel  to  it  in  Fig.  2  gives  L. 

The  reaction  at  A  is  Lo  acting 
down,  and  at  J3,  oL  acting  up. 

Between  £  and  F9  the  moment 
is  constant.  This  is  the  graphic 
interpretation  of  the  principle,  page 
185,  that  the  moment  of  a  couple  is  constant  for  any  point  in  its  plane. 

Ex.  9.  A  UNIFORMLY  DISTRIBUTED  LOAD. — Let  the  load  be  uniformly  distributed. 
We  might  consider  it  as  a  system  of  equal  and  equidistant  weights  very  close  together. 

Thus  in  Fig.  I  the  load  area,  which  is  a  rectangle  of  uniform  density,  whose  height  is 
the  load  per  unit  of  length,  and  whose  length  is  AB,  may  be  divided  into  any  number  of 

equal  parts.  The  weight  on  each 
of  these  parts  acts  at  its  centre  of 
mass.  We  can  then  lay  off  the  force 
polygon  Fig.  2.  Since  the  reactions 
at  A  and  B  are  equal,  we  take  the 
pole  in  a  horizontal  through  the 
middle  point  of  the  force  line.  The 
closing  lineA'B'  will  then  be  parallel 
to  AB  (page  4 1 7).  We  can  then  draw 
s0 ,  slt  s2,  etc.,  and  construct  the 
equilibrium  polygon.  It  is  evident  that  the  points  a,  b,  c,  d,  etc.,  will  enclose  a  curve 
tangent  to  ab,  be,  cd,  etc.,  at  the  points  midway  between,  that  is,  where  the  lines  of  division 
of  the  load  area  meet  the  sides  of  the  equilibrium  polygon. 

The  ordinates  to  this  curve,  multiplied  by  the  pole  distance  H,  give  the  moment  at 
any  point  on  the  ordinates. 

It  will  be  seen,  however,  that  this  method  is  deficient  in  accuracy,  because  the  lines  ab, 
be,  cd,  etc.,  are  so  short  and  there  are  so  many  of  them.  If,  however,  we  can  find  what  the 
curve  A' abed,  etc.,  is,  we  could  draw  the  curve  at  once. 

Suppose  we  divide  the  load  area 
into  only  two  portions  of  lengths  x  and 
/  —  x,  where  /  =  AB,  Fig.  3.  The  en- 
tire weight  over  the  portion  x  can  be 
considered  as  acting  at  the  centre  el  of 
the  load  area.  The  same  holds  good 
for  the  portion  /  —  x.  We  thus  have 
two  forces  Fl  and  F2. 

Taking  the  pole  as  before,  so  that  the  closing  line  A' B'  shall  be  parallel  to  AB, 
construct  the  equilibrium  polygon  A'abB'.  The  curve  of  moments  will  be  tangent  at  A', 
c  and  B'. 


k 

FIG. 

i. 

—  i  — 

i  

Fie 
B                    Or 

r.    2. 

T 

;i 

s     1 

;  i 

6 

7 

V 
8                      2' 
^B'                3 

\i 

1   41 

H  ~~'~  "-ilj, 

i 

^^ 

^ 

^ 

5 

FIG.  4. 
0 

\ 

L 


420 


ST4TICS  OF  RIGID  BODIES. 


[CHAP.  in. 


Now  we  see  that,  no  matter  where  the  load  area  is  supposed  to  be  divided,  we  shall 
always  have  for  the  distance  */,  between  Ft  and  Ft 


That  is,  no  matter  where  the  line  of  division  is  taken,  the  horizontal  projection  of  the 
line  ab  of  the  equilibrium  polygon  is  constant  and  equal  to  -/.     But  ab  is  a  tangent  to  the 

curve  required.      But  if  from  any  point  on  the  line  A'd  we  draw  a  line  ab  limited  by  the  line 
B'd,  so  that  the  horizontal  projection  is  constant,  the  line  ab  will  envelop  a  parabola. 

This  may  easily  be  proved  as  follows:   Let  the  load  per  unit  of  length  be/.     Then  the 

pi 
entire  load  is //and  the  reaction  at  each  end  is  —  • 

The  moment  at  any  point  distant  x  from  the  left  support  is  then 


Pi         r^x 

*-**-**? 


But,  since  Fl  is  equal 


This  is  the  equation  of  a  parabola.  At  the  centre  x  =  -,  and  we  therefore  have  the 
centre  ordinate  -r-. 

o 

COR.  i.  We  see,  therefore,  that  when  a  string  is  suspended  from  two  points  A',  B'  and 
sustains  a  load  uniformly  distributed  over  the  horizontal,  the  curve  of  equilibrium  is  a 
parabola. 

Also,  the  horizontal  component  of  the  stress  at  any  point,  as  is  evident  from  the  force 
polygon,  is  constant  and  equal  to  H.  Also,  the  vertical  component  of  the  stress  at  any  point, 
as  c,  Fig.  3,  is  ^  —  Flf  or  equal  to  the  total  load  between  the  lowest  point  and  the  point 
considered. 

COR.  2.   We   have   the  following  construction  for  the   equilibrium  curve.      Lay  off  a 

perpendicular  eK  at  the  centre  e  and  make  it  equal  by  scale  to^-.     Through  A,  K  and  B 

o 

construct  a  parabola  having  its  vertex  at  K.  The  ordinate  to 
this  parabola  through  any  point  will  give  the  moment  at  that 
point. 

The  distance  Kd  is  also  equal  to  -=-,  because  the  moment 

o 

of  the  reaction  with  reference  to  e  is 

-Pl~*P* 


and  Kd=ed-eK  = 


CHAP.  III.] 


STATICS   OF  RIGID  BODIES. 


421 


COR.  3.   How  to  Draw  a  Parabola. — Since  we  know,  then,  the  distance  ed  =~    we 

4  ' 

can  always  draw  the  lines  Ad  and  Bd.      If  then  we  divide  Ad  and  Bd  into  any  number  of 
equal  parts  and  number  these  parts  along  one  line  away  from 
d  and  along  the  other  towards  d,  we  have  only  to  draw  lines 
joining  any  two  points  having  the  same  number,  and  these  lines 

will  all  have  the  same  horizontal  projection  — . 

They  will  therefore  enclose  the  parabola  required.     Tan- 
gent to  these  lines  we  may  sketch  the  curve. 

A  better  method  is  to  plot  the  ordinates  to  the  curve  from 

its  equation, 


Methods  of  Solution  of  Framed  Structures. — In  Chapter  I  we  have  given  and  illus- 
trated two  methods  of  computation  for  framed  structures: 

1st.   By  Resolution  of  Forces  (page  400). 

2d.   By  Moments  or  the  "  Method  by  Sections  "  (page  401). 
In  the  present  Chapter  we  have  the  corresponding  graphic  methods: 

1st.    By  Resolution  of  Forces  (page  404). 

2d.    By  Moments  (page  415). 

Examples. — (i)  A  roof-truss  has  a  span  of  50  ft.  and  a  centre  height  of  12.$  ft.  Each  rafter  is  divided  into 
four  equal  panels,  and  the  lou>er  horizontal  tie  is  divided  into  six  equal  panels.  The  bracing  is  as  shown  in  the 
figure.  Find  the  stresses  in  the  members,  by  Hie  graphic  method  of  moments,  for  a  weight  of  800  Ibs.  at  each 
upper  apex. 

ANS.  We  have  computed  the 
stresses  (page  402),  by  the  two 
methods,  resolution  of  forces  and 
moments.  We  have  also  found 
the  stresses  by  the  graphic 
method  of  resolution  of  forces 
(page  410,  ex.  (i)). 

We  can  construct  the  force 
polygon  Fig.  2,  and  then  the 
equilibrium  polygon  Fig.  i.  This, 
however,  is  not  advisable  for  rea- 

f-jt/  sons  already  given.     It   will   be 

FIG  2.  FIG.  i.  more  accurate  to  assume  the  pole 

distance  as  unity,  thus  discarding  the  force  polygon  altogether,  and  construct  points  in  a  parabola  from  the 
equation 


A 

B 
C 

D  -._ 

L 

E 

FT 
G 


X 


^•o 


In  the  present  case  the  load  per  foot  is,  if  we  suppose  half  weights  of  400  at  the  ends,  — —  =  128  Ibs.  =  p. 


Taking  x  —  =-1,  ^-/,  etc.,  we  have 


ll. 


ll. 


/=  17500,     30000,     37500,     40000  Ib.-ft. 

Laying  these  off  to  any  convenient   scale,  we  determine  very  accurately  the  points  a,  b,  c,  d  of  the 
equilibrium  polygon.     The  other  half  of  the  polygon  is  precisely  similar. 


422  GRAPHICAL  STATICS— NON-CONCURRING  FORCES.  [CHAP.  III. 

The  ordinates  to  this  polygon  will  give,  to  the  scale  adopted,  the  moment,  for  any  point  of  the  truss,  of 
the  outer  forces  left  or  right.  Thus  the  moment  with  reference  to  k  of  all  forces  right  or  left  is  km,  Fig.  i. 
We  find  by  scale  km  =  2i666|  Ib.-ft.  In  the  same  way  for  the  next  lower  apex  we  find  the  moment  35000 
Ib.-ft.  The  moment  at  the  next  lower  apex  or  centre  of  the  span  is  40000  Ib.-ft. 

Now  by  the  method  of  sections  (page  401)  we  have  for  any  member 

Stress  x  lever-arm  +  2  moments  of  outer  forces  =  o. 

The  second  term  is  given  by  the  ordinates  of  the  equilibrium  polygon  to  scale. 

As  regards  the  centre  of  moments  for  any  member,  we  must  observe  the  rule  (page  401),  viz.  :  Cut  the 
truss  entirely  through  by  a  section  cutting  only  three  members  the  strains  in  which  are  unknown.  For  any 
one  of  these  take  the  point  of  moments  at  the  intersection  of  the  other  two. 

For  the  proper  sign  for  the  first  member  of  the  equation  place  an  arrow  on  the  cut  member  pointing 
away  from  the  end  belonging  to  the  left-hand  portion,  and  take  the  moment  (  +  )  or  (— )  according  as  the 
rotation  indicated  by  this  arrow  is  counter-clockwise  or  clockwise. 

If  the  stress  comes  out  positive,  it  indicates  tension  ;  if  negative,  compression. 

Take,  for  instance,  the  first  lower  panel,  La.  The  centre  of  moments  must  be  taken  at  the  first  upper 
apex.  The  moment  for  this  point  is  given  by  the  ordinate  na  of  the  equilibrium  polygon,  or  —  17500  Ib.-ft. 
We  take  the  minus  sign,  because  the  rotation  is  clockwise.  We  have  then 

La  x  3.125  —  17500  =  o,     or     La  =  +  5600  Ibs., 

where  3.125  ft.  is  the  lever-arm  of  La. 
In  similar  manner  we  have 

Lc  x  6.25     —  30000  =  o,     or     I^c  =  +  4800  Ibs., 

where  6.25  ft.  is  the  lever-arm  of  Lc. 
For  Le  we  have 

Le  x  9.375  —  37500  =  o,     or     Le  =  +  4000  Ibs., 

where  9.375  ft.  is  the  lever-arm  of  I.e. 

For  the  first  upper  panel,  Aa,  take  the  centre  of  moments  at  k  The  moment  f<  r  this  point  is  given  by 
the  ordinate  from  k  to  the  first  line  of  the  polygon  produced.  It  is  therefore  larger  tlian  km.  which  gives  the 
combined  moment  of  the  reaction  and  first  weight.  We  find  it  by  scale  to  be  —  23333^  Ib.-ft. 

We  have  then 

—  Aa  x  3.727  —  23333$  =  o,     or     Aa  =  +  6260  Ibs., 

where  3.727  ft.  is  the  lever-arm  for  Aa. 

In  like  manner  for  Bb  we  have  centre  of  moments  at  k,  and  moment  km  =  —  21666}.      Hence 

-  Bb    x  3.727  —  21666}  =  o,     or     Bb  =   +  5813  Ibs. 
For  Cd  we  have 

—  Cd  x  7.454  —  35000    =  o,     or     Cd  —  +  4691  Ibs., 

where  7.454  ft.  is  the  arm-lever  for  Cd. 
For  Df  we  have 

—Df  x   11.151  -  40000    =  o,     or     Df  =  +  3587  Ibs. 

For  all  the  braces  the  point  of  moments  is  at  the  left-hand  end.  Taking  a  section  through  Bb,  ab  and 
La,  we  have  acting  on  the  left-hand  portion  only  the  weight  AB  and  the  reaction.  The  moment  of  the  weight 
relative  to  the  left  end  is  the  ordinate  a'b',  or  by  scale  —  5000  Ib.-ft.  The  lever-arm  for  ab  is  6.934  ft. 
Hence 

—  ab  x  6.934  —  5000  =  o,     or     ab  =  —  721  Ibs. 
For  be  we  have 

+  ab  x  6.934  —  5000  =  o,     or    ab  ==  +  721  Ibs. 

For  cd  the  moment  is  a'V  +  b'<? ,  or  —  1 500.     We  have  then 

—  cd  x    13.869  —   15000  =  o,     or    cd  =  —  1081  Ibs., 
and  so  on.     All  lever-arms  can  be  scaled  off  the  frame  or  must  be  computed. 


CHAP.  III.] 


STATICS   OF  RIGID  BODIES. 


423 


The  present  method  is  not  to  be  recommended  for  the  braces.  In  prolonging  the  sides  ab,  be,  etc.,  of 
tlie  equilibrium  polygon,  a  slight  variation  in  direction  will  make  considerable  error  in  the  ordinate  at  the 
end.  Also  as  the  sides  ab,  be,  etc.,  are  short  they  do  not  give  direction  accurately  enough. 

Of  all  our  four  methods,  the  graphic  method  by  resolution  of  forces  (page  404)  is  the  easiest  of  application 
to  such  cases. 

The  more  irregular  the  frame  the  more  advantageous  it  is. 

(2)  A  bridge-girder,  as  shown  in  the  figure,  10  feet  deep,  80  feet  long,  eight  equal  panels  in  the  lower  chord 
and  seven  equal  panels  in  the  upper  chord,  has  a  load  of  5  tons  at  each  lower  apex.  Find  the  stresses  by 
diagram  and  by  moments. 

FIG.   i. 


1 


ANS.  The  panel  length  is  10  ft.,  sec  6  =  1.117. 
Aa  x   10  —  17.5   x    10  =  o, 


~s 

,  </  ,  \<       A       . 

\  A  •„ 

/ 

V             V       W 

&        -       d     -  f~h 

FIG.  2. 

By  moments,  then, 
or     Aa 


=  + 


Be  x  10  —  17.5  x   15  +  5  x   5  =  0, 

Ce   x  10  —  17.5  x  25  +  5(5  +  15)  =  °. 

Dg  x  10  —  17.5  x  35  +  5(5  4-  15  +  25)  =  °. 

—  Ib  x   10  —  17.5  x    10  =  o, 

-  Id  x   10  -  17.5  x  20  4-   5  x   10  ==  o, 

-  If  x    10  -  17.5  x  30  +  5(10  +  20)  =  o, 

—  Ih  x   10  —  17.5   x  40  4-  5(10  +  20  4-  3°)  =  °. 

la  =  -  17.5  x   1.117  =  -  19-55.  de  =  —  7.5   x 

ab  =  +  19.55,  ef  —   +  8'48' 

be  —  -   12.5  x   1.117  =  -  J3-96«  fg  —  —  2-5   x 

cd  —  +  13.96,  gh  =  4-  2.79. 


=  + 


1. 117 
1.117 


17.5  tons. 
23-75  " 
33-75  " 
38.75  " 

17-5 
30 

37-5   " 
40 

=  -  8-38, 
=  -  2.79, 


CHAPTER    IV. 


WALLS.     MASONRY    DAMS. 


Definitions  of  Parts  of  a  Wall. — The  face  of  a  wall  is  the  front  surface,  or  outside 
surface,  or  the  surface  farthest  from  the  pressure.  The  back  is  the  rear  surface,  or  inside 
surface,  or  the  surface  which  sustains  pressure. 

The  stone  which  forms  the  face  is  called  the  facing ;  that  which  forms  the  back,  the 
backing;  that  which  forms  the  interior,  the  filling. 

A  horizontal  layer  of  stone  in  a  wall  is  called  a  course.  If  the  stones  in  each  layer  are 
of  the  same  thickness,  we  have  regular  courses ;  if  they  are  not  of*  the  same  thickness,  we  have 
irregular  or  random  courses. 

The  mortar  layer  between  the  stones  is  the  joint.      The  horizontal  joints  are  bed-joints. 

Cut  stone  or  squared  masonry  is  called  ashlar.  Unsquared  masonry  or  rough  stone  is 
called  rubble. 

The  inclination  of  the  face  or  back  of  a  wall,  measured  by  the  ratio  of  its  horizontal 
to  its  vertical  projection,  is  called  the  batter  of  the  face  or  back.  The  batter  is  then  the 
tangent  of  the  angle  which  the  face  or  back  makes  with  the  vertical. 

Weight  and  Friction  of  Masonry. — We  give  here  a  short  table  of  average  values  of  the 
coefficient  of  static  sliding  friction  /^,  the  density  or  mass  per  cubic  foot  6,  and  the  allowable 
compressive  unit  stress  C  in  tons  per  square  foot,  taking  2000  Ibs.  to  a  ton,  for  different 


We  also  give  the  specific  mass  (page    16)  — ,  where  y  is  the  mass  of  a 


kinds  of  masonry. 

cubic  foot  of  water,  or  62.5  Ibs. 

In  discussing  the  stability  of  walls,  the  influence  of  the  mortar  is  neglected,  both  because 
of  its  uncertain  character  and  because  such  neglect  is  on  the  side  of  safety. 


Kind  of  Masonry. 

Coefficient  of 
Friction 
p. 

Density  in  Ibs.  per 
Cubic  Foot 
1. 

Specific  Mass 

y 

Allowable  Com- 
pressive Unil  Stress 
Cm  tonspersq.lt 

Limestone  and  granite  : 

.  .  ', 

jAc 

2  6d 

o  6 

o  6 

o  6 

Sandstone  : 
Ashlar  masonry  
Large  mortar  rubble  

0.6 
0.6 
o  6 

ISO 
130 

2.40 
2.08 
i  ?6 

20  to  25 
10  to  15 

o  6 

1  60 

6  t°    ° 

424 


CHAP.  IV.] 


STABILITY  OF   A  MASONRY  JOINT. 


425 


Stability  of  a  Masonry  Joint.— Let  AB  represent  a  masonry  joint,  and  R  the  resultant 
of  all  the  external  forces  acting  upon  it  at  the  point  P. 

Then  we  must  have  the  following  conditions  for  stability : 

1.  The  resultant  R  of  all  the  external  forces  must  intersect 
the  joint  at  some  point  P  within  the  joint ;  otherwise  we  have 

no  rotation.  A ^ B 

2.  The  resultant  R  must  make  an  angle  RPN  with  the 

normal  to  the  joint  whose  tangent  is  less  than  the  coefficient  of  friction;  otherwise  we  have 
sliding. 

3.  The  greatest  unit  pressure  at  any  point  of  the  joint  must  not  exceed   the  allowable 
compressive  unit  stress  C  for  the  material ;  otherwise  the  joint  is  overloaded. 

Determination  of  this  Greatest  Unit  Pressure. — Let  N  be  the  normal  component  of  the 

resultant  R  of  all  the  external  forces  acting  at  the  point  P, 
and  let  e  be  the  distance  of  P  from  the  nearest  edge  B. 

Then  the  least  unit  pressure /t  will  act  along  the  farthest 
edge  A,  and  the  greatest  unit  pressure  p  will  act  along  the 
nearest  edge  B.  If  we  lay  off  Aa  =  p^  and  Be  =/,  the 
unit  pressure  at  any  point  will  be  given  by  the  co-ordinate 
to  the  straight  line  ac.  Let  A  be  the  area  of  joint,  and  b  the 
breadth  of  joint  AB. 

We  have  then  the  mean' unit  pressure  ,  and  hence  the  total  normal  pressure 


_  A£_I 


(A 


2N 


or     p.  =  —3 


The  entire  load  area  is  made  up  of  the  rectangular  area  aABb  and   the  triangular  area 
abc.      The  load  represented  by  the  rectangular  area  is/j/4,  and  its  centre  of  action  is  at  c  at 
1)                                                                                                                                   (p  —  p\A 
a  distance  -  from  the  edge  B.      The  load  represented  by  the   triangular  area  is 

acting  at  E  at  a  distance  —  from  the  edge  B. 

We  have  then,  taking  moments  about  the  edge  B, 


(2) 


Substituting  the  value  of  pl  from  (i),  we  have  for  the  greatest  unit  pressure 


(3) 


where  N  is  the  normal  pressure  on  the  joint  of  area  A  and  breadth  b,  and  e  is  the  nearest 
edge  distance  from  N. 

From  (3)  we  can  find  in  any  case  the  greatest  unit  pressure,  and  this  must  not  exceed 
the  allowable  compressive  unit  stress  C,  otherwise  the  joint  is  overloaded. 


426 


STATICS   OF  RIGID  BODIES. 


[CHAP.  IV. 


b  N 

11  Middle  Third  "  Rule.  —  From  (3)  we  see  that  when  e  =  —  we  have/  =  -T- 

2  A 


"\ 

N 

\ 

That  is, 

when  the  resultant  R  of  all  the  external  forces  acts  at  the  centre 
of  mass  of  the  joint  the  load  N  is  uniformly  distributed  and  the 


p  N 

unit  pressure  at  every  point  is/  =  -~t  so  that/,  =  /. 


T  As  the  point  of  application  P  of  R  approaches  B,  /  increases 

and  /,  decreases,  and  when'  e  —  —  we  have  /  =  —r-  and  /,  =  o.     That  is,  when  the  resul- 


tant R  acts  at  —  b  from  the  nearest  edge,  the  unit  pressure  at  the 

farthest  edge  is  zero,  and  the  greatest  unit  pressure  at  the  nearest 
edge  is  twice  as  great  as  if  the  load  were  uniformly  distributed. 

b  A  p    "j     a 

If  then  e  is  less  than  — ,  the  entire  joint  is  not  brought  into 

action.  The  effective  breadth  of  joint  is  then  3^  =  DB 
and  the  portion  AD  affords  no  resistance,  if  we  disregard 
the  tensile  strength  of  the  mortar. 

If  /  is  the  length  of  joint,  the  area  in  action  is  ^el 

2N 

B    and  the  greatest  unit  pressure/  =  — - 

We  see,  then,  that  in  order  to  have  the  entire  joint  in  action  the  resultant  R  of  all  the 
external  forces  must  intersect  the  joint  inside  the  middle  third. 

This  is  called  the  "  middle  third  rule,"  and  in  a  well-proportioned  masonry  structure  it 
should  be  complied  with. 

If  then  the  point  P  lies  within  the  middle  third,  we  have  the  greatest  unit  pressure 


2N 


If  the  point  /Ms  just  at  the  limit  of  the  middle  third,  we  have 

_2_N_ 

If  the  point  Pis  outside  the  middle  third,  we  have 

2N 


In  any  case  the  value  of  /  must  not  exceed  the  allowable  compressive  unit  stress  C, 
otherwise  the  joint  is  overloaded. 

Stability  of  a  Wall  in  General. — We  can  now  investigate  the  stability  of  a  wall  in 
general 

NOTATION. — Let//1  be  the  height  of  the  wall,  bl  the  top  base,  £2  the  bottom  base,  fa  the 
batter  angle  of  the  back.  0  the  batter  angle  of  the  face,  tf  the  density  of  the  wall. 

Let  there  be  a  horizontal  thrust  T'per  unit  of  length  of  wall  acting  at  the  distance  A  from 


CHAP.  IV.] 


STABILITY   OF  A   WALL   IN  GENERAL. 


427 


the  bottom,  and  let  there  be  a  pressure  P  per  foot  of  length  of 
tion  at  a  distance  dl  from  the  bottom.*  Let  H  and  V 
be  the  horizontal  and  vertical  components  per.  ft.  of 
length  of  the  wall  of  this  pressure  P.  Let  Wbe  the 
weight  per  ft.  of  length  of  the  wall  acting  at  the  centre 
of  mass  O,  and  let  R  be  the  resultant  of  H,  V,  Wand 
T,  intersecting  the  base  at  a  point  G  distant  <?  from  the 
front  edge.  Let  N  be  the  total  normal  pressure  upon 
the  base  per  foot  of  length  of  the  wall. 

The  weight  W  per  foot  of  length  of  the  wall  is  given 
then  by 


G  B 


(I) 


and  the  total  pressure  per  foot  of  length  of  the  wall  upon  the  base  is 

N  =  W  +  V.     ...........     (2) 

STABILITY  FOR  SLIDING.  —  If  u  is  the  coefficient  of  friction  as  given  in  the  table  page 
424,  we  have  the  friction  per  foot  of  length  of  the  wall 


V) 


(3) 


If  T+  H  is  greater  than  this  and  the  joint  AB  extends  through  without  break,  we  have 
sliding  on  the  base;  if  T-{-  His  less  than  this,  there  is  no  sliding.  For  safety  let  us  take 
the  friction  equal  to  n  times  T-\-  H,  so  that  n  is  the  factor  of  safety  for  sliding.  We  have 
then 


pN=n(T+ff),     or     «  = 


(I) 


T+H   • 

If  then,  in  any  case,  n  is  less  than  unity,  we  have  sliding  for  a  through  joint  AB.  If 
«  =  I,  the  wall  is  on  the  point  of  sliding.  If  ;/  is  greater  than  unity,  there  is  no  sliding  and 
the  greater  n  is  the  greater  the  security. 

In  practice  we  should  have  n  at  least  2  or  even  more  if  shocks  are  to  be  apprehended. 
If  the  joint  AB  does  not  extend  through,  or  if  the  masonry  courses  are  irregular,  so  that  no 
joint  extends  through,  there  is  no  possibility  of  sliding  and  equation  (I)  need  not  be  applied. 

STABILITY  FOR  ROTATION. — The  distance  ~x  of  W  from  the  front  edge  B  is,  from  page  2  5 , 

^,  (4) 


or,  putting  h^  tan  ft  —  b2  —  bv  —  /^  tan  ft, 

*  =  ?£± 


-  bf  - 


,  +  *,)  tan  ft 


(5) 


The  distance  of  W  from  G  is  then  x  —  e.  The  distance  of  Ffrom  G  is  bt  —  e  —  d^  tan  ft. 
The  distance  of  H  from  G  is  dr  The  distance  of  T  from  G  is//.  If  then  we  take  moments 
about  G,  we  have,  if  the  resultant  R  passes  through  G, 

-  Th  -  Hdl  +  V(b2  -e-d,  tan  ft)  +  W(x  -  e]  =  O. 


*  For  both  water-pressure  and  earth-pressu~e  /•  is  the  distance  to  the  water  or  earth  surface,  and  </,  =  —h 
(pages  431,  454). 


428  STATICS  OF  RIGID  BODIES.  [CHAP.  IV. 

Hence  the  distance  e  from  the  front  edge  B  of  the  point  G  where  the  resultant  R  cuts 
the  base  is  given  by 

HSr  +  V(b,  -  d,  tan  ft,)  -  Hd,  -  Th 
W+V 

If  in  any  case  e  as  given  by  (II)  is  negative,  the  point  G  falls  outside  of  the  base  Afiand 
we  have  rotation.  If  e  is  positive,  the  point  G  is  within  the  base  and  there  is  no  rotation. 

STABILITY  FOR  PRESSURE. — We  have  still  to  test  as  to  whether  the  base  is  overloaded 
or  not.  We  have  N  —  W -\-  V.  From  page  426  we  have  for  the  greatest  unit  pressure  on 
the  base,  provided  we  take  e  as  the  distance  from  G  to  the  nearest  edge, 


when  e  is  less  than         —62 ./  = 


when  e  is  equal  to 


(III) 


when  e  is  greater  than  -b2 /  =  — — ^-  —  \2  —  ~\ . 

In  these  equations  e  is  always  the  distance  of  G  from  the  nearest  edge.  In  any  case  the 
value  of/  should  be  less  than  the  allowable  unit  stress  C  given  in  the  table  page  424,  or  else 
the  base  is  overloaded. 

It  is  then  a  very  simple  matter  to  test  by  equations  (I),  (II)  and  (III)  the  stability  of  a 
wall  whose  dimensions  are  given  in  any  case  where  T,  H  and  V  are  given. 

High  and  Low  Wall — The  distance  e  from  the  front  edge  B  of  the  point  G  where  the 
resultant  cuts  the  base  (figure,  page  427)  is  given  by  equation  (II).  As  we  have  seen  (page 

425),  if  e  is  less  than  -bi>  the  entire  base  is  not  brought  into  action.       In  such  case  the 

joints  on  the  inside  tend  to  open,  and  in  a  well-designed  wall  this  should  not  occur.  Also, 
the  greatest  unit  pressure/  should  never  exceed  the  allowable  unit  stress  C  as  given  by  the 
table  page  424. 

In  a  properly  designed  wall,  then,  e  must  be  equal  to  or  greater  than  —  b2,  and  /  must 
be  less  than,  or  at  most  equal  to,  C. 

If,  in  any  case  when  e  =  -£2,  /  is  less  than,  or  at  most  equal  to,  C,  the  wall  is  said  to  be 

low.      If,  when  /  =  C,  we  have  e  greater  than  —  bz ,  the  wall  is  said  to  be  high. 
If  then  in  the  second  of  equations  (III),  we  make/  —  C,  we  have  for  e  =  -b^ 

_2(W+V) 

2  ~    — c — * 

For  trapezoid  section 


CHAP.  IV.]  DESIGN  OF  LOW  AXD  HIGH  WALLS—  TRAPEZOID  SECTION.  429 

Hence  we  have  for  the  limit  of  low  wall 


Equation  (I)  gives  the  limit  of  /^  for  low  wall  of  trapezoid  section,  for  given  top  base 
bl  and  bottom  base  bz.  For  /^  greater  than  this  limit  the  wall  is  high.  For  hv  less  than  this 
the  wall  is  low. 

For  rectangular  section  we  have  bz  =  b^  ,  and  in  this  case  we  have 


b.C  - 


Equation  (i)  gives  the  limit  of  hv  for  low  wall  of  rectangular  section  for  given  base  br 
For  hv  greater  than  this  limit  the  wall  is  high.  For  /il  less  than  this  the  wall  is  low. 

Design  of  Low  Wall — Trapezoid  Section. — For  both  water  and  earth  pressure  h  (figure, 
page  427)  is  the  distance  from  base  to  water  or  earth  surface,  and  d^  as  we  shall  see  hereafter 

(page  43 1),  isj/*. 

BOTTOM  BASE. — If  then  we  make  e~  ~bz  inequation  (II),  page  428,  put  d^  =  -h,   and 

substitute  the  values  of  Wand  ^  as  given  by  equations  (i)  and  (5),  page  427,  and  solve  for 
b2,  we  have 

'*2=-^+  VBf  +E,* (II) 

where  the  quantities  Sl  and  El  are  given  by 

*i  =  2L(^  +  ^  -  *i  tan  £),  E,  =  bfa  +  2^  tan  fa  +  ~  (  Ftan  ftl  +  H  +  3  7). 

Equation  (II)  gives  the  bottom  base  b2  for  a  properly  designed  low  wall  of  trapezoid 
section.  If  the  value  of  &2,  as  given  by  (II),  substituted  in  (I),  gives  the  limit  h^  greater 
than  the  actual  height,  the  wall  is  low.  If  less,  the  wall  is  high  and  equation  (II)  does  not 
apply. 

TOP  BASE. — For  economy  the  top  base  b^  should  be  assumed  as  small  as  possible  con- 
sistent with  practical  and  local  considerations.  We  should  not  in  any  case  assume  bl  greater 
than  by  If  then  we  make  bz  =  bv  in  (II),  we  have  for  the  greatest  allowable  value  of  ^ 

max.  bl=  —  B+  V £  +  E, (2) 

where  the  quantities  B  and  E  are  given  by 

B =^(lf- 3^ tan  &     E  =  w,(v  tan  fr+ff+*T)' 

Equation  (2)  gives  then  the  largest  allowable  value  of  bl  for  low  wall,  that  is,  when  Al  is 
less  than  or  equal  to  limit  k^  as  given  by  (i). 

Design  of  High  Wall — Trapezoid  Section. — As  we  have  seen,  if  the  value  of  £2  as  given 
by  (II),  substituted  in  (I),  gives  the  limit  hv  greater  than  the  actual  height,  the  wall  is  low. 

If  less,  the  wall  is  high.     In  this  case  e  is  greater  than  -£2,  and/  =  C. 


43°  ST/tTICS  Of  RIGID  BODIES.  [CnAi-.   IV. 

LOWER  BASE.  —  From  the  third  of  equations  (III),  page  428,  we  have 


For  /  =  C  we  have 

_  *  A       fiy.... 

-  3^        <KW  +  J/y 

If  we  equate  this  to  the  value  of  e  given  by  equation  (II),  page  428,  make</,=-A, 

substitute  the  values  of  Wand  x  from  equations  (i)  and  (5),  page  427,  and  solve  for  b2,  we 
have  for  the  value  of  bt 

bz=-B2  +  VB^+E2  ..........     (Ill) 

where  the  quantities  B2  and  Ez  are  given  by 


,  +  2A,  tan  /?,)  + 

TOP  BASE.  —  For  economy  the  top  base  b^  should  be  assumed  as  small  as  possible  con- 
sistent with  practical  and  local  considerations.  We  should  not  in  any  case  have  b^  greater 
than  bz-  If  then  in  (III)  we  make  b2  =  blt  we  have 

max.  bv  =  -  B  +    V&  +  E,       ........     (3) 

where  the  quantities  B  and  E  are  given  by 


Equation  (3)  gives  then  the  largest  allowable  value  of  £,  for  high  wall,  that  is,  when  h^ 
is  greater  than  limit  //,  as  given  by  (i). 

Design  of  Wall  in  General  —  Trapezoid  Section.  —  In  general,  then,  in  order  to  design 
a  wall  of  trapezoid  section  for  given  dl  and  //,  when  V,  H  and  T  are  known,  we  first  find 
from  (II),  page  429,  the  value  of  bz  for  low  wall.  If  this  value  of  b2  substituted  in  (I),  page  429, 
gives  the  limit  /^greater  than  the  actual  height,  the  wall  is  low,  and  the  value  of  bz  given  by 
(I)  is  the  value  required.  If  limit  /il  is  less  than  the  actual  height,  the  wall  is  high,  and  we 
must  find  £2  from  (III),  page  430. 

In  the  case  of  low  wall  the  value  of  dl  assumed  should  not  be  greater  than  max.  £,*given 
by  equation  (2),  page  429.  In  the  case  of  high  wall  the  value  of  ^  assumed  should  not  be 
greater  than  max.  bv  given  by  equation  (3),  page  430.  In  any  case,  for  economy  w*:  should 
take  £j  as  small  as  local  or  practical  considerations  permit. 

Water-pressure.  —  It  is  a  well-known  principle  of  Physics  that  the  direction  of  water- 
pressure  upon  a  submerged  surface  is  always  perpendicular  to  that  surface.  Also,  upon  a 
plane  surface  the  pressure  is  equal  to  the  weight  of  a  prism  of  water  whose  base  is  the  sub- 
merged area  and  whose  height  is  the  distance  from  the  water-level  to  the  centre  of  mass  of 


CHAP.  IV.] 


WATER-PRESSURE.—  EXAMPLE. 


43  l 


the  submerged  area.       This  pressure   acts  at    a    point  at  a  distance  from  the  water-level 
equal  to  —  of  the  depth  of  the  water,  or  at  a  distance  dv  from  the  bottom  equal  to  dv  =  —A, 

where  h  is  the  depth  of  water. 

Thus,  in  the  figure,  let  AlDl  be  the  water-level  and  h  the  depth  of  water  above  the  base 
AB.     Then  the  pressure  P  per  foot  of  length  of 
the  wall  upon  the  submerged  area  ADl  X  i  ft.  is 
perpendicular  to  that  area.       This    pressure  P 
acts  at  a  point  at  a  distance  from   the    bottom 

equal  to  —h.  The  centre  of  mass  of  the  sub- 
merged area  is  at  a  distance  of  —h  below  the 
water-surface.  The  pressure  P  is  then  equal  to  the  weight  of  a  prism  of  water  whose  base 
isADl  X  i  ft.,  and  whose  height  is  —A.  If  y  is  the  density  of  water,  the  pressure  P  per  foot 
of  length  of  the  wall  is  then 


Let  fil  be  the  batter  angle   of   the   back.      Then  ADl .  cos  fil  =  A,   or  AD  = 
Hence  the  pressure  P  per  foot  of  length  of  the  wall  is 


cos 


The  vertical  component  F"  of  P  per  foot  of  length  of  the  wall  is  then 


=       -  tan 


(2) 


and  the  horizontal  component  Hoi  Pper  foot  of  length  of  the  wall  is 


(3) 


Both  these  components  act  at  a  point  at  a  distance  from  the  bottom  equal  to  -//. 

Example. — A  dam  20  feet  high  whose  back  has  a  batter  of  i  to  2  sustains  water-pressure.      The  water- 
level  is  2  feet  below  the  top.     Find  the  horizontal  and  vertical  pressure  per  foot  of  length. 

ANS.  We  have  Al  —  20  ft.,  h  =  18  ft.,  /=  i  ft.,  tan  fii  =  — .     For  water  y  =  62.5  Ibs.  per  cubic  ft. 
Hence  the  horizontal  pressure  is 

yh*      62.5x18x18 
H  =  - —  =  —  —  =  10125  pounds, 

and  the  vertical  pressure  is 


:i8xi8    i 

=  5062.5  pounds. 


These  pressures  act  at  -h  —  6  ft.  from  the  bottom. 


432  STATICS  OF  RIGID  BODIES.  [CHAP.  IV. 

Ice  and  Wave  Pressure.  —  A  dam  has  to  sustain,  in  addition  to  the  water-pressure,  a 
horizontal  pressure  per  foot  of  length  at  the  water-level  due  to  waves  or  to  the  pressure  of 
ice.  We  denote  this  horizontal  thrust  per  foot  of  length  by  T.  For  waves  we  may  take,  on 
the  basis  of  experiments  made  by  Stevenson,  T  =  24000  pounds  per  foot  of  length,  and  for 
ice,  on  the  basis  of  the  Report  of  the  Aqueduct  Commission  on  the  Quaker  Bridge  Dam, 
1889,  we  may  take  7^=40000  pounds  per  foot  of  length.  Since  both  these  do  not  act 
together,  we  have  only  to  consider  T  for  ice  in  cold  climates  and  T  for  waves  in  warm 
climates,  if  the  body  of  water  back  of  the  dam  is  sufficiently  extensive  for  wave-action 
to  occur. 

Stability  of  a  Dam  —  Trapezoid  Section.  —  We  have  then  only  to  use  this  value  of  T 
and  the  values  of  V  and  //  given  by  equations  (2)  and  (3),  page  431,  in  the  general  equa- 

tions (I),  (II)  and  (III),  page  428,  and  make  d^  —  —  //,  in  order  to  investigate  the  stability  of 

any  given  dam  of  trapezoidal  section. 

Examples.  —  (l)  Investigate  the  stability  of  a  dam  of  granite  ashlar  20  feet  high,  -with  vertical  back  and 
face.      The  water-level  is  2  feet  below  the  top  of  the  dam.      The  top  base  is  2  feet.     Section  trapezoid. 

ANS.  We  have  b\  —  2  ft.,  6,  =  2  ft.,  tf,  =  o,  ft  =  o,  //i  =  20  ft..  //  =  18  ft.,  y  =  62.5  —  r—  j-    Also,  from  the 

Ibs.  pounds 

table  page  424,  S  =  165  —  j—  y-  ,  C  =  30  x  2000  --•  60000  -  -  p-  ,  n  =  0.6. 

Hence  we  obtain  V  =  o,  //  =  —  —  =  10125  pounds  per  ft.,  W  =  165  X  2  x  20  =  6600  pounds 

per  ft.     From  (I),  page  427,  then,  disregarding  ice  and  wave  thrust,  we  have 

nW        6  x  6600 
«  =     .,   =  —  •  —  =  0.39. 

H         10  x   10125 

The  dam  is  therefore  not  secure  against  sliding  even  if  we  disregard  ice  and  wave  thrust. 

We  have  from  (4),  page  427,  ~x  =  \  ft.,  and  from  (II),  page  428,  disregarding  ice  and  wave  thrust, 

W-™ 

3        6600  —  60750 


W  6600 

Since  e  is  negative,  the  resultant  R  falls  outside  the  base,  and  we  have  rotation  even  when  ice  and  wave 
thrust  are  neglected.     The  dam  is  therefore  unstable  both  for  sliding  and  rotation. 
(2)  Let  us  change  the  bottom  base  to  12  feet,  the  rest  being  the  same  as  before. 

ANS.    We  have  fa  =  2  ft.,  fa  =  12  ft.,  ft,  =  o,   tan    ft  =  *-,  /;,  =  20  ft.,    //  =  18  ft.,  y  =  62.5    ^St      . 

Ibs.        _      ,         pounds 
5  =  165  —  r-—f-,  C  =  60000  -  -  ~t  —  ,  u  =  0.6. 
•'cub.  ft.  sq.  ft 

Hence  we  have    W  =  165  x  7  x  20  =  23100  pounds  per  ft.,    V  =  o,  and  H  =  10125  pounds  per  ft.,  as 
before. 

From  (I),  page  427,  then,  disregarding  ice  and  wave  thrust, 

6  x  23'QQ  _  , 
10  x  10125        '* 
We  ought  to  have  «  =  2  at  least. 

The  dam,  then,  is  not  secure  against  sliding  even  when  we  disregard  ice  and  wave  thrust.     Let  us  sup- 
pose. however,  that  the  courses  are  irregular,  so  that  sliding  cannot  take  place. 
We  have  from  (4),  page  427, 

=  6      6  fu 


•  3  *  '" 

From  (II),  page  428,  disregarding  ice  and  wave  thrust,  we  have 

=  23100  x  6.9  -  60750 
23100 


CHAP.  IV.]  STABILITY  OF  DAMS— EXAMPLES. 


433 


wave  thrust.     Since  e  is  greater  than  —6*  =  4  and  less  than  -fa  =  8,  we  have,  from  the  third  of  equations 


Since  e  is  positive,  the  resultant  7?  falls  within  the  base  and  there  is  no  rotation  if  we  disregard  ice  and 
e  thrust.     Since  e  is  greater  than  —6*  =  4 
(III),  page  428,  for  the  greatest  unit  pressure 

f  =  '-JLa!S?,  -  =  36,9  pounds  per  sq.  ft. 


This  is  much  less  than  the  allowable  compressive  unit  stress  C  =  60000  pounds  per  square  foot. 
If  we  take  into  account  wave-pressure,  we  have  T=  24000  and 

_  23100  x  6.9  —  60750  —  432000 

23100  ~  I4'4' 

Since  e  is  negative,  the  resultant  R  now  falls  outside  the  base  and  we  have  rotation. 
The  dam,  then,  if  constructed  so  that  sliding  is  not  possible,  is  stable  when  we  neglect  wave-pressure, 
but  is  unstable  if  we  take  wave-pressure  into  account. 

(3)    The  San  Mateo  dam,  California,  is  built  of  concrete  having  a  density  S  =  150  Ibs.  per  cubic  foot.     The 

height  hi  =  770  feet,  top  base  fa  =  20  feet,  bottom  base  fa  =  176  feet.     The  back  batter  is  i  to  4,  or  tan  fti  =  ^. 
Investigate  the  stability  for  depth  of  water  h  =  165  feet.     (See  example  (6),  page  438.) 

ANS.  We  have  fa  =  fa  +  /&,  tan  /ff,  +  hi  tan  ft-  hence  tan  B  =  —  . 

340 
Also,  from  table  pape  424, 

C=  17  x  2000=  34000  P°Unds,        p  =  0.6. 
sq.  ft. 

For  a  section  one  foot  in  length  we  have 

,r       yh?         0        62.5  x  165  x  161; 

V  =  -  —  tan  fti  =  —  !  -  ?  =  212700  pounds  per  ft., 

2  2x4 

....       yh*       62.5  x  165  x  165 

H  =  ~-  =  —  2  -  ?—      -2  =  850780  pounds  per  ft., 

W  =  —   -  -  ^—  =  150  x  98  x  170  =  2499000  pounds  per  ft. 

For  the  climate  of  San  Mateo  we  have  no  ice.  For  wave-thrust,  taking  T  =  24000,  we  have,  from  (I). 
page  427,  for  security  against  sliding 

_  n(W  +  V}  _  0.6  x  2711700  _ 
"  T+  H  874780 

There  are  no  through  joints,  and  sliding  is  impossible  by  construction.  But  even  if  it  were  not,  the 
dam  would  be  safe  even  if  wave-thrust  is  taken  into  account,  although  the  coefficient  of  safety  is  not  as 
high  as  2. 

From  (4),  page  427,  we  have  ~x  =  101  feet.    From  (II),  page  428,  we  have  then,  taking  T  =  24000,  for  wave- 

pressure  e  =  87  feet.     This  is  greater  than  -fa,  =  58!  ft.  and  less  than  —  fa  =  1  17^  ft.    The  resultant  R  of  the 

•j  3 

weight,  pressure  and  wave-pressure  cuts  the  base,  then,  within  the  middle  third. 

From  the  third  of  equations  (III),  page  428,  then,  we  have  for  the  greatest  unit  pressure 

P  —  I593°  pounds  per  sq.  ft, 

This  is  much  less  than  the  allowable  unit  stress  C  =  34000  pounds  per  sq.  ft. 

If  the  dam  is  empty,  we  have  H  =  o,  V  =  o,  T  =  o,  and  e  =  ~x  =  101  ft.  The  weight  then  cuts  the 
base  within  the  middle  third,  and  we  have 

p  =  8579  pounds  per  sq.  ft.  :  • 

The  dam  is  then  stable,  never  overloaded,  and  safe  even  for  through  joints  and  wave-pressure. 


434  STATICS  OF  RIGID  BODIES.  [CHAP.  IV. 

Design  of  Dam—  Trapezoid  Section.—  If  we  insert  the  value  of  T  as  given  (page  432), 
and  the  values  of  Kand  //given  by  equations  (2)  and  (3),  page  431,  in  the  general  equa- 
tions (I),  (II)  and  (III),  pages  429  to  430,  we  can  design  a  dam  of  trapezoid  section. 

LIMIT  br  —  From  (I),  page  429,  we  have  then 


(I) 


Equation  (I)  gives  the  limit  of  //,  for  low  dam  of  trapezoid  section  for  given  top  base 
and  bottom  base  br     For  //,  greater  than  this  limit  the  dam  is  high. 
For  rectangular  section  we  have  b2  =  b^  ,  and  hence 


Equation  (i)  gives  the  limit  of  /*,  for  low  dam  of  rectangular  section.     If  hv  is  greater 
than  this,  the  dam  is  high. 

LOW  DAM— BOTTOM  BASE.— From  (II),  page  429,  we  have  for  low  dam 

where  the  quantities  Bl  and  £l  are  given  by 


Equation  (II)  gives  the  bottom  base  bz  for  a  properly  designed  low  dam  of  trapezoid 
section.  If  the  value  of  bz  as  given  by  (II),  substituted  in  (I),  gives  the  limit  ^  less  than 
the  actual  height,  the  dam  is  high  and  (II)  does  not  apply. 

Low  DAM—  TOP  BASE.  —  For  economy  the  top  base  should  be  assumed  as  small  as 
possible  consistent  with  practical  and  local  considerations.  We  should  not  in  any  case 
assume  £,  greater  than  br  If  then  we  make  b^  =  bl  in  (II),  we  have  rectangular  section,  and 
for  the  greatest  allowable  value  of  &l 


max.  £,  =  —  B+  V&+£,       ........      (2) 

where  the  quantities  B  and  E  are  given  by 


_  yjl\  tan  P\[  2  A*       3  A  F  _  y^(l  +tan2/?t)  +  67* 

26      \Xf  ~  ~y )'  *A, 

Equation  (2)  gives  the  largest  allowable  value  of  ^  for  low  dam,  trapezoid  section ;  that 
is,  when  A,  is  less  than  or  equal  to  limit  A,  as  given  by  (l). 

Low  DAM — BEST  VALUE  OF  /?,. — We  see  from  the  table    page  424   that  -  is  greater 

y  2 

than  2   for  all  materials  except  brickwork  and  small  dry  rubble.      The  ratio  j-2  can  never 

1 

exceed  unity.      For  all  materials,  then,  except  brickwork  and  small  dry  rubble  the  quantity 


CHAP.  IV.]  DESIGN  OF  DAM—TRAPEZOID  SECTION.  435 

I  r-j- )  is  always  negative,  and  even  for  the  last  two  materials  it  is  also  negative  if  h  is 

not  greater  than  — hr 

In  general,  then,  as  ftl  decreases  the  value  of  Bl  increases  and  the  value  of  E  decreases. 
Also,  Bv  increases  more  rapidly  than  VB*  +  E.  Hence  £2  in  equation  (II)  has  its  least 
value  when  fiv  =  o,  or  the  most  economic  section  for  low  dam,  trapezoid  section,  is  that  for 
which  the  back  is  vertical. 

Low  DAM — ECONOMIC  TRAPEZOID  SECTION. — If  then  we  make  fa  =  o  in  (II),  we  have 
for  low  dam,  economic  trapezoid  section, 


and  for  the  greatest  allowable  value  of  b^,  making  b2  =  b^  we  have  for  rectangular  section 

• 

/yk3  -\-  6Tk  .  ^ 

max.  b^  =  A  / —, v.2  / 

* 

Equation  (2')  gives  the  value  of  bl  for  low  dam,  which  makes  the  economic  section 
rectangular. 

'HIGH  DAM — BOTTOM  BASE. — Inserting  the  values  of  V  and  H  in  (III),  page  43, 
we  have  for  high  dam 

where  the  quantities  J32  and  £2  are  given  by 


Equation  (III)  gives  the  bottom  base  £2  for  a  properly  designed  high  dam  of  trapezoid 
section. 

HIGH  DAM  _  TOP  BASE.  '—  For  economy  the  top  base  should  be  assumed  as  small  as 
possible  consistent  with  practical  and  local  considerations.  We  should  not  in  any  case 
assume  bl  greater  than  by  If  then  we  make  b^  =  bl  in  (III),  we  have  for  the  greatest 
allowable  value  of  £, 

max'.  6l=-JS  +  VB*  +  E,      .........      (3) 

where  the  quantities  B  and  E  are  given  by 


Equation  (3)  gives  the  largest  allowable  value  for  ^  for  high  dam,  trapezoid  section. 

8 
HIGH  DAM  —  BEST  VALUE  OF  ft^—  We  see  from  the  table  page  424  that  -is  greater 

than  unity  for  all  materials.      The  ratio   %  can  never  exceed  unity.      For  all  materials,  then, 

*i 

the  auantitv  f—  -  -}    is  always  negative,  and  -  B2  is  therefore  always  positive  and  decreases 
3  \h*       v> 


43<5  STATICS  OF  RIGID  BODIES.  [CHAP.  IV. 

as  /?,  decreases.     So  also  does  E2.     Hence  £2  in  (III)  has  its  least  value  when  /?,  =  o,  or  the 
most  economic  section  for  high  dam  of  trapezoid  section  is  that  for  which  the  back  is  vertical. 

HIGH  DAM—  ECONOMIC  TRAPEZOID  SECTION.—  If  then  we  make  fa  =  o  in  (III),  we 
have  for  high  dam,  economic  trapezoid  section, 


_i_ 

c- 

and  for  the  greatest  allowable  value  of  bl 


6Th 

............     (30 


CONDITIONS  FOR  RECTANGULAR  SECTION.—  If  we  make  £a  =  bl  in  (I),  and  fa  =  o,  we 
have  for  limit  of  /^  for  low  dam  of  rectangular  sectron 


Ts W 


and  from  (2^)  for  the  value  of  bv 


From  this,  if  we  make  h  =  hv   and  solve  for  Jilt  we  have  for  water  level  with  top  of  dam 


M!_^. 
1    V   r       Y 


We  see  from  this  that  if  ^  is  equal  to\/-^~  »  we  have  h^  —  o,  and  if  ^  is  less  than  this, 
//,  is  imaginary.     We  have  then 


mn.       = 


and  from  (5),  making  /*  =  7/j  ,  and  inserting  the  value  of  h^  from  (4), 


Local  and  practical  considerations  must  control  the  choice  of  top  base  b^  and  we  should 
take  it  as  small  as  such  considerations  will  allow.  We  should  not  take  it  greater  than  given 
by  (7).  If  we  take  it  greater  than  given  by  (6),  the  section  can  be  rectangular. 

If  we  take  it  less  than  given  by  (6),  the  section  cannot  be  rectangular. 

Examples.  —  (i)  Design  a  rectangular  dam  of  granite  ashlar  20  feet  high,  the  water-level  to  be  2  feet  below 
the  top,  -with  and  without  ice-pressure. 

ANS.  We  have  h\  =  20,  h  =  18,  bi  =  b\  ,  ft\  —  o,  P  =  40000,  and,  from  table  page  424,  C  =  60000 
d  =  165. 

From  equation  (I),  page  434,  making  ft,  =  o  and  £1  =  £1,  we  have  for  the  limit  h\  for  low  dam 


2<S         330 


CHAP.  IV.]  DESIGN  OF  DAMS-EXAMPLES.  437 

The  dam  is  therefore  low;  that  is,  when  e  =  -bi  ,p  is  less  than  C. 

From  equation  (IT),  page  435,  making  6-,  =  bi ,  we  have  for  the  value  of  b\  in  order  that  e  shall  be  just 
equal  to  -b\ ,  without  ice-pressure, 


.  ft. 

With  ice-pressure 


A         i/rh*  +  6TA       V62.5  x  5832  +  6  x  40000  x  18 

*'  -  V  —JZT-    =  r  -  I65  x  20    -  -  =  37-67  ft.,       A  =  753-4  sq.  It. 

We  see  that  the  breadth  must  be  quite  large  in  order  to  bring  the  entire  base  into  action. 

(2)  Design  a  trapezoidal  datn^of  granite  ashlar  20  feet  high  with  vertical  face,  the  tangent  of  the  back  batter 


angle  being  —  and  the  water-level  2  feet  below  the  top,  with  and  without  ice-pressure. 


ANS.  We  have  hi  =  20,  h  —  18,  ft  =  o,  tan  fti  =  —,  6t  =,61  +  2,  T  =  40000,  and,  from  table  page  424, 

8  =  165,  C  =  60000. 

From  equation  (II),  page  434,  making  6,  =  61  +  2  and  solving  for  fa,  we  have  for  low  dam 


without  ice-pressure  bi  =  —  0.61  +  4/109.48  =    9.85  ft.,     hence     6,  =  11.85  ft-  '• 


with  ice-pressure  bi  =  —  0.61  +  4/1418.55  =  37.05  ft.,     hence    b*  =  39.05 -ft. 

If  now  in  (I)  page  434,  we  insert  these  values  of  bi  and  6,,  we  find  that  the  limit  of  hi  for  low  dam  in 
both  cases  is  much  greater  than  the  height  hi  =  20.  The  dam  is  therefore  low  in  both  cases,  and  the 
values  of  b\  and  b-t  just  found  hold  good. 

We  have  then  for  the  area  of  cross-section 

without  ice-pressure  A'=  217  sq.  ft., 

with  ice-pressure  A  =  761  sq.  ft. 

We  see  by  reference  to  the  preceding  example  that  we  have  saved  no  material  by  having  a  back  batter. 

We  could,  however,  save  material  by  having  a  back  and  a  face  batter,  as  we  shall  see  from  the  next  example. 

(3)  Design  a  trapezoidal  dam  of  granite  ashlar  20  feet  high,  the  tangent  of  the  front  and  back  batter 

angles  being  —  and  the  water-level  2  feet  below  the  top,  with  and  without  ice-pressure. 

ANS.  We  have  hi  =  20,  h  =  18,  tan  ft\  —  tan  /?  =  — ,  69  =  61+4,   T=  40000,  S  =  165,  C  =  60000   (table 

page  424). 

From  equation  (II),  page  434,  making1  bi  =  b\  +  4  and  solving  for  bi,  we  have  for  low  dam  without  ice- 
pressure 

6,  =  —  3.52  +  4/126.65    =    7.73  ft.,     and  hence     6*  =  11.73  : 
and  with  ice-pressure 

bi  =  —  3.52  +  4/I435.65  =  34.36  ft.,     and  hence     £,  =  38.36. 

From  (I),  page  434,  inserting  these  values  of  6t  and  6,,  we  have  for  the  limit  of  hi  for  low  dam 
without  ice-pressure  limit  hi  =  218.5  ft., 

with  ice-pressure  limit  hi  =  191.6  ft. 

Both  these  limits  are  greater  than  20  ft.  The  dam  is  then  low  in  both  cases,  and  the  values  of  bi  and 
6'y  just  found  hold  good. 

We  have  then  for  the  area  of  cross-section 

without  ice-pressure  A  =  194.6  sq.  ft., 

with  ice-pressure  A  =•  727.2  sq.  ft. 

We  see  by  reference  to  the  preceding  examples  that  we  have  saved  material  by  giving  the  face  and  back 
a  batter, 


438  STATICS  OF  RIGID  BODIES.  [CHAP.  IV. 

We  might,  however,  have  saved  more  still  by  making  the  back  vertical,  as  we  shall  see  from  the  next 
example. 

(4)  Design  a  trapezoidal  dam  of  granite  ashlar  20  feet  high  with  vertical  back,  the  tangent  of  the  front 

batter  angle  being  —  and  the  water-lei>el  a  feet  below  the  top,  with  and  without  ice-pressure. 

ANS.  We  have  hi  =  20,  h  —  18,  fti  =  o,  tan  ft  =  — ,  b%  =  bi  +  2,  T  =  40000,  and,  from  table  page  424, 

6  =  165,  C=  60000. 

From  equation  (II),  page  434,  making  bi  =  <$,  +  2  and  solving  for  £, ,  we  have  for  low  dam  without  ice- 
pressure 

bi  =  —  3  +  ^115.45    =    7.74  ft.,     and  hence     £a  =    9.74  ft.; 
and  with  ice-pressure 

bi  =  —  3  +  ^  \  424. 54  =  34.74  ft.,    and  hence    bt  =  36.74  ft. 

From  (I),  page  434,  inserting  these  values  of  b\  and  b*,  we  have  for  the  limit  of  h\  for  low  dam 
without  ice-pressure  limit  h\  =  268.5  ft-> 

with  ice- pressure  limit  h\  =  188.5  ft- 

Both  these  limits  are  greater  than  20.  The  dam  is  then  low  in  both  cases,  and  the  values  of  b\  and  bi 
just  found  hold  good. 

We  have  then  for  the  area  of  cross-section 

without  ice-pressure  A  =  174.8  sq.  ft., 

with  ice-pressure  A  =  714.8  sq.  ft. 

This  is  the  most  economic  section  for  the  given  proportions. 

We  see,  then,  that  there  is  a  saving  of  material  over  the  preceding  examples  by  making  the  back  vertical. 
We  could  save  still  more,  however,  by  increasing  the  front  batter  angle  or  decreasing  the  top  base,  as  in  the 
next  example. 

(5)  Design  a  trapezoidal  dam  of  granite  ashlar  20  feet  high,  top  base  2  feet,  water-level  2  feet  below  top, 
for  economic  section. 

ANS.  We. have  h\  =  20,  /*=i8,  bi  =  2,  T=  40000,  5  =  165,  C  =  60000,  and  for  economic  section  fti  =  o. 
From  (IT),  page  435,  we  have  for  low  dam 


without  ice-pressure  £»  =  —  i  +  I/  5  H  --  ^  —  -  —  -'  -•=  10.16  ft., 

6   x  20 


165  x  2 

/       62.5x;8-?2       6x40^00x18 
w.th  .ce-pressure  *  =  -  i  +  |/  5  +  ~^^  +        ,65  x  20 

From  (I),  page  434,  inserting  these  values  of  b*,  we  have  for  the  limit  of  h\  for  low  dam 

b*C  10  1  6  x  60000 

without  ice-pressure          limit  hi  =  -r-r  —  2"  =  3°3-8  «i 

8(6t  +  6,)         165  x  1  2.16 

with  ice-pressure  limit  hi  =  —  —  =  344.8  ft. 

105  x  30.74 

Both  these  limits  are  greater  than  20  ft.  The  dam  is  then  low  in  both  cases  and  the  values  of  A,  just 
found  hold  good. 

We  have  then  for  the  area  of  cross-section 

without  ice-pressure  A  =  121.6  sq.  ft., 

with  ice-pressure  A  =  387.4  sq.  ft. 

We  see  that  there  is  a  large  saving  of  material  over  the  preceding  cases.  This  is  the  most  economic 
section  for  the  given  dimensions.  We  could  save  more  only  by  reducing  the  top  base  still  more. 

(6)  Design  the  San  Mateo  dam  (page  433).  taking  the  top  base  at  20  feet  and  the  back  baiter  i  to  4  as 
actually  built,  and  taking  into  account  wave-pressure. 

ANS.  We  have  h\  =  170,  h  =  165,  b\  =  20,  tan  fti  =  -•  y  —  62.5,  5  =  150,  C  =  34000,  T=  24000. 


CHAP.  IV.]  .          DESIGN  OF  DAMS-EXAMPLES.  439 

From  equation  (II),  page  434,  we  have  for  low  dam 


fa  =  —  5.432  +  ^29.51  +  15461.38  =  119.03  ft. 
From  equation    I),  page  434,  we  have  for  the  limit  of  hi  for  low  dam 


limit  hi  =  "          x  34000  _  4  x  150  x  139.03 
159  x  139.03  62.5  x  165' 

This  is  greater  than  the  height  170  ft.     The  dam  is  therefore  low  and  the  value  of  £,  just  found  holds 
good. 

We  have  then  the  area  of  cross-section 

A  =  11818  sq.  ft. 
The  weight  per  foot  of  length  is 

W  =  SA  =  150  x  11818  =  i  772  700  pounds  per  ft., 
the  vertical  pressure  V  (page  431)  is 

Y  IK*  62.5  x  170* 

V  =  '—£-  tan  fti  =  —      g      —  =  22578  pounds  per  ft., 

and  from  the  second  of  equations  (III),  page  428,  the  greatest  unit  pressure  is 


V) 
p  —  -  -yj  -  =  33580  pounds  per  sq.  ft., 

or  less  than  the  allowable  stress  C  =  34000  pounds  per  sq.  ft.,  the  value  of  e  being  —bi. 

The  dam  as  actually  built  (see  page  433)  has  a  base  b*  —  176  ft.,  an  area  A  =  16660  sq.  ft.,  the  value  of  e 
is  greater  than  —  hi,  and  the  greatest  unit  pressure^  =  15930  pounds  per  sq.  ft. 

We  see,  then,  that  by  properly  designing  we  have  for  e  =  —bi,  so  that  the  entire  base  is  still  in  action, 

a  saving  of  over  29  per  cent,  and  the  dam  is  still  safe  against  rotation  and  crushing. 
For  sliding  we  have,  as  on  page  427,  for  the  coefficient  of  safety 

_n(W+  V)  _  0.6  x  1998480  _ 
H  +  T  874780 

Since  n  is  greater  than  unity,  the  resistance  to  sliding  for  through  joint  at  base  is  greater  than  the  thrust  * 
H  +  T;  but  if  friction  only  is  relied  upon,  the  coefficient  is  not  large  enough.     By  breaking  joints,  however, 
sliding  is  impossible. 

By  making  the  back  vertical  we  could  make  a  still  greater  saving,  as  we  see  from  the  next  example. 

(7)  Design  the  San  Mateo  dam  for  economic  section,  taking  into  account  wave-pressure  and  the  top  base 
20  ft,  as  actually  built. 

ANS.  We  have  b\  =  20,  hi  =  170,  h  =  165,  y  =  62.5,  8  =  150,  c  —  34000,  T  =  24000,  and  for  economic 
section  fit  ~  o. 

From  equation  (IT),  page  435,  we  have  for  low  dam 


./      62.1;  x  (165)'   6  x  24000  x  165        , 

b*  =  —  10  +  4/  500  +  — -^ : — iL  +  ±     — 2  —  101.54  ft. 

150  x  170  150  x  170 

From  equation  (I),  page  434,  we  have  for  the  limit  of  hi  for  low  dam 

limit  hi  =  ™l'*>   X  34000=i89ft. 
150  x   121.54 

This  is  greater  than  the  height  170  ft.     The  dam  is  therefore  low  and  the  value  of  bi  just  found  holds 
good. 

We  have  then  the  area  of  cross-section 

A  =  10331  sq.  ft., 


44°  ST/tTICS  OF  RIGID  BODIES.  [CHAP.  IV 

the  weight  per  foot  of  length 

W  —  8 A  =  \y>A  =  i  549650  pounds  per  ft., 

and  from  the  second  of  equations  (III),  page  428,  the  greatest  unit  pressure 

2/F 

p  =  -j-  =  30523  pounds  per  sq.  ft.,  f 

or  less  than  the  allowable  unit  stress  C  =  34000  pounds  per  sq.  ft. 

The  dam  as  actually  built  (see  page  433)  has  a  base  <J,  =  176  ft.,  a  back  batter  of  i  to  4,  and  an  area 

A  =  16660  sq.  ft.     The  value  of  e  is  greater  than  -t>»,  and  the  greatest  unit  pressure  is  only/  =  15930  pounds 
per  sq.  ft. 

By  making  the  back  vertical  and  e  =  -6,,  so  that  the  entire  base  is  still  in  action,  we  save  over  38  per 

cent  on  the  section,  and  the  dam  is  safe  against  rotation  and  crushing. 
For  sliding  we  have,  as  on  page  427,  for  the  coefficient  of  safety 

I*W         0.6  x   1549650 

-7TTT=    —87-4780- 

Since  ft  is  greater  than  unity,  the  resistance  to  sliding  for  through  joint  at  base  is  greater  than  the 
thrust  //  +  T,  but  the  coefficient  is  not  large  enough  if  friction  only  is  relied  upon.  By  breaking  joints, 
however,  sliding  is  impossible. 

(8)  Suppose  the  height  in  the  last  example  to  be  taken  at  200  ft.  ;  the  -water  level  5  ft.  from  the  top. 

ANS!  We  have  £,  =  20,  //,  =  200,  h  =  195,  y  =  62.5,  S  —  150,  C  =  34000.  T  =  24000,  and  for  economic 
section  /Si  =  o. 

From  equation  (IT),  page  435,  we  have  for  low  dam 


=  _  ,o  +  4/  SOP  +  62.5  x  (195)*  +  6  x  24000  x    ,95  f 

10    X    2OO 


150 

From  equation  (I),  page  434,  we  have  for  the  limit  of  hi  for  low  dam 


limit  h,  =  I2°  =  I9?  ft. 

150  x  140 

This  is  less  than  the  height  200  ft.     The  dam  is  therefore  high  and  the  value  of  bt  just  found  does  not 
hold  good.     We  should  then  use  equation  (III'),  page  436.     We  obtain,  then, 


./I5o*x   200  x  (20)*  +  62.5  x(i95)»  +  6  x  24000  x   195 

*  =  \  - 


=  120.30  it. 
34000 


The  area  of  cross-section  is  then 

A  =  14038  sq.  ft. 

Design  of  Dam—  Economic  Section.—  We  have  seen,  pages  435  and  436,  that  for  low 
and  high  dam  and  trapezoid  section  the  greatest  economy  is  for  back  vertical.  We  have 
also  seen,  page  436,  that/or  rectangular  section  b^  must  not  be  greater  than 

*      67- 
+~* 
nor  less  than  |    .     ,     .......     (i) 

6T 


Local  and  practical  considerations  must  control  the  choice  of  top  base  £,  ,  and  we  should 
take  it  as  small  as  such  considerations  will  allow. 


CHAP.  IV.] 


DESIGN  OF  DAM— ECONOMIC  SECTION. 


441 


FIRST  SUB-SECTION.  —  For  given  top  base  blt  then,  less  than  max.  bl  and  greater  than 
min.    bv   as   given   by    (i),    the  section  should   be  a  rectangle,  A^B^DE.  D      &, 

We  can  run  this  rectangular  section  down  to  a  distance  /tl  below  the  top,       d 


until   e  shall  be  just  equal  to  —b^   so    that  the    entire    joint  AlBl    acts, 

3 
provided  this  joint  is  not  overloaded. 

We    have  from  equation  (2'),   page  435,  for   rectangular  section  and 


A 


__± 


Let  d  be  the  distance  of  the  water-level  below  the  top.    Then  h  =  h^  —  d.      Substituting 
this  value  of  h  and  solving  for  //1,  we  have 


X  V  - 


(2) 


Equation  (2)  will  give  the  value  of  h^  for  the  first  sub-section. 

From  equation  (I),  page  434,  we  have  for  the  limit  of  h^  for  rectangular  section 


limit  h,  = 


(3) 


We  can  then  run  the  first  rectangular  sub-section  A^B^ED  down  for  the  distance  /^  given 
by  equation  (2)  without  overloading,  provided  that  /^  as  given  by  (2)  is  less  than  the  limit 
of  h^  given  by  (3)- 

If  b±  is  taken  less  than  */$£!,    there   can   be   no   rectangular   sub-section,    but  the  first 

sub-section  should  be  a  trapezoid,  as  given  in  the  next  article. 

SECOND  SUB-SECTION. If   the  height  of   dam    is  greater  than  the   value  of  h^  as  given 

by  equation  (2),  we    still  continue    the    back    vertical,    but   the 

breadth   must    increase    so    that  e^  shall   be   equal  to  -6r      We 

have  then  a  second  sub-section,  A^B^B^A^  with  front  batter  and 
vertical  back.  Since  the  vertical  pressure  remains  unchanged 
for  dam  empty,  we  can  run  this  second  sub-section  down  to  a 
distance  hz  below  the  top,  until  the  back  edge  distance  s2  for 

dam    empty  is   also   equal  to   --bz,    provided   the    greatest  unit 
stress/  is  less  than  or  equal  to  C.      Or  we  can  run  it  down  until 
p  =  C,  provided  s2  is  greater  than  or  equal  to  -br 

We  have  then  two  cases,  sz  =  e2  =  -3*2,  and  /  less  than  or  equal  to  C,  and  e2  =  -£2, 

p  —  C,  and  s2  greater  than  or  equal  to  er 

The  height  of  the  second  sub-section  is  then  (hz  -  k^.     It  is  required  to  find  //,  and  bv 


442  ST4T1CS  OF  RIGID  BODIES.  [CHAP.  IV. 

Since  the  resultant  for  full  dam  is  to  cut  the  base  b2  =  AyB9  in  both  cases  at  a  distance 
^a=  -£,  from  £3,  we  have  in  both  cases 


From  equation  (5),  page  427, 


We  have  also 


Inserting  these  values,   making  e2  =  —  £2,  and  solving  for  bz  ,  we  have  for  the  base  b^  of 
the  second  sub-section  in  both  cases 


where  the  quantities  B2  and  Ez  are  given  by 


If  we  have  bv  less  than      / — .(  there  is  no  rectangular  section  (page  436),  and  we  have 
A,  =  o  in  (4).  V 

The  weight  Wl  per  foot  of  length  of  the  first  sub-section  acts  at  a  distance  —bv  from  A2 

and  \\\  at  the  distance  (b2  —  ^2)  from  A2.      The  resultant  Wl  -f-  W*  acts  then  at  the  distance 
st  from  Ay  given  by 


Inserting  the  values  of  W7, ,  W^2  and  ^2,  we  have 

-'tj 


When  s2  =  —&2  we  have,  solving  for  //2  for  the  limit  of  //2t 

i 
'when  en  =  S9  =  —b9  and  2//.(£.  —  £,) 

2        2       7   2  >  limit  //,  =  — 1M i (6) 

(  /^, 

/  less  than  or  equal  to  C,  ' 

and  when  e  —  -b   and/  =  C,  we  have  from  the  second  of  equations  (III),  page  428, 


CHAP.  IV.]  DESIGN  OF  DAM— ECONOMIC  SECTION. 

or,  inserting  the  values  of  Wl  and  W2  and  solving  for  //a,  we  have  for  the  limit  of  A2 


when     s2  >  —  b*  —  e2     and 
/  equal  to  C, 


..    .    . 
limit  h   = 


443 


(7) 


If  we  substitute  the  value  of  £2  given  by  (4)  in  equations  (6)  and  (7),  the  least  of  the  two 
values  for  limit  h2  thus  obtained  will  be  the  limiting  value  of  hz  for  the  second  sub-section. 
We  can  then  find  bz  from  (4)  for  any  value  of  hz  up  to  this  limiting  value. 

If  we  have  bl  less  than  \  /  — — ,  there  is  no  rectangular  section,  and  we  make  hl  =  o  in 
(6)  and  (7).  The  limit  h2  in  this  case  is 


THIRD  SUB-SECTION. — If  the  height  of  the. dam  is  greater  than  the  limiting  value  of 
/i2  just  found,  we  have  a  third  sub-section.  The  limit  h2  for  the  second  sub-section  may  be 
given  by  equation  (6)  or  by  equation  (7)  as  we  have  seen,  and  we  thus  have  two  cases  to 
consider. 

ist  Case. — If  the  limit  hz  is  given  by  equation  (6),  that  is,  if  e2  =  s2  —  -  b2  and  p  less 
than  or  equal  to  C,  we  must  batter  both  face  and  back,  so 
that  e3  shall  be  greater  than  —6S  and  /  =  C  for  dam   full. 

We    have   then   the    sub-section  A2B2£3AS  with  front  and 
back  batter.      The  back  batter  angle  we  denote  by  fly 

We  can  run  this  sub-section  down  to  the  base  of  the 
dam,  so  that  h^  is  the  given  height  of  dam ;  and  since  the 
vertical  pressure  is  unchanged  for  dam  empty,  we  should 
have  the  edge  distance  s3  from  As  for  dam  empty  equal  to 
the  edge  distance  e^  from  B3  for  dam  full.  It  is  required 
to  find  b^  and  fiy 

The  resultant  Wl  +  J^  is  at  the  distance  s2  =  -&2  from 

A2  for  dam  empty.     The  weight  W3  is  at  a  distance  -?3  from  B^  given  from   equation  (5), 
page  427,  by 


•      f, 

d 

&J  +  &7 

V 

1' 

"\        ' 

A 

y 

V 

i 

"T 

T-^A- 

tan 


(9) 


There  is  a  vertical  water-pressure  V  on  the  inclined  back  A2A3.  For  the  sake  of  sim- 
plicity we  neglect  V.  This  omission,  which  is  on  the  side  of  safety,  involves  no  prac- 
tical error. 

Neglecting  V,  then,  we  have  the  edge  distance  s3  from  A3  given  by 


+  (A,  - 


tan 


(10) 


ST/1TICS  OF  RIGID  BODIES.  [CHAP.  IV. 

For  dam  full,  neglecting  Vt  we  have  the  edge  distance  e^  from  B^t 


*8  -      -  (h,  -  kj  tan  A    +  Wf,  -  -  T(h,  -  d) 


where  Wv  and  W^  are  already  known  and 

„,_*(*!  +  W»-  A.)  rr 

w-          ~~ 


From  the  third  of  equations  (III),  page  428,  we  have  when  es  is  greater  than  -£3  and 
p  =  C,  neglecting  F, 

Cbj 


2  +  wy 


If  we  equate  (n)  and  (12),  insert  the  values  of  Wy  H  and  ^3,  and  solve  for  £3,  we  have 
for  the  base  £3 


where  the  quantities  B3  and  £3  are  given  by 

2(  yrt  +»;)-*(*,-*,)»  tan  /?, 

~" 


If  we  have  bl  less  than  *  /-_.,  there  is  no  rectangular  section  (page  436),  and  we  make 

Wl  =  o,  //,  =  o. 

If  we  equate  (10)  and  (12),  insert  the  value  of   W^  and  solve  for  /?3,  we  have  for  B^ 


_  2(^.  +   ^(2b,  -  b^  +  6(h,  -  k&b*  +  V,  -  ^)   -   Cb} 

(*.  -  *,)[*(*,  -  ^2)(^3  +  2^2)  +  6(^  H-  H/)] 

If  we  have  bv  less  than  A  /  —  ,  there  is  no  rectangular  section  (page  436),  and  we  make 

PF,  =  o,  //,  =  o. 

From  (13)  and  (14)  we  can  find  £3  and  /?3. 

2d  Case.  —  If  the  limit  h^  is  given  by  equation  (7),  that  is,  if  s2  is  greater  than  ea  =  —bz 

and  /  equal  to  C,  and  if  the  height  of  the  dam  is  greater  than  this  limit,  we  have  also  a  third 
sub-section. 

In  this  case  we  still  continue  the  back  vertical,  but  the  breadth  must  increase  so  that 

f3  =  ^3  shall  be  greater  than  —  #3  and  /  =  C.     We  have  then  the  third  sub-section  A2  B2  B3  A3 
with  vertical  back. 


CHAP.  1V.J                                       DESIGN  OF  DAM-ECONOMIC 

Since  the  vertical  pressure  remains  unchanged   f 
section  down   with   vertical    back    until  the  edge  dist 
from  the  back  for  dam   empty   is  equal  to.  the  edge 
e3  from    Bz  for  dam  full.     The  height  of  this  sub-s( 

SECTION.                                                445 

or  dam  empty,  we  can  run  this  sub- 
ance    s^              D                E 

distance        d 

•    x 

n 

action  is 

.A, 

/ 

=   S2   I*1 

A* 
(15) 

7/3  —  hy      It  is  required  to  find  7/3  and  by 
Evidently  we  have  only  to  make  /J3  =  o  and  — 
equations  (13)  and  (14)  and  we  have 

\ 

X 

\ 

bz  =  —  B  +  VB*  +  E     .     .     .     . 

where  the  quantities  B3  and  E3  are  given  by 

A, 

6,                   B» 

^             C       '         E*                 C                          C 

,     C&*  —  2(W.+ 

limit  k.  =  h-A  .  ..  *     . 

+                   C 

If  we  substitute  the  value  of  £3  given  by  (15)  in  (16)  and  solve  for  7/3,  we  shall  have  the 
limiting  value  of  hz  for  the  third  sub-section  in  this  case.  We  can  then  find  bz  from  (15)  for 
any  value  of  h^  up  to  this  limiting  value. 

/' sr  *r* 

If  we  have  bl  less  than  A  I  ,  there  is  no  rectangular  section  (page  436),  and  we  make 

W^=.  o  and  h^  =  o.  , 

FOURTH  SUB-SECTION. — If  the  height  of  the  dam  is  greater  than  the  limiting  value  of 
h3,  when  we  have  the  third  sub-section  with  vertical  back,  we  must  batter  both  face  and  back 

so  that  e±  shall  be  greater  than  —  b±  and  p  =  C  for  dam  full. 


We  have  then  a  fourth  sub-section,  A^ByB^A^,  with  front 
and    back    batter.      We    denote    the    back  batter    angle 

by  A- 

We  can  run  this  sub-section  down  to  the  base  of  the 
dam,  so  that  //4  is  the  given  height  of  dam  ;  and  since  the 
vertical  pressure  is  unchanged  for  dam  empty,  we  should 
have  the  edge  distance  s4  from  A±  for  dam  empty  equal  to 
the  edge  distance  e^  from  Bt  for  dam  full.  It  is  required 
to  find  b±  and  /?4. 


In  this  case  we  have  from  (9)  and  (10),  by  making  /33  =  o, 


and  from  (13)  and  (14) 


07) 


446  ST/tTlCS  OF  RIGID  BODIES. 

where  the  quantities  B±  and  £t  are  given  by 

2(  W,  H-  W%  +  **g  -  6(A<-  h,  y  tan  ft 
~ 


[CHAP.  IV 


From  (17)  and  (18)  we  can  find  bt  and  /?4. 

If  we  have  b,  less  than  A  /-—  ,  there  is  no  rectangular  section  (page  436)  and  we  make 
V     " 


^  =  o,  A,  =  o. 

For  dam  full  and  empty/  =  C,  and  in  both  cases  et  =  s4,  or,  from  (12), 


(19) 


Examples,— (t)  The  height  of  the  proposed  QuakerBridge  Dam,  New  York,  is  170  ft.,  top  thickness  20  ft., 
density  of  masonry  150  pounds  per  sq.  ft. ,  depth  of  water  163  ft.,  allowable  compressive  stress  20000  pounds 
per  sq.  ft.  Find  the  economic  section  without  ice-pressure  or  wave-thrust. 

ANS.  We  have  d  =  7,  £,  =  20,  C  —  20000,  8  =  150,  y  =  62.5,  and  without  ice-pressure  or  wave-thrust 
T  —  o. 

The  top  base  b\  =  20  is  less  than  maximum  b\  and  greater  than  minimum  b\  as  given  by  equations  (i) 
page  440,  viz., 

max.  bi  =  i/  — -r-s  =  43  ft.     and      min.  b\  =  y  — —  =  o. 

The  first  sub-section  is  then  rectangular. 

First  Sub-section.— For  the  height  of  this  rectangular  sub-section  we  have,  from  equation  (2),  page  441, 

//is  -  2i//ia  —  813^1  =  343,     or    //,  =  41.02  ft. 
This  value  of  hi  is  less  than  the  limit  //i  given  by  equation  (3),  page  441,  viz., 


limit  *,  =        =- 


300 


=  66*  ft. 


The  base  is  then  not  overloaded.     The  unit  stress  is  then,  from  the  second  of  equations  (III),  page  428, 

D     6,     E 
d 


/  B'\ 

A    »-       X 


2Wj 

bi    ' 


We  have  for  the  wejght  W^  and  cross-section  A, 
Wi  —  Sbihi  =  123060  pounds  per  ft.         Ai  —  820  sq.  ft., 


^  and  hence  for  the  greatest  unit  stress  at  front  edge  B  \ 


P  =  12306  pounds  per  sq.  ft., 
and  for  the  greatest  unit  stress  at  back  edge  At  ,  for  dam  empty, 


P  =  —r-  =  61530  pounds  per  sq.  ft. 


V  B, 

Second  Sub-section. — If  in  equation  (4),  page  442,  we  assume  for  a  first  approximation  h*  =  78,  we  have 

#*  =  54-54,     £"1  =  5760.32     and 
A  =  -  54-54  +  ^8734.92  =  38.91- 


CHAP.  IV.] 


EXAMPLES. 


447 


Substituting  this  value  of  bi  in  equations  (6)  and  (7),  page  443,  we  find  the  least  value  of  limit  hi  given  by 
(6),  and  from  (6)  we  have  limit  hi  =  77.5,  or  less  than  we  assumed  it. 

If  we  assume  again  hi  =  78.5  in  equation  (4),  we  have  bi  =  39.33,  and  from  (6),  hi  =  79.2,  or  greater  than 
we  assumed  it. 

The  value  of  hi  is  then  between  78  and  78.5.  If  we  try  again  for  h*  =  78.1,  we  have,  from  (4),  hi  =  39.07. 
and,  from  (6),  hi  —  78.187,  or  slightly  larger  than  we  assumed  it.  The  value  of  hi  is  then  between  78  and  78.1, 

It  will  then  be  quite  accurate  if  we  take 

hi  =  78.1  ft.     and     bi  =  39.07  ft. 
f 

The  weight  per  ft.  of  the  second  sub-section  and  the  area  of  cross-section  are  then 

—  =  164262  pounds  per  ft.,     Ai  =  1095  sq.  ft. 

Since  the  least  value  of  limit  Jii  is  given  by  equation  (6),  page  442,  we  have  s-i  =  bi  =  —  bi,  and  hence  from 
the  second  of  equations  (III),  page  428,  we  have  for  the  greatest  unit  stress  for  both  dam  full  and  empty 

2(  Wi  +  Wi} 
p  -        — ^ —       =  14715  pounds  per  sq.  ft. 

The  tangent  of  the  front  batter  angle  is  given  by  ~ =  0.51. 

Third  Sub-section. — Since  the  least  value  of  limit  hi  is  given  by  equation  (6),  page  442,  we  have  the  first 
case,  given  on  page  443.  The  third  sub-section  then  extends  to  the  base  of  the  dam  and  has  a  front  and 
back  batter.  We  have  the  base  ba  and  back  batter  /3S  from  equations  (13)  and  (14),  page  444. 

If  in  (13)  we  assume  tan  ft3  —  0.2,  we  have  £a  =  8.02,  Et  =  18288.38  and 

£3  =  -  8.02  +  4/18352.7  =  127.45. 

If  we  substitute  this  value  of  b*  in  (14),  we  have  tan  /J3  =  0.16,  or  less  than  we  assumed  it. 
If  we  try  again  for  tan  /3t  =  0.15,  we  have,^  =  9.61,  E»  =  17643.69  and 


ft  =  —  9.61  +  4/17736.04=  123.56. 

With  this  value  of  t>,  in  (14)  we  have  tan  fia  =  0.17,  or  greater  than  we  assumed  it. 

The  value  of  tan  ySs  is  then  between  0.2  and  0.15.     If  we  assume  tan  /33  =  0.17,  we  have  B\  =  8.97 
E  =  17901.57  and 

b*  =  —  8.97  +    4/17982.03  =  125.12. 

With  this  value  of  bs  in  (14)  we  have  tan  /$3  =  0.169,  or  almost  exactly  what  we  assumed  it. 
We  have  then  for  the  third  section 

//a  —  hi  =  91.9,         bt  =  125.12,         tan  /33  =  0.17. 
The  weight  per  foot  of  the  third  sub-section  and  the  cross-section  are  then 

h^  =  i  131  670  pounds  per  ft.,         A,  =  7544  sq.  ft. 


The  tangent  of  the  front  batter  angle  is  given  by 


=  o  76. 


The  unit  pressure  back  and  front  is  20000  pounds  per  sq.  ft. 

The  front  edge  distance  is,  from  (12),  ;,   =  46.64,  and  the  back  edge  distance  the  same. 

We  have  then  the  following  table: 


h 

b 

A 

tan  P  back. 
O 

O 
0.17 

tan  /3  front. 

e 

• 

/  back. 

/    front. 

41.02 

78.1  . 

170 

20 
39-07 
125.12 

820 
1095 

7544 

O 
0.51 
0.76 

~     6.6 
13.02 
46.64 

10 

13.02 
46.64 

61530 
?47T3 
2OOOO 

125060 

M7I3 
20OOO 

9459 

44^  STATICS  OF  RIGID  BODIES.  [CHAP.  IV. 

In  this  table  the  first  column  contains  the  distance  in  feet  from  the  top  of  the  dam  to  the  bottom  of  each 
sub-section,  the  second  the  base  in  feet  of  each  sub-section,  the  third  the  area  in  square  feet  of  each  sub- 
section, the  fourth  and  fifth  the  tangent  of  the  back  and  front  batter  angles,  the  sixth  and  seventh  the  back 
and  front  edge  distances  in  feet,  the  last  two  the  greatest  unit  stress  back  and  front  in  pounds  per  square  foot. 

The  total  area  is  9459  sq.  ft.  We  see,  from  example  (3),  page  433,  that  in  the  case  of  the  San  Mateo  dam, 
which  has  the  same  height  and  density  and  an  even  greater  allowable  unit  stress  C,  the  area  is  15810  sq.  ft. 
There  is  a  saving  in  this  case  of  40  per  cent,  due  to  economic  section. 

(2)  Design  tht  proposed  Quaker  Bridge  Dam,  Neu  York,  taking  ice-pressure  into  account,  dimensions  and 
data  as  given  in  the  preceding  example.  , 

ANS.  We  have  T  =  40000  and,  as  before,  d  =  j.  t*  •=  20,  C  =  20000,  8  =  150,  y  =  62.5. 

Th«  top  base  £,  =  20  ft.  is  less  than  y  -r-  **  t°  ft-,  and  therefore  (page  436)  there  is  no  rectangular 

sub-section.     We  have  then  to  make  (f  i  =  o  and  ^1=0  wherever  they  occur  in  our  equations. 

First  Sub-section.— The  first  sub-section,  Ihen,  is  given  by  equations  (4)  and  (7),  page  443,  remembering 
to  wake  \V\  =  o,  h\  =  o. 

If  we  assume  h*  =  102  ft.  in  (4),  we  hs»ve  .#,  =  10,  E,  =  5239.48  and 

£,  -  -  10  -f-   4/5339.98  =  63.07  ft. 

Substituting  this  in  (7).  we  fcav<  /*,  =  99.6  ft.,  or  less  than  we  assumed. 
If  we  assume  h*  =  101  ft.  in  (4),  we  have  2?»  =  10,  E*  =  5315.61  and 


fa  =  -  10+   4/5415-61  =  63.59  ft. 

Substituting  this  in  ft),  we  have  h*  =  101.43  ft.,  or  greater  than  we  assumed.     The  value  of  ht  is  then 
between  101  and  102  *t. 

Assuming  //t  =  101.5  ft.  in  (4),  we  have  Bt  =  10,  Et  =  5353.97  and 


Substituting  this  in  (7),  we  have  //j  =  101.47  i*->  or  almost  exactly  what  we  assumed. 
We    have    then  for  the  first  sub-section  a  trapezoid  with  vertical  back  and  height  and  lower  base 
given  by 

»  ht  =    IOI.5  ft"  *»  =  63.7  ft. 

We  have  then  the  weight  W9  per  foot  of  length  and  the  area  of  cross-section  At  , 
W,  =  S^  *  W*  =  633166  pounds  per  ft.,        A*  =  4214  sq.  ft. 

We  have  also  for  the  dam  full  e3  =  —t>,  =  21.33   fc-»  an{^  tne  greatest  unit    pressure  p  =  C=  20000 

pounds  per  sq.  ft. 

For  the  dam  empty  we  have  from  (5),  page  428,  making  ^,  =  o,  s*  =  22.82  ft.,  and  from  the  third  of 
equations  (III),  page  442,  the  greatest  unit  pressure  p  =  18388  pounds  per  sq.  ft. 

The  tangent  of  the  front  batter  angle  is  given  by  —      —  =  0.43. 

h* 

Second  Sub-section.  —  Since  the  limit  of  //»  is  given  by  equation  (7),  page   443,  we  have  the  second  case, 
given  on  page  445  by  equations  (15)  and  (16),  page  445,  after  making  //i  =  o,  W\  =  o. 
If  we  assume  h%  =  107  ft.  in  (15),  we  have  £»  =  31.66,  Et  =  8827.015,  and 

bt  =  —  31.66  +  4/9829.365  =  67.48. 

Substituting  this  in  (16),  we  have  h\  =  111.04,  °r  greater  than  we  assumed. 
If  we  assume  At  =  110.5  ft-  >n  ^5)«  we  nave  B*  =  31.66,  E*  =  9315.26  and 

bt  =  —31.66  +  4/10317.61  =  69.91. 

Substituting  this  in  (16),  we  have  As  =  110.81,  or  greater  than  we  assumed. 
If  we  assume  h%  =  1  10.8  in  (15),  we  have  Bt  =  31.66,  Et  =  9358.2  and 

b*  =  —  31.66+  4/10360.55  =  70.12. 


CHAP.  IV.]    , 


EXAMPLES. 


449 


Substituting  this  in  (16),  we  have  h*  =  110.81,  or  almost  exactly  what  we  assumed. 

We  have  then  for  the  second  sub-section  a  trapezoid  with  vertical  back  and  height  and  lower  base 
given  by 

7it  —  h  =  9.3  ft.,        b»  =  70.  12  ft. 
The  weight  W*  per  ft.  of  length  and  the  area  of  cross-section  A*  are  then 


We  have  for  the  dam  full  and  empty/  =  C  =  20000  pounds  per  square  foot,  and  in  both  cases  e»  =s  st, 
or  from  equation  (12),  page  444,  making  W^  =  o, 

<?,  =  s3  =  24.18  ft. 
The  tangent  of  the  front  batter  angle  is    *  ~     '  =  0.69. 

Third  Sub-section.—  We  have  the  last  sub-section  given  by  equations  (17)  and  (18),  page  446,  after 
making  Wi  =  o  and  h^  =  o. 

If  we  assume  tan  fi<  =  a  i  in  (17),  we  have  B^  —  35.06,  Et  =  23439.6  and 

bi  =  —  35.06  +  ^24668.  8  =  122. 

Substituting  this  in  (18),  we  have  tan  /34  =  0.29,  or  greater  than  we  assumed. 
If  we  assume  tan  /34  =  0.3,  we  have  B4  =  32.47,  E<  —  24434.8  and 


£4  =  -  32.47  +  4/25489.1  =   127.18. 

Substituting  this  in  (18),  we  have  tan  /54  =  0.29,  or  less  than  we  assumed. 
If  we  assume  tan  ft*  =  0.2,  we  have  £t  =  32. 56,  E*  =  24385  and 


<$4  =  —  32.56  +  ^25445.15  =  126.94. 

Substituting  this  in  (18),  we  have  tan  dt  =  0.29,  or  what  we  assumed.    We  have  then  for  the  last  sub- 
section a  trapezoid  whose  back  batter  is  given  by 

tan  dt  =  0.29, 

whose  height  hi  -  h*  =  59.2  ft.  and  whose  base  0«  =  126.94  ft.    The  tangent  of  the  front  batter  angle  then  is 
b\  —  b*  —  (hi  —  hi)  tan  /34 

—      _    —  :         —  =  0167.     The  weight  per  foot  of  length  and  the  area  of  cross-section  are 


We  have  for  dam  full   and  empty  p  —  C  =  20000   pounds  per  sq.  ft.,  and  in  both  cases  /4  =r  j4,  or 
from  equation  (19),  page  446,  making  W\  =  o, 

ei  =  St  =  51.08. 
We  have  then  the  following  table  : 


h 

b 

A 

tan  (3  back. 

tan  /3  front. 

e 

* 

/  back. 

/  front. 

101.5 
no.  8 
170 

63.7 

7O.  12 
126.94 

4214 
622 

5833 

O 
O 

0.29 

0-43 
0.69 
0.67 

21-33 
24.18 
51.08 

22.82 
24.18 
51.08 

18388 
2OOOO 
2OOOO 

20000 
2OOOO 
20000 

10669 

In  this  table  the  first  column  gives  the  distance  in  feet  from  the  top  of  the  dam  to  the  bottom  of  each 
sub-section,  the  second  the  base  in  feet  of  each  sub-section,  the  third  the  cross-section  in  square  feet  of  each 
sub-section,  the  fourth  and  fifth  the  tangent  of  the  back  and  front  batter  angles,  the  sixth  and  seventh  the 
back  and  front  edge  distances  in  feet,  the  last  two  the  greatest  unit  stress  back  and  front  in  pounds  per 
square  foot. 

The  total  area  is  10669  S<1-  ft-  We  see  from  example  (3),  page  443,  that  in  the  case  of  the  San  Mateo 
dam,  which  has  the  same  height  and  density  and  an  even  greater  allowable  C,  the  area  is  isSiosq.  ft.  There 
is  then  a  saving  in  the  present  case,  due  to  economic  section,  of  over  32  per  cent. 


450 


STATICS  OF  RIGID  BODIES. 


[CHAP.  TV. 


FIG. 


Arch  Dam.  —  When  the  dam  is  made  in  the  form  of  an  arch,   so  that  it  supports  the 
water-pressure  back  of  it  wholly  by  reason  of  its  action  as  an  arch,  it  is  called  an  arch  dam. 

The  water-pressure    upon  the  back  must  always  be   normal  to   the   surface,    and  the 
pressure  upon  a  unit  area  always  the  same  at  the  same  depth. 

Let  aaa,  Fig.  I,  be  the  centre  line  of  a  horizontal  cross-section  of  the  dam,  one  foot  in 

height.  Let  />,  and  P%  be  the  equal 
normal  pressures  upon  the  equal  por- 
tions a'a!  ',  a'a',  and  H  the  horizontal 
pressure  at  the  crown. 

In  Fig.  2,  lay  off  H  from  O  to  o 
horizontally,  and  let  Oo  represent  the 
magnitude  of  H.     Then  lay  off  o  i  and 
I  2  parallel  and  equal  in  magnitude  to 
jP,  and  />,,  and  draw  the  rays  Oi,  O2. 
In  Fig.  I,  let  H  act  at  a,  and  prolong  its  direction  till  it  meets  /^  at  b.      From  b  draw 
be  parallel  to  O\  till  it  meets  P3  at  c.      From  c  draw  ca  parallel  to  O2. 

Then    (page  413)  abca,  Fig.    I,    is    the    equilibrium    polygon.      We    have    by    similar 
triangles 


P'.H::cb'.bCor  cC\ 


cb       cC 


The  same  holds  true  no  matter  how  many  equal  portions  a'a'  we  take.      But  as  we 
increase  the  number  of  portions,  the  polygon  approaches  a  curve.      For  an  indefinitely  great 

p 
number  of  portions  we  have  for  the  curve  of  equilibrium  cb  =  ds,  also  -±  =  p  =  the  unit 

pressure  and  cC  =  r  =  the  radius  of  curvature.      Hence 


or 


H  —  rp  X  ds. 


But/  must  be  constant,  and  we  see  from  the  construction  that  H  is  constant.  There- 
fore r  is  constant  and  the  equilibrium  curve  is  a  circle. 

If,  then,  we  make  the  dam  circular  in  cross-section,  as  shown  in  Fig.  i,  the  curve  of 
equilibrium  will  coincide  with  the  centre  line,  and  the  horizontal  pressure  H  at  the  crown 
acts  at  the  centre  line  and  is  given  by 


ds. 


(0 


Also,  since  in  Fig.  2  the  force  polygon  012  becomes  a  .circle  of  radius  //  when  the  seg- 
ments of  the  arch  are  indefinitely  great  in  number,  and  since  any  ray,  as  o  i  in  Fig.  2,  gives 
the  stress  in  the  corresponding  segment,  cb,  Fig.  I,  of  the  equilibrium  polygon  (page  413), 
it  is  evident  that  the  pressure  at  every  point  of  the  centre  line  is  tangent  to  the  centre  line  at 
that  point  and  equal  to  H. 

If,  then,  C  is  the  allowable  stress  per  square  foot,  we  have  for  the  area  A  of  cross- 
section  of  the  arch  one  foot  in  depth  at  any  distance  below  the  surface 


X  ds 


CHAP.  IV.] 


ARCH  DAM. 


45' 


where/  is  the  unit  pressure  at  any  distance  below  the  surface  of  the  water,  and  C  is  given 
by  our  table  page  424. 

FIRST  SUB-SECTION.  —  Let  the  water-level  be  at  a  distance  d  below  the  top  (Fig.  3). 
Let  the  top  base  be  b^  ,  and  let  the  dam  be  rectangular  for  a 
distance  /il  below  the  top. 

At  this  depth,  then,  we  have  \~f~ 


-    -          - 
*~  ds      C' 

But  the  unit  pressure  of  the  water  at  the  depth  til  —  d  is 
y(h-i  —  d},  where  y  is  the  mass  of  a  cubic  foot  of  water,  or 
62.5  Ibs. 

If  T  is  the  ice-thrust  at  the  surface  per  unit   of  length, 

—  is  the  unit  pressure  due  to  ice-thrust  if  h  is  the  depth  of 
water.      The  total  unit  pressure  is  then 


We  have  then 


or  the  height  kl  of  the  first  rectangular  sub-section   in  order  that  the  allowable  unit  stress 
C  may  be  just  attained  is  given  by 


J,        J+. 

"i  —  d  ~\ —  ~T. 

r  yr        yh 


(2) 


TOP  THICKNESS. — The  choice  of  top  thickness  bl  must  in  general  be  determined  by 
local  and  practical  considerations.  We  can  scarcely  take  b^  =  o,  or  have  no  rectangular 
portion,  as  it  would  not  be  allowable  to  bring  the  top  to  an  edge.  Some  breadth  must  be 
assumed. 

We  have,  from  (2), 

.    ..../.,,.     .     .     .'.    .     (3) 


Equation  (3)  will  give   bl  for  any  value  of  h^  and  'd  we  choose.       Thus   if  we  take 

\  =  -j-^,,  we  shall  have  h*  =  d.      We  ought  not  to  take  b  less  than  this.      If  we  take 
hi. 


we  shall  have  hv  equal  to  the  entire  height  h2  of  the  dam,  and  the  entire  cross-section  will 
be  rectangular.      We  cannot  take  bl  greater  than  this. 
Between  these  two  values  of 


rT  yr 

and   *  =        ~ 


rT 


(4) 


we   can  assume  £x  what  we  wish,  and   will  have  then  /^  greater  than  d  and   less  than 
There  will  then  be  a  second  sub-section  not  rectangular. 


452 


ST/tTKS   OF  RIGID  BODIES. 


[CHAP.  IV. 


SECOND  SUB-SECTION. — For  any  distance  y  below  the  top  greater  than  //,  and  less  than 
//2  (Fig.  3)  we  have  the  thickness 


and  the  total  unit  pressure 


Hence  we  have  for  the  thickness  at  any  distance  y  from  the  top  when  y  > 


(5) 


The  arch  dam  requires  far  less  masonry  than  the  gravity  dam,  but  the  pressure  on  the 
arch  stones  increases  with  the  depth  and  span,  and  so  does  the  thickness.  When  the  thick- 
ness becomes  great  we  cannot  be  sure  that  each  arch  stone  will  take  its  own  share  of  the 
pressure.  The  distribution  of  the  pressure  over  the  cross-section  is  then  uncertain.  For 
such  reasons  the  arch  dam  is  best  restricted  to  short  and  low  dams.  It  is  also  evidently 
unwise  to  make  the  stability  of  a  dam  depend  wholly  upon  its  action  as  an  arch,  except 
under  the  most  favorable  conditions  as  to  rigid  side  hills  for  abutments,  and  the  most 
unfavorable  conditions  as  to  cost  of  masonry. 

Although  it  is  not,  then,  generally  wise  to  make  the  stability  of  a  dam  depend  wholly 
upon  its  action  as  an  arch,  it  is  well  to  make  a  gravity  dam  curved  up-stream,  so  that  the 
arch  action  may  give  additional  security. 

There  are  but  two  examples  of  the  pure  arch  type:  the  Zola  dam  in  the  city  of  Aix  in 
France,  and  the  Bear  Valley  dam  in  the  San  Bernardino  Mountains,  Southern  California. 
The  first  is  of  rubble  masonry,  height  120  ft.,  radius  158  ft.,  thickness  at  top  19  ft.,  at  base 
42  ft.  The  Bear  Valley  dam  is  of  granite,  height  64  ft.,  radius  300  ft.,  thickness  at  top 
3.16  ft.,  at  base  20  ft. 

Examples. — (i)  The  Bear  Valley  dam  in\the  San  Bernardino  Mountains,  Southern  California,  is  an  arch 
dam  of  granite  ashlar,  radius  r  =  300  ft.,  top  base  b\.  =  j.  /7  ft. ,  bottom  base  bi  —  20  ft.,  depth 
of  water  h  =  60  ft.,  height  hi  =  64  ft.,  and  therefore  d  =  4  ft.,  face  vertical  and  other 
dimensions  as  shown  in  the  following  figure.  Examine  its  stability, 

ANS.  We  have  from  the  given  dimensions  and  from  equation  (5),  neglecting  the  ice- 
thrust  T,  for 

y  =  12          24  36  48  64  ft. 

[/  =    448      5-79        7-1  8.42         20  " 

C=  —  (y  —  d)  =  16.72     32.22      42.25        49.  28.12  tons  per   square  foot. 

From  page  424,  the  allowable  unit  stress  fought  not  to  exceed  30  tons  per  square  foot. 
The  dam  as  built  then  has,  except  at  top  and  bottom,  a  much  higher  unit  stress  than  ordi- 
nary practice  would  consider  allowable. 

(2)  Design  an  arch  dam  of  the  same  height  and  radius  as  the  Bear  Valley  dam,  for 
same  depth  of  water  and  for  an  allowable  compressive  stress  of  30  tons  per  square  foot. 

ANS.  We  have  h  =  60,  hi  =  64,  </  =  4«  r  =  300,  y  =  62.5,  C=  60000.  Taking  the 
ice-thrust  T  =  40000  pounds  per  ft.,  we  have,  from  equations  (4),  page  451,  b\  =  3^  ft.  and  b\ 
=  22.08  ft.  We  should  not  take -<Ji  less  than  the  first  value,  in  which  case  7/i  will  equal 
d,  nor  greater  than  the  second,  in  which  case  h\  will  equal  hi.  Let  us  then  take  b\  =  4  ft. 
ist.  Without  Ice-thrust. — For  the  first  sub-section  we  have,  from  equation  (2),  page  451, 


hl  = 


62.5  x  300 


^  =  16.8  ft. 


CHAP.  IV.] 


EXAMPLES. 


453 


The  cross-section  is  then  rectangular  for  a  distance  hi  —  16.8  ft.  from  the  top. 
Below  this  point  the  thickness  should  increase.  We  have  then,  from  equation  (5),  page 
452,  for  any  value  of  y  greater  than  16.8  the  corresponding  thickness 


If  we  take 

y  —  16.8       20 


44 


64  ft., 


we  have 


8.75       12.5       18.75 


If  we  make  the  face  vertical  and  batter  the  back,  we  have  then 
the  cross-section  shown  in  the  left-hand  figure,  4  ft.  thick  for  the 
first  1 6. 8  ft. 

2d.  With  Ice-thrust. — For  the  first  sub-section  we  have,  from 
equation  (2),  page  451, 

60000  x  4  40000        _ 

*'  ~  4  +  62.5  x  3~So      ~627Tx  60    - 

The  cross-section  is  then  rectangular  for  a  distance  ki  =  6. 14 
ft.  from  the  top.     Below  this   point  we  have,  from  equation  (5), 
page  452,  for  any  value  oiy  greater  than  6.14  the  corresponding  thickness 

300  x  40000 


62.5  x  300 

60000 


If  we  take 
we  have 


y  =  6.14 


20 
8.33 


32 
12.08 


44 
15.83 


64ft., 
22.08  ft. 


CHAPTER  V. 

RETAINING  WALLS.      EARTH    PRESSURE.     EQUILIBRIUM   OF   EARTH. 

Retaining  Wall. — A  wall  designed  to  resist  the  pressure  of  earth  back  of  it  is  called  a 
retaining  wall. 

In  the  case  of  a  dam  we  know  that  the  water-pressure  is  normal  to  the  submerged  sur- 
face, and  its  magnitude  is  as  given  on  page  431.  This  pressure  acts  at  a  distance  from  the 

bottom  of  dl  =  -A,  where  //  is  the  depth  of  water.      Its  vertical  and  horizontal  components 

Fand  //"are  then  known  in  magnitude  and  point  of  application,  as  given  on  page  431,  and 
in  the  preceding  chapter  we  have  seen  how  to  investigate  the  stability  and  design  a  dam.. 

In  the  case  of  a  retaining  wall  the  earth-pressure  is  not  in  general  normal  to  the  surface, 
and  the  rule  for  magnitude  of  earth-pressure  no  longer  holds  for  earth.  A  special  investi- 
gation is  therefore  necessary  in  order  to  find  Fand  H  in  the  case  of  a  retaining  wall.  When 
once  these  are  known  in  magnitude  and  point  of  application  the  general  principles  and 
equations  already  given  in  the  preceding  chapter,  apply  at  once  and  we  can  investigate  the 
stability  or  design  a  retaining  wall. 

It  then  only  remains  necessary  to  find  Fand  //and  the  point  of  application  for  earth- 
pressure. 

Point  of  Application  of  Earth-pressure. — In  treating  retaining  walls  it  is  allowable  to 
neglect  the  cohesion  of  the  earth.  We  can  therefore  consider  the  earth-pressure  as  zero  at 
the  earth-surface,  and  increasing  for  any  point  of  the  back  of  the  wall  directly  as  the  depth 
of  that  point  below  the  earth-surface.  The  point  of  application  of  the  pressure  is  then  just 

the  same  as  for  water,  viz.,  at  a  distance  of  */,  =  —h  from  the  bottom. 

We  have  already  used  this  value  of  </,  in  finding  the  general  equations  of  pages  429  to 
432.  These  general  equations  apply,  then,  directly  as  they  stand  to  retaining  walls,  if  we 
make  the  ice-  or  wave-thrust  T  zero  wherever  it  occurs. 

It  only  remains  to  determine  the  magnitude  and  direction  of  the  pressure  P  in  the  case 
of  earth-pressure. 

Magnitude  and  Direction  of  Earth-pressure— Graphic  Determination.  — We  shall  first 
discuss  some  cases  of  equilibrium  of  a  prism  in  general. 

•    CASE  i. — Let  abc,  Fig.  I  (a),  be  any  small  prism  of  thickness  /,  and  let  +/,  be  the 
FIG.  i  (a).  normal  pressure  per  unit  of  area  upon  the  faces  ac  and  be  at  right 

angles,  the  (-(-)  sign  indicating  direction  up  and  to  the  right. 

We  shall  prove  that  for  equilibrium  the  pressure  per  unit  of 
ana  upon  the  third  face  ab  is  also  normal  and  equal  to  pr 

For  if  we  multiply  the  area  nc  X  /  of  the  face  ac  by  +/r 
we  have  the  total  horizontal  force  -(-  H  =  +  px  X  ac  X  /;  and  if 
we  multiply  the  area  be  X  /of  the  face  be  by  /p  we  have  the  total 
vertical  force  -\-  F  =  +A  X  be  X  /. 

454 


CHAP.  V.]      MAGNITUDE  AND  DIRECTION  OF  EARTH-PRESSURE-GRAPHIC  DETERMINATION.      455 


If  we  lay  these  forces  off  in  Fig.  i  (b)  from  A  to  Hand  H  to  N,  so  that  AH  -  -\- H 


and  HN  —  +  V,  the  resultant  for  equilibrium  is  NA,  making 
with  the  vertical  an  angle  ANH  whose  tangent  is 

H _pl  X.oc  X  /      oc 

v~  p,  x  fox  i~Tc' 

The  angle  of  NA  with  the  vertical  A  V  is  then  the  same  as 
the  angle  of  ab  with  the  horizontal  be.      Since  A  V  is  perpen- 
dicular to  qc,  NA  must  be  perpendicular  to  ab. 
We  have  also 


FIG. 


X/2+  A 


X  P  = 


or  for  equilibrium  the  pressure  per  unit  of  area  upon  the  face  ab  is  normal  to  the  face  and 
equal  to  pr 

CASE  2. — Suppose  now  the  normal  unit  pressure  on  the  face  ac  to  be  reversed™  direction, 
so  that  it  is  -/^  Fig.  2  (a).  FIG.  2  («). 

We  shall  prove  that  for  equilibrium  the  pressure  per  unit  of 
area  upon  the  third  face  ab  is  also  p^  as  before,  but  its  direction  is  no 


FIG.  2  (6). 


longer  normal  to  ab  but  makes  the 
angle  N'A  V  on  the  other  side  of  AV 
equal  to  the  angle  NA  V  in  the  first  case 
of  Fig.  i  (b]. 

For  we  have  the  total  pressure  on  the 
face  be  equal  to  +/,  X  be  X  /  =  +  V 


•P, 


the  same  as  before,  and  the  total  pressure  on  the  face  ac  is  now 
—  A  x  ac  X  I  =  ~  H,  or  just  the  same  as  before  in  magnitude 
but  opposite  in  direction. 

If  we  lay  these  forces  off,  in  Fig.  2  (£),  from  A  to  H' ,  and  H' 
to  N' ' ,  so  that  AH'  —  —  H,  and  HN'  =  -f  V,  the  resultant  for 
equilibrium  is  N'A.  It  is  at  once  evident  that  the  magnitude  of  N' A  is  the  same  as  NA 
before,  but  its  direction  makes  the  angle  N'A  Fon  the  other  side  of  A  F  equal  to  the  angle 
NA  V  in  the  first  case  of  Fig.  I  (a). 

In  this  second  case,  then,  for  equilibrium  the  pressure  per  unit  of  area  upon  the  face  ab  is  still 
pl ,  but  makes  an  angle  N'  AV  on  the  other  side  of  AV  equal  to  the  angle  NA  V  in  the  first  case. 
If  then  we  lay  off  in  Fig.  2  (b)  AN  —  pl  normal  to  ab,  and  with  N  as  a  centre  describe 
an  arc  of  a  circle  intersecting  the  vertical  AV  3.1.  S,  then  the  line  .SAT  will  give  the  direction 
and  magnitude  for  equilibrium  of  the  unit  pressure  on  the  face  ab  in  the  second  case  of 
Fig.  2  (b). 

CASE  3. — Suppose  now  that  the  normal  pressures  per  unit  of  area  on  the  two  faces  ac 
and  be,   Fig.    5,   are    unequal  and    given  by  -f-/2    an<^  "HA- 
Required  to   find   the   magnitude   and  direction   of   the  unit 
pressure/  on  ab  for  equilibrium. 

We  can  divide  the  normal  pressure  +  A  on   tne  ^ace  ^c 

the  other  equal  _^(p~ 


into  two  parts,  one  equal  to  -j (A~fA)  ai 

to  -j — (A  — A)'  as  indicated  in  Fig.  3  (a). 

We  can  also  divide  the  normal  pressure  +  A  on  ^e  ^ace 
ac  into  two  parts,  one  equal  to  -\ — (A  +A)  anc*  t^ie  otner  equal  to  —  j(A  ~~A)- 


456  STATICS  OF  RIGID  BODIES  [CHAP.  V. 

Now,  by  Case  I,  the  unit  pressure  normal  to  the  face  ab  which  balances  -f-  -(^  -\- p^  on 

the  face  be  and  -\ — (/>,  -f-/>2)  on  the  face  ac  is  just  the  same,  or  NA  =  -(pl  -j-/2),  Fig.  3  (b], 

FIG.      £.  laid  off  normal  to  ab. 

Also,  by  Case  2,  the  unit  pressure  on  the  face  ab  which 

balances  -\ — (pl  —  p^  on  the  face  be  and  —  ~(pl  —  p^)    on   the 

face  ac  is  just  the  same,  or  -(/,  —  A)'    but    ^  makes    an  angle 
N'A  Fon  the  other  side  of  A  F  equal  to  NA  V. 

If  then  we  lay  off,  in  Fig.  3  (b),  NA  normal  to  ab  and  equal 

to  —  (A~t~A)»    an<^  with  N  as  a   centre    and   NA   as  a    radius 
describe  the  arc  of  a  circle  intersecting  the  vertical  A  V  at  the 

point  S,  then  SN  will  give  the  direction  of  -(pl—  /2)  acting  on 
the  face  ab. 

If  then  we  lay  off  along  this  line  SN  the  distance  NR  =  -(pl  —  /2)  and  draw  RA,  the 

line  RA  will  give  the  magnitude  and  direction  of  the  resultant  unit  pressure  p  on  the  face  ab 
when  the  nmoral  unit  pressures  p^  and  p^  on  the  faces  be  and  ac  are  unequal. 

It  is  also  evident  that  the  angle  RNF  is  equal  to  twice  the  angle  VAN,  or,  if  we  bisect 
the  angle  RNF,  we  shall  have  the  direction  of  AV or /,. 

Suppose,  now,  the  faces  ac  and  be,  Fig.  3  (b],  to  remain  invariable  in  direction  and  the 
normal  unit  pressures  pl  and/2  on  these  faces  to  remain  unchanged,  but  let  the  third  face,«£, 

vary  its  inclination.      Then  the  magnitudes  of  NA  =  —  (pl  -j-/2)  and  of  RN  —  —(pl  —  p^ 

in  Fig.  3  (b}  will  be  unchanged,  but  their  directions  will  change  according  as  the  face  ab 
changes  its  inclination.  It  is  evident  that  the  greatest  possible  value  of  the  angle  RAN 
which  the  resultant  pressure/  =  RA  on  the  face  ab  makes  with  the  normal  NA  to  that  face 
will  be  when  RN  is  perpendicular  to  RA,  or  when  the  angle  ARN  =  90°. 

Now  for  earth-pressure  the  greatest  possible  value  of  the  angle  RAN  is  the  angle  of 
friction  or  of  repose  0°  for  earth  on  earth. 

The  angle  RNF  is  then  equal  to  90°  +  0°,  and,  as  we  have  just  seen,  if  we  bisect  this 
we  have  the  direction  of  /,.  Hence  the  angle  VAN  of  />t  with  the  normal  NA  is 

Y. 

CASE  4. — In  Fig.  4,  let  ab  be  any  earth-surface,  and  let  the  resultant  unit  pressure 
RA  =/  on  this  surface  be  given  in  direction  and  magnitude.  It  is  required  to  find 

~(A~I~A)'  ~(P\  ~~  A)  and  tne  direction  of  pr 

Draw  AF  normal  to  the  surface  ab,  and  AR'  making  the  angle  of  friction  0  with  this 


normal.  Then  find  by  trial  a  point  N  in  this  normal  AF  such 
that  if  we  take  TV  as  a  centre  and  NR  as  a  radius,  the  arc  ARR' 
will  be  just  tangent  to  AR'.  When  this  point  N  is  thus  found  by 


FIG.  4. 


trial,  then,  by  Case  3,  the  distance  AN  will  be  -(/>,  -f~ 


and 


R'N  =  RN  will  be  -(pl  —  /,).     Also,  if  we  bisect  the  angle  RNF 

by  the  line  NS',  we  have  the  direction  of  />,  . 

APPLICATION    TO    THE    RETAINING    WALL.  —  Let    AD, 
Fig.    5,  be   the  back  of   the  wall,  and  D^FI  the  earth-surface 


CHAP.  V.]        MAGNITUDE  AND  DIRECTION  OF  EARTH-PRESSURE—GRAPHIC  DETERMINATION.       457 


Pass  a  plane  AAl  through  the  foot  of  the  wall 
FIG.  5. 


making  the  angle  OL  with  the  horizontal. 
A  parallel  to  the  earth-surface,  and 
draw  AJ  vertical  and  A^F  normal  to 
the  earth-surface.  The  pressure  upon 
every  square  foot  of  this  plane  AA^  is 
vertical  and  equal  to  the  weight  of  a 
column  of  earth  of  vertical  height  AJ 
and  cross-section  I  sq.  ft.  X  cos  a. 

If  y\  is  the  mass  of  a  cubic  foot  of 
earth,  then  we  have 

Yl  X  A~JX   I  sq.  ft.  X  cos  a 

for  the  mass  of  this  column.  But 
/ij/cos  a  =  A^F,  hence  the  mass  of  this 
column  is 

yt  X  A^F  X  I  sq.  ft. 

If  then  we  draw  A^F  perpendicular  to  the  earth-surface  and  revolve  A^F  about  Al  as 
centre  to  the  vertical  A^ ,  we  have  for  the  vertical  unit  pressure/ 

p  =  yv  X  AlRl  pounds  per  sq.  ft., 

where  Y\  is  the  mass  in  Ibs.  of  a  cubic  foot  of  earth,  a.ndAlRl  is  measured  in  feet.      We  have 
then,  as  in  Case  4,  /  given  in  magnitude   and   direction,    and   we    can    find,  as  in  Case  4, 

~(A~t~A)»   "(A  — A)>  and  the  direction  of/r 

Thus,  as  in  Fig.  4,  draw  A^R'  (Fig.  5)  making  with  the  normal  A^the  angle  R'A^F , 
equal  to  the  angle  of  friction  or  repose  0  for  earth  on  earth.  Find  by  trial  a  point  JVl  on  the 
normal  A ^  such  that  the  arc  of  a  circle  with  Nl  as  a  centre  passes  through  Rl  and  is  tangent 
to  A^R' ' .  Then,  as  in  Case  4, 

;( 


i  -A)  =n  -^Pi- 


Bisect  the  angle  R^N^F  by  the  line  NVS'  .  Then  the  line  N^'  gives  the  direction  of 
/j  ,  as  in  Fig.  4. 

Now  lay  off  at  the  foot  of  the  wall  A  (which  may  be  considered  as  identical  with  A^) 
the  distance  NA  —  NlAl  in  a  direction  perpendicular  to  the  back  of  the  wall  at  A.  Draw 
the  line  AS  parallel  to  TV^S',  the  direction  of/j  already  found.  Then,  as  in  Fig.  3  (b},  with 
N  as  a  centre  and  NA  as  a  radius  describe  an  arc  of  a  circle  intersecting  AS  at  S,  and  lay  off 
along  NS  the  distance  NR  =  N'lRr  Then,  as  in  Case  3,  RA  represents  the  magnitude  and 
direction  of  the  pressure  per  square  foot  at  the  foot  of  the  wall.  Thus,  if  ^  is  the  density 
of  earth  and  we  measure  RA  in  feet,  the  pressure  per  square  foot  at  the  foot  A  of  the  wall 
is  given  in  magnitude  by 


and  its  direction  is  the  direction  of  RA. 


STATICS  OF  RIGID  BODIES. 


[CHAP.  V. 


Since  the  pressure  is  zero  at  the  top  Dv  and  greatest  at  the  foot  A,  and  varies  for  any 
point  directly  as  the  distance  of  that  point  below  /?, ,  the  average  pressure  per  square  foot  is 


2ri  X 
The  total  pressure  P  is  then  for  a  wall  of  one  foot  in  length 

P—  -yl  X  RA  X  D^A, 


where  y\  *s  tne  mass  of  a  cubic  foot  of  earth  and  RA,  D^A  are  taken  in  feet,  and  Pis  the 
pressure  per  foot  of  length. 

This  pressure  P  acts  (page  454)  at  a  point  A!"  at  a  distance  above  the  base  of  the  wall 

equal  to  -//,  where  h  is  the  distance  D^O  of  the  earth-surface  above  the  base  of  the  wall,  and 

is  parallel  to  the  direction  of  RA  already  found. 

We  can  thus  find  by  a  simple  graphic  construction,  in  any  given  case,  the  magnitude, 

direction  and  point  of  application  of  the  earth-pressure  P  on  the  back  of  the  wall.       The 

vertical  and  horizontal  components  Fand  //"of  Pare  then  easily  found,  and  then  the  general 

principles  and  equations  already  given  in  the  preceding  chapter,   apply  at    once,   and   we 

can  investigate  the  stability  of,  or  design,  a  retaining  wall. 

Magnitude   and   Direction   of  Earth-pressure— Analytic    Determination. — From    the 

graphic  construction  just  given  we  can  easily  derive  the  corresponding  general  formulas  for 

the  magnitude  and  direction  of  the  earth-pressure  P. 

NOTATION. — Let  h  =  DXO  be  the 
height  of  the  earth-surface  at  Dl  above 
the  base  AB  of  the  wall ;  the  angle  of 
the  earth-surface  with  the  horizontal 
is  «;  the  batter  angle  of  the  back 
of  the  wall  with  the  vertical  is  /?, ;  the 
earth-pressure /"makes  the  angle  0  with 
the  normal  to  the  back  of  the  wall ; 
the  angle  R'A1N1  =  0  is  the  angle  of 
repose  for  earth  on  earth ;  the  angle 
R1N1F=  7;,  and  the  angles  R^N^  and 

StNtF are  each  |;  the  angle  RA S=  e; 

the  angle  RSA  =  GO;  yl  is  the  mass 
A  0  B    of  a  cubic  foot  of  earth. 

SOLUTION. — Now  by  the  graphic  construction  we  have 


-(A  +  A)  sin 


-(A  -A)- 


We  have  also 


CHAP.  V.]    MAGNITUDE  AND  DIRECTION  OF  EARTH-PRESSURE— ANALYTIC  DETERMINATION.        459 
and  since  A^R^  =  A^F,  we  have  by  construction 

I  V  /I 

^A  -  A)  sin  r?  —  ^-p  cos  (ft  —  a)  sin  a.      .......     (2) 

We  have  also  by  construction 

[^  A  +  A)  +  j(A  -  A)  cos  V]2  +  [J( A  -  A)  sin  T?]   =  \j£-ft  cos  (ft  -  «)]  ,       (3) 

and  also 

^(  A  +  A)  +  ^(  A  -  A)  cos  7/  =  -^j-jjj  cos  (ft  -  a)  cos  or (4) 

From  (i),  (2)  and  (3),  eliminating -(pl  -\-  /2)  and -(A  —  A)»  we  ODtain 


siir  <*           /              .          /         sin*  «\ 
cos  77  =  —  — h  \  /  (i   —  sin*  ar)^l  —        2     j (1) 

We  have  also  directly  from  the  figure  GO  =  angle  NSA  —  angle  NAS,  or 

fi,=  9o-ft_^  +  a (II) 

From  (2)  and  (i)  we  have 

yJi  cos  (#  ^-  ft")  sin  «(i  --1-  sin  0)  .  , 

^,    =  —  ^ : - — : \S) 

ri  cos  ft  sin  0  sin  ?/ 

_  ^//  cos  (a  —ft)  sin  Q'd  —  sin  0)  ,^ 

A  ~  cos  ft  sin  0  sin  /; 

We  have  also  from  the  figure 

RS  sin  co 

tan  e  =  -- 


—  RS  COS  r«3 

But  YI  •  RS  =  A  »  and  y^  AS=  (A  ~r  A^  cos  «•     Therefore 


Substituting  the  values  of  A  and  A  from  (5)  and  (6)> 

i  —  sin  0 

tan  e  =  —    — . — -; 

I  -\-  sin  0 

We  have  also  directly  from  the  figure 


tan  e  =  L^_s          ,an  «  =  tan'     45  -        tan  « •      (III) 

i  -}-  sin  0 


(IV) 


460  STATICS  OF  RIGID  BODIES.  [CHAP.  V. 

Also,  __  __ 

r,  .  RA  =  V/28  sin8  GJ  +  (ri  .  ZS  -  /2  cos  a?)8  =  V/28  sin8  a?  -f/>,8  cos2  GO, 


or,  substituting  the  values  of  pl  and/2  from  (5)  and  (6),  we  have  for  the  earth-pressure  P 


or 

Yl#  cos  (ft  -  „)  sin  a  ^          ^         _       .  ..... 

2  cos8  /?!  sin  0  sin  77 

From  (i)  and  (4)  we  obtain 

_  y^k  cos  (ftl  —  a)  cos  a  (i  -f-  sm  0) 
*  ""  cos  ytfj  (i  +  sin  0  cos  ^) 


Comparing  this  with  (5),  we  have 

sin  a  cos  a 


sin  0  sin  77       i  -\-  sin  0  cos  ;/' 


(7) 


Making  this  substitution  in  (V),  we  obtain  an  equivalent  expression  for  P  which  can  be 
used  when  a  =  o,  viz., 

v,//2  cos  (8,  —  a)  cos  a     ./.  — 

P  —  -*-*-  -.  —  4  —  —  r  V(i  -j-  sm  0r  —  4  sin  0  sin2  GO.  .     .     .       (VI) 

2  cos2  /^(i  +sm  0  cos  rj) 

We  see  from  the  figure    page  458    that  the  earth-pressure  P  makes  the  angle  6  -\-  ftl 
with  the  horizontal.      We  have  then  for  the  vertical  and  horizontal  components  of  P 

V  =  P  sin  (0  +  A)' 

'  (VII) 


The  values  of  Fand  //  being  thus  known,  the  principles  and  general  equations  already 
given  in  the  preceding  chapter,  apply  at  once. 

The  case  of  water-pressure  is  a  special  case  of  the  equations  just  deduced.  Thus  for 
water  we  have  y  in  place  of  y{  ,  since  the  surface  is  level  we  have  a  =  O,  and  since  there  is 
no  friction  0,  =  o.  We  have  then,  from  (\),  rj  —  o;  from  (II),  ca  =  90  —  /?,  ;  from  (III), 
6=0?;  from  (IV),  6  —  o.  The  pressure  of  water  is  therefore  normal  to  the  surface.  From 
(VI)  we  have 


2  COS 

and  from  (VII) 


These  are  precisely  the  values  of  Fand  H  given  on  page  431  for  water-pressure.  We 
see,  then,  that  water-pressure  is  but  a  special  case. 

Surface  of  Rupture.  —  If  there  were  no  wall  and  we  were  to  disregard  cohesion,  a  prism 
of  earth  AD^G  would  tend  to  slide  off  along  a  plane  AG  which  would  make  with  the  hori- 
zontal the  angle  of  repose  0  for  earth  on  earth. 


CHAP.  V.]    MAGNITUDE  AND  DIRECTION  OF  EARTH-PRESSURE—  ANALYTIC  DETERMINATION.        461 

But  on  account  of  the  wall  this  plane  AG  makes  with  the  horizontal  an  angle  ip  greater 
than  0. 

This  angle  fy  we  call  the  angle  of  riipture,  the  plane 
A  G  is  the  plane  of  rupture,  and  the  prism  AD^G  which 
thus  tends  to  separate  along  AG  and  force  the  wall  is  the 
prism  of  rupture. 

If,  in  the  figure  page  458,  pv  remains  unchanged  in 
direction  and  magnitude,  while  AA^  is  resolved  about  A 
until  the  pressure  upon  AAl  makes  with  the  normal  AlNl 
the  angle  0,  then  this  new  position  of  AAl  is  the  plane  of  rupture.  But  for  this  new  position 

pl  makes  the  angle  45°  +  "~  W1*-h  t^ie  normal.     The  normal  A^Nlt  and  hence  the  plane  AAlt 

have  then  been  revolved  through  the  angle  45  -f  -  —  -  .     The    angle  which  the  plane  of 
rupture  A  G  makes  with  the  horizontal  is  then 


In   the  case  of  water,    a—  o,  0  =  o  and,  from    (I),  77  =  0.     We  have  then   for   water 

#  =  45°- 

General  Metkod.—Wz  have  in  any  case  the  following  general  method  ; 

1st.    Find  rf  from  (I). 
2d.     Find  GD  from  (II). 
3d.     Find  e  from  (III). 
4th.   Find  0  from  (IV). 
The  angle  ft  gives  the  inclination  of  the  earth-pressure  with  the  normal  to  the  back  of 

the  wall. 

5th.   Find  P  from  (V)  or  (VI). 
6th.   Find  V  and  H  from  (VII). 

If  desired  we  can  find  the  angle  of  rupture  from  (VIII). 
Knowing    now  J^and  H,  we  can  use  the  general   equations  pages  426  to  430,  making 

in  them  T=  o. 

Special  Cases.—  The  formulas  (I)  to  (VIII)  just  given  are  general  and  admit 

cation  for  special  cases. 

CASE  i.  EARTH-SURFACE  HORIZONTAL—  If  the  earth-surface  is  horizontal,  we  have 
a  =  o,  hence,  from  (I),  r,  =  O  and,  from  (II),  «  =  9o°  -  &•        We  have  then,  from  (    [I), 

tan  e  =  tan2(45  -  f  )  cotan  ft  ,....'...-.-        (8) 

and  then,  from  (IV), 

0=9o°  -A  -  e;     .......... 

from  (VI), 


D_ri/z2     /     r  4  sin  0      . 

:~7~Vcos2/?i      (i  +  sin  0)2' 

and  from  (VII), 


=  Psin  (90°  -  e),          H  =  P  cos  (90°  -  e)  ......     (l  0 


462  STATICS  OF  RIGID  BOOTES.  [CHAP.  V. 

From  (VIII)  the  surfact  of  rupture  makes  with  the  horizontal  the  angle 

V>  =  45°  +  f.    •     .     .     .       :  .     .'  .   ~.     .     •     (12) 

CASE  2.  EARTH-SURFACE  HORIZONTAL  —  BACK  VERTICAL.  —  If  in  the  preceding  case 
we  make  fa  =  o,  we  have  e  =  90°,  0  =  o  and  therefore  the  earth-pressure  is  normal  to  the 
back  of  the  wall,  or  horizontal. 

From  (10),  then, 


and  from  (u)  V=  o. 

The  surface  of  rupture  makes  as  before  the  angle  if)  with  the  horizontal  given  by 


CASE  3.   EARTH-SURFACE  HORIZONTAL—  BACK   BATTER   ANGLE  EQUAL  TO  90°—^. 

0 
—  In  this  case  a  =  o,  fa  =  90°  —  ^.     We  have,  from  (I),  r/  =  o  and,  from  (VIII),  $  =  45°+  J- 

Hence  #=45°-  ^,  and,  from  (II),  GO  =  *f>  =  45°-f  ^;    from  (III),  e  =  45°  ~  ^<t>\  from  (IV), 

0=0, 

or  the  pressure  makes  the  angle  of  'friction  0  with  the  normal  to  the  back. 
From  (VI)  we  have 


p= 


2  cos  ^45°  —  |j 
But  by  Trigonometry 


— \/(l  +  sin  0)2  -  4  sin  0  sin2  (45°  -  |). 


Hence 

sin  0  =  I  —  2  cos2  (45°  -f-  -J,          I  -f  sin  0  =  2  —  2  cos2  (45' 


cos2  0  =  4  cos2  (450  +  |)  ~  4  cos*  (45°  +  j). 
Inserting  these  values  and  reducing,  we  have 


, 

cos  0  cos  ^45°  —  -j 

From  (VII), 


(15) 


CHAP.  V.]   MAGNITUDE  AND  DIRECTION  OF  EARTH-PRBSSURE—  ANALYTIC  DETERMINATION.        463 

CASE  4.  EARTH-SURFACE  INCLINED  AT  THE  ANGLE  OF  REPOSE.  —  In  this  case  a—<p. 
We  have,  from  (I),  cos  rf  =  —  sin  0,  or  77  =  90  -{-  0;   from  (II),  GO  =  45°  —  &  -f  -  ;  from  (III), 


'.     (16) 
From  (IV), 


From  (V), 


-  *  si-  *  si-2  (45°  -  A  +  )• 


From  (VII), 
From  (VIII), 

*/>    =    0> 

or  the  surface  of  rupture  is  parallel  to  the  earth-surface. 

CASE  5.  EARTH-SURFACE  INCLINED  AT  THE  ANGLE  OF  REPOSE — BACK  VERTICAL. — 
In  this  case  we  have  only  to  make  /31  =  o'm  the  preceding  case,  and  we  then  have  r/  =  90  +  0* 

GO  =  45  -f-  -,  and,  from  (16), 

tan  e  = 
or,  since,  by  Trigonometry, 

1  +  S!"  0  =  tan2  (45°  4-  ^j,          '  ~  sin  ^  =  tan2  (45' 
i  —  sin  0  V        T2J>          j  _^   sin  ^  ^3 

i  —  sin  0     /I  -|-  sin  0  /I  —  sin  0  /     0       0\ 

tan ^  =•  x  +  sin 0y  rz  1HT0  =  V i  + sin 0  =   n^45  ~^' 

Hence  e  =  45°  —  ^,  and,  from  (17), 


or  the  pressure  makes  the  angle  of  friction  0  with  the  normal  to  the  back. 
From  (i  8), 


P  =      p*(l  +  sin  0)2  -  4  sin  0  sin2    45° 
or,  reducing  as  in  Case  3, 

cos  0 


__  r  (20) 


STATICS  OF  RIGID  BODIES— RETAINING   WALLS. 


[CHAP.  V 


From  (19), 


The  surface  of  rupture  makes,  as  before,  the  angle  0  with  the  horizontal. 
Values  of  0,  ^  and  yr — We  give  in  the  following  table  the  value  of  the  angle  of  friction 
0,  of  the  coefficient  of  friction  /*  =  tan  0,  and  of  the  density  y\  f°r  earth,  sand  and  gravel. 


Material. 

Angle  of 
Repose 
*. 

Coeffcient 
of  Friction 
P* 

Density  in 
Ibs.  per  cu.  ft. 
Yi« 

30° 

o  58 

40 

o  84 

Sand    dry  

si 

o.  70 

0.84 

o  <;8 

Earth    dry 

0.84 

4C 

Examples.— (i)  At  Northfield,  Vt.,  on  the  line  of  the  Central  Vermont  R.K.,  is  a  retaining  wall  13  ft.  high, 
top  base  2  ft.,  bottom  base  b  ft.  The  wall  is  composed  of  large  blocks  of  limestone  without  cement.  The  density 
of  the  masonry  is  about  170  Ibs.  Per  cubic  foot.  The  face  of  the  wall  has  a  batter  of  /  inch  horizontal  for  every 
foot  of  height.  The  wall  is  over  30  years  old  and  in  as  good  condition  as  when  laid.  Investigate  the  stability, 
taking  the  angle  of  repose  38° ,  the  density  of  the  earth  90  Ibs.  per  cubic  foot,  and  the  coefficient  of  friction  for 
the  masonry  //  =  0.66  (page  424). 


—  —  ,    or 


ANS.  We  have  h  =  hi  =  15,  di  =  2,  b*  =  6,  5  =  170,  tan  fii 

=  0.66. 
Take  a  section  of  the  wall  i  foot  in  length.     Then  the  weight  of  this  section  is 


=  10"  23',  0  =  38, 


90. 


IV  = 


10200  pounds  per  ft. 


From  Case  i,  page  461,  we  have 

p_  90  x  225 


, 

=  2983  pounds  per  ft>> 


tan  e  =  tan*  26°  cot  10°  23',     or     e  =  52°  26', 

V  =  P  sin  (90  —  €)  =  1814  pounds  per  ft.,         H  =  P  cos  (90  —  e)  =  2364  pounds  per  ft. 

Now  that  we  know  Kand  H,  we  can  proceed  as  on  pages  426  to  430,  making  T  =  o  in  all  equations  in 
which  it  occurs. 

Thus  for  stability  for  sliding  we  have  from  (I),  page  427,  for  the  factor  of  safety  for  sliding 

n  _  o.66(to2oo  +  1814)  _ 
2364 

There   is  therefore  ample  security  against  sliding  even  for  through  joints.     If  there  are  no  through 
joints,  there  is  in  any  case  no  possibility  of  sliding. 

For  stability  for  rotation  we  have,  from  equation  (5),  page  427, 


72  +  24  -4  -  15(4  +  6) 


2.75 


-5-  =  2.646  ft., 


CHAP.  V.]  RETAINING   WALLS— EXAMPLES.  465 

and  from  (II),  page  428,  since  d\  =  —  h  =  5, 

10200  x  2.646  +  1814^6 -5  X^'75V  2364  x  5 

10200  +  1814  "  =  2'03- 

Since  e  is  positive,  the  resultant  pressure  on  the  base  falls  within  the  base  and  there  is  no  rotation. 
For  stability  for  pressure,  since  e  is  greater  than  —  b*  =  2,  we  have,  from  the  third  of  equations  (III), 
page  428,  for  the  greatest  unit  pressure 

2(10200  +  i8i4)/         3  x  2.03\ 
p  =  — g —         \2  ~    — 6 — J  =  3944     P°unds  Per  ^  ft« 

From  our  table  page  424  we  see  that  the  allowable  unit  stress  is  C  =  from  50000  to  60000  pounds  per 
sq.  ft.  The  wall  is  then  abundantly  secure  against  sliding,  rotation  and  crushing.  We  also  see  that  the 
bottom  base  might  have  been  considerably  less,  thus  saving  much  material,  without  being  insecure. 

Check  by  Graphic  Construction,  page  457. 

(2)  Design  the  preceding  wall  properly  for  the  same  top  base,  height  and  back  batter,  for  an  allowable  stress 
of  C  =  40000  pounds  per  sq.  ft. 

ANS.  We  have  h  =  hi  =15,  b\  =  2,  8  =  170,  tan  /?i  —  -L-^-,  u  =  0.66,  and  of  course  the  same  values  for 
V  and  H  as  before,  viz., 

V '  —  1814,        H  =  2364. 

From  equation  (II),  page  429,  making  T  =  o,  we  have  for  the  value  of  bi  for  low  wall 


£,  =  —  1.065  +    i/I-I34  +  5°-95  *•  6.16  ft. 
Substituting  this  value  of  <5a  in  equation  (I),  page  429,  we  have  for  the  limit  of  hi  for  low  wafl 

„  .        6.16  x  40000-  3628  . 

ltmtt  hl  =  --  --  =  I75  £t' 


Since  this  is  greater  than  the  actual  height,  the  wall  is  low  and  the  value  of  £,  just  found  is  the  value 
ired.      We  have  for  this  base  e  = 
We  have  then  the  weight  per  ft. 


required.      We  have  for  this  base  e  ==  —t>,  and  p  less  than  C. 


I7Q  x  816  x  15 


From  (I),  page  427,  we  have  the  coefficient  of  safety  for  sliding 

/  .     _  0.66(10494  +  1814)  _ 

•••*    I  2364 

There  is  no  danger  of  sliding  even  for  through  joints. 

Check  by  Graphic  Construction,  page  457. 

(3)  Design  the  -wall  of  the  preceding  example  for  the  same  top  base  and  height  and  vertical  back,  for  an 
allowable  stress  C  =  40000  pounds  per  square  foot,  and  show  that  there  is  a  saving  of  over  17  per  cent  due  to 
vertical  back. 

ANS.  We  have  h  =  hi  =  15,  &  =  2,  S  =  170,  /?i  =  o,  0  =  38°,  y,  =  90,  //  =  0.66. 

From  Case  2,  page  462,  we  have 


466  STATICS  OF  RIGID  BODIES—  RETAINING   WALLS.  [CHAP.  V. 

• 

From  equation  (II),  page  429,  making  T  =  o,  we  have 

fa  =  -  i  +  */33/J3  =  4.77  ft. 


Substituting  this  value  of  b*  in  equation  (I),  page  429,  we  find  the  limit  of  h\  for  low  wall  greater  than 
the  given  height  of  dam.     The  wall  is  then  low  and  the  value  of  fa  just  found  is  the  value  required.     For 

this  base  we  have  e  =  -fa  and  p  less  than  C. 

We  have  then  the  weight  per  ft. 

<5(£,  +  fa)t       170  x  6.77  x   15 

---  —  —  8632  pounds  per  ft., 

as  against  10404  pounds  per  ft.  in  the  preceding  example,  or  a  saving  of  17.03  per  cent. 
From  (I),  page  427,  we  have  the  coefficient  of  safety  for  sliding 


There  is  then  no  danger  of  sliding  even  for  through  joints. 

Check  by  Graphic  Construction,  page  457. 

(4)  Find  the  bottom  base  of  a  trapezoidal  retaining  wall  of  granite  ashlar  with  vertical  back,  20  feet  high, 
earth  surface  horizontal  and  level  with  the  top,  <£  =  33"  40'  ',  y  —  100  pounds  per  cubic  foot. 

ANS.  In  this  case  0,  =  o,  V  =  o,  //  =  //,.  From  Case  (2),  page  462,  we  have  the  pressure  P  normal  to  the 
back  and  given  by 

P  =  H  =      —  -  tan"  28°  1  5'  =  5774  pounds  per  ft. 
From  equation  (II),  page  429,  we  have 


If  we  take  the  top  base  <5,  =  2  ft.  and  S  =  165  Ibs.  per  cubic  ft.  (page  424),  we  have  fa  =  7.66  ft. 
Check  by  Graphic  Construction,  page  457. 

(5)  Same  as  preceding  example,  with  back  batter  ft\  =  8°. 

ANS.  P  =  6420  pounds  per  ft.,   0  =  18°  9',  H  =  5758  pounds  per  ft.,    V  =  2825  pounds  per  ft.,  fa  =  7.9  ft. 

(6)  Design  a  retaining  wall  20  ft.  high,  back  batter  fti  =  8°,  S  =   770  Ibs.  per  cubic  foot,  earth-surface 
inclined  to  horizontal  at  angle  of  repose  <f>  =  33°  40' ,  y^  =  100  Ibs.  per  cubic  ft.,  earth-surface  at  top  of  wall. 

ANS.  From  Case  4,  page  463,  we  have   e  =  21°  22',        6  =  32°  28',         P  —  21740  pounds  per  ft., 
H  =  16522  pounds  per  ft.,         V  =  13230  Ibs.  per  ft. 

If  we  take  the  top  base  fa  =  2  ft.,  we  have,  from  equation  (II),  page  429,  fa  =  9.6  ft. 
Check  by  Graphic  Construction,  page  457. 

(7)  A  wall  15  ft.  high  retains  an  earth  filling  which  supports  a  double-track  railway.      The  top  base  is 
3.5  ft.     Find  the  bottom  base  for  y,  =  100,  <f>  =  jj°  40',  fti  =  8°,  S  =  170. 

ANS.  If  we  take  the  train-load  at  6000  pounds  per  linear  ft.  and  top  of  fill  15  ft.,  the  pressure  per 
square  ft.  on  top  is  400  pounds,  which  is  equivalent  to  a  column  of  earth  4  ft.  high.  We  have  then  ^,  =  15. 
h  =  19,  and  from  Case  i,  page  461, 

P  =  5795  pounds  per  ft.,         6  =  18°,         H  =  5200  pounds  per  ft.,          V  =  2540  pounds  per  ft. 

From  equation  (II),  page  429,  fa  =  7  ft. 
Check  by  Graphic  Construction,  page  457. 

COHESION  OF  EARTH. — Cohesion  is  that  resistance  to  motion  which  occurs  when  two 
surfaces  of  the  same  kind  are  in  contact.  It  is  found  by  experiment  that  cohesion  is  directly 
proportional  to  the  area  of  contact,  varies  with  the  nature  of  the  surfaces,  in  contact,  and  is 
independent  of  the  pressure. 


CHAP.  V.]  EQUILIBRIUM  OF  A  MASS  OF  EARTH. 

For  any  given  surfaces,  then,  the  cohesion  is 


467 


where  A  is  the  area  of  contact  and  c  is  the  coefficient  of  cohesion. 

EQUILIBRIUM  OF  A  MASS  OF  EARTH. — Let  ADG  be  a  mass  of 
earth,  the  batter  angle  of  the  face  being  fi. 

If  there  were  no  cohesion,  a  prism  of  earth  ADG  would  tend  to 
slide  off  along  a  plane  AG  which  would  make  with  the  horizontal 
the  angle  of  repose  0.  But  if  there  is  cohesion,  this  plane,  which  is 
called  the  plane  of  rupture,  will  make  an  angle  ^  with  the  horizontal 
greater  than  the  angle  of  repose  0.  We  call  rp  the  angle  of  rupture. 

Let  the  angle  of  the  earth-surface  DG  with  the  horizontal  be  «,  and  the  weight  of  the 
prism  ADG  per  unit  of  length  be  W. 

The  weight  W  acting  at  the  centre  of  mass  C  can  be  resolved  into  a  force  N  normal  to 
the  surface  of  rupture  AG  and  a  force  P  parallel  to  the  surface. 

We  have  then 

P=  W  sin  ^,          N=  W  cos  $ (i) 

The  force  P  tends  to  cause  sliding.  This  force  is  resisted  by  the  friction  and  the 
cohesion.  The  friction  is  pN,  where  p  =  tan  0  is  the  coefficient  of  static  sliding  friction,  and 
the  cohesion  is  c .  AG,  where  c  is  the  coefficient  of  cohesion. 

We  have  then  for  equilibrium 


P—  tN—  c  .  AG  = 


or 


P  — 


—  c.  AG, 


Now,  for  any  plane  which  makes  an  angle  with  the  horizontal  greater  or  less  than  ^ 
there  will  be  no  sliding,  and  for  that  plane  P  —  pN  will  be  less  than  c  .AG.  For  the  plane 
of  rupture,  then,  we  must  have 


AG 


=  a  maximum 


Let  the  vertical  height  of  the  prism   be  h.     Then  AD  =  ^r^>    the    angle    DAG   is 
_  (£  _j_  ^  the  area  of  the  prism  ADG  is  then 


AG 
— 


2cos/?  > 

and  the  weight  W  per  foot  of  length  is,   if  yl  is  the  density  of  the  earth, 

'.  co.  (/»+») 


2   COS  ft 


(4) 


468  STATICS  OF  RIGID  BODIES—  EARTH  EQUILIBRIUM.  [CHAP.  V. 

Inserting  this  value  of    W  in  (i)  and  the  corresponding  values  of  P  and  N  in   (3),  we 
have,  since  p  =  tan  0, 

yh  cos  (8  +0) 

—  .  sin  (tb  —  0)  =  c  =  a  maximum  ......     (5) 

2   COS  ft  COS  0 

ANGLE  O'F  RUPTURE.  —  Equation  (5)  is  a  maximum  when 

cos(/?  +  ^)  =  sin  fy  -  0)  =  cos  [90°  -  ty  -  0)], 
or  when 

*  =  45°  +  (--^  .....      (6) 


Equation  (6)  gives,  then,  the  angle  of  rupture,  or  the  angle  which  the  plane  of  rupture 
AG  makes  with  the  horizontal. 

COEFFICIENT  OF  COHESION  —  If  we  insert  this  value  of  tf>  in  (5),  we  have 

yh  cos  [45  +  :k0  +  /?)]  sin  [45  -  ^(0  +  /?)]  =  2c  cos  ft  cos  0, 


^[i  —  sin  (0  +  /?)]—  4^  cos/?  cos  0  ........      (7) 

Let  ft  =  o,  and  let  //  in  this  case  be  //0.      Then  we  have,  from  (7), 


c  = 


4  cos  0 


From  (8)  we  can  determine  c  by  experiment.      Thus  if  a  trench  with  vertical  sides,  of 
R         D  considerable  length  as  compared  to  its  width,  is  dug  in  the  earth, 

with  a  transverse  trench  at  each  end,  so  that  lateral  cohesion  may 
not  prevent  rupture,  it  will  be  observed  after  a  few  days  to  have 
caved  in  along  some  plane,  as  AG.      Let  the  observed  depth  AD  be 
7/0.     Then  from  (8)  we  can  compute  the  coefficient  of  cohesion  c. 
HEIGHT  OF  SLOPE. — If  we  substitute  the  value  of  c  from  (8)  in  (7),  we  obtain 

//0(i  -  sin  0)  cos/? 
i-sin(0  +  /?)    ' 

From  (6)  and  (9)  we  see  that  the  angle  of  rupture  tf>  and  the  height  of  slope  h  are 
independent  of  the  inclination  «  of  the  earth-surface. 

Equation  (9)  gives  the  limiting  height  //  of  slope  when  sliding  is  about  to  take  place. 
Let  «  be  a  factor  of  safety,  so  that  if  n  is  2  or  3,  the  height  can  be  taken  two  or  three  times 
the  safe  height  before  sliding  begins.  Then  we  have  for  the  safe  height 

*  =  %'"!?!  ,*>!?/-..  -co) 


Equation  (10)  gives  the  safe  height  of  slope  for  any  given  batter  angle  ft. 


CHAP.  V.]  ANGLE  AND  CURVE  OF  SLOPE.  469 

ANGLE  OF  SLOPE.  —  If  h  is  given  and  the  corresponding  batter  angle  /3  is  required,  we 
have,  from  (10), 

i  -  sin  (0  +  ft)  __  hQ(i  -sin  0)  = 
cos  ft  nh 

where  the  second  member,  being  a  known  quantity,  is  denoted  by  a. 

If  we  develop  the  numerator  in  the  first  member  and  substitute  for  sin  ft  and  cos  ft  their 

values  in  terms  of  tan  —ft,  viz., 

I  „  I 

2  tan  -ft  i  —  tan2  -ft 

sin  ft  =  —        -—  ,  cos  ft  =  -       —  , 

I  +  tan2  -ft  i  +  tan2  -ft 

2  2 

we  obtain  a  quadratic  whose  solution  gives 


Equation  (u)  gives  the  batter  angle  for  a  factor  of  safety  »  when  the  height  is  given. 

CURVE  OF  SLOPE.  —  Let  a  be  any  point  of  the  slope  Da  A,    whose  vertical  distance 
below  D  is  da  =  y,  and  let  aG  be  the  plane  of  rupture  at 
the  point  a,  making  the  angle  ip  with  the  horizontal. 

Then  the  prism  DGa  of  weight  W  per  foot  of  length 
tends  to  slide  along  aG  and  is  prevented  by  friction  and 
cohesion.  Let  ^V  and  Pbe  the  components  of  W  normal 
and  parallel  to  aG.  Then  if  n  is  the  factor  of  safety  and  jw 
is  the  coefficient  of  static  sliding  friction,  we  have 


(12) 


Let  A  be  the  area  daD  between  the  curve  of  the  slope  and  any  ordinate  da  —  y.     Then 
Y^A  is  the  weight  of  the  prism  daD  per  foot  of  length,  if  Yl  is  the  mass  of  a  unit  of  volume 

of  earth.      The  area  daG  is  ^—  —  per  foot  of  length.      Hence  the  weight  IV  of  the  prism 


DaG  is 


If  we  insert  this  value  of  W  in  the  equations  (i)  for  P  and  N,  and  then  substitute  in  (12), 

-  —  ^Vsin  ^  —  p  cos  ^)  —  *-j--7  =  O, 
y v        '        '  sin  w 


V 
we  obtain,  since  aG  =  ^— - 


47°  STATICS  OF  RIGID  BODIES—  EARTH  EOJJ1LIBR1UM.  [CHAP.  V. 

or,  dividing  by  sin  ^, 


cot       =  ° 


If  <*£  makes  an  angle  with  the  horizontal  greater  or  less  than  0,  we  have,  from  (12), 
n(P  —  )*N)  less  than  c  .  aG,  or  the  left  side  of  equation  (13)  less  than  zero.  The  value  of  ^> 
must  then  make  (13)  a  maximum. 

If  then  we  differentiate  (13)  with  reference  to  cot  ^  and  put  the  first  derivative  equal 
to  zero,  we  obtain 

o.        .     .     .     (14) 


Eliminating  cot  $  from  (13)  and  (14),  we  obtain 


A  =  \-n^^y  +  V  -  *  */2c(n«ri}>  +  2^(1  +  ^)].       ...     (15) 


Equation  (15)  gives  the  area  A  between  the  curve  of  the  slope  and  any  ordinate  da  =y. 

Examples.  —  (i)  A  bank  of  earth  without  cohesion  stands  3°  ft.  high  with  a  slope  of  jo  ft.     Find  the  coef- 
ficient of  friction  and  the  angle  of  repose. 

"?O 

ANS.  The  horizontal  projection  of  the  slope  is  40  ft.     Hence  u  =  tan  <f>  =  —  =0.75,  and  <p  is  about  35,, 

(2)  A  bank  of  earth  with  vertical  face  is  found  to  cave  for  a  distance  of  3  ft.  below  the  stir  face.      The  same 
earth  loose  and  without  cohesion  takes  a  slope  <>f  1.25  to  f  horizontal.     Find  the  slope  after  rupture.     Also,  if 
the  mass  of  a  cubic  foot  is  100  Ibs.,  find  the  coefficient  of  cohesion. 

ANS.  We  have  ft  =  o  and,  from  equation  (6),  page  468,  ^>  =  45°  +  —  .     The  tangent  of  the  angle  of 

repose  is  n  =  tan  <f>  =  0.75.     Hence  0  is  about  35°  and  il>  is  about  62°. 

From  equation  (8),  page  468,  since  //„  =  3  ft.,  y\  =  100  Ibs.  per  cubic  ft.,  <f>  =  35°, 

100x3(1  —  sin  35°)       128 

c  =  —  -    =  39  pounds  per  sq.  ft. 

4  cos  35°  3.28 

(3)  A  bank  of  earth  the  same  as  in  the  preceding  example  has  a  height  of  30  ft.  and  a  batter  of  45'  '.     Find 
the  limiting  height  for  the  same  slope,  also  the  factor  of  safety. 

ANS.   From  equation  (9),  page  468,  since  A»  =  3,  ft  =  45°,  <f>  =  35°,  the  limiting  height  is 

h  =  3('~  sin  35')  cos  45°  = 
I  —  sin  80° 

or  the  factor  of  safety  is  2. 

(4)  A  bank  of  earth  the  same  as  in  example  (2)  is  required  to  have  a  height  of  30  ft.  and  a  factor  of  safety 
of  a.     Find  the  batter  of  the  face. 

ANS.   From  equation  (i  i),  page  469,  ft  =  45°. 

(5)  A  bank  of  earth  -with  vertical  face  caves  for  a  distance  of  5  ft.  below  the  surface.      The  same  earth 
loose  and  without  cohesion  takes  a  slofie  of  1.25  to  i  horizontal.      The  mass  of  a  cubic  foot  is  too  Ibs.      Find  the 
angle  of  rttpture  and  the  coefficient  of  cohtsion.     If  the  batter  of  the  face  is  made  45°  and  the  height  jo  ft  ,  find 
the  factor  of  safety. 

ANS.  The  angle  of  repose  <f>  is  about  35°.     The  angle  of  rupture  ^  is  about  62°.     The  coefficient  of 
cohesion  is  c  =  65  pounds  per  sq.  ft.     From  equation  (10),  page  468, 

5(1  -  sin  35°)  cos  45°  _ 
30(1  -  sin  80°) 


CHAP.  V.]  EARTH  EQUILIBRIUM- EXAMPLES.  47 1 

(6)  Find  the  batter  angle  of  the  slope  in  the  preceding  example  for  a  height  of  30  ft.  and  a  factor  of  safety 

ANS.    From  equation  (u),  page  469,  ft  =  45°. 

(7)  Find  the  curve  of  the  slope  in  example  (5)  for  a  factor  of  safety  of  3  and  a  height  of  40  feet. 
ANS.  We  have  jj,  =  0.75,  c  =  65,  n  =  3,  yi  =  100,  and  equation  (i  5),  page  470,  becomes 


If  we  take 
we  have 


^=10 


30 
413 


40  ft., 
777  sq.  ft. 


Considering  the  area  between  the  slope  and  any  ordinate  as  made  up  of  trapezoids  as  shown  in  the 
figure,  we  have 

i 


—  .  loxDa'  =  33,       or     Da'—    6.6ft.; 


DC' 6'       c' 


10  +  20 

33  +  —  — 


167+ 


=  167,     or     a'tf  =    9  ft.; 
=  413,     or      b'c1  =    9.8ft.; 
=  777,     or     c"d'  =  10.4  ft. 


40 


We  see  from  equation  (15),  page  470,  that  for  small  values  of  y,  A  is  negative.  The  equation  should  not 
be  used  for  y  less  than  //o,  and  the  upper  part  of  the  slope  should  be  rounded  off  as  shown  in  the  figure. 

(8)  //  is  desired  to  cut  a  bank  30  ft.  high  into  three  terraces  as  shown  in  the  figure,  with  a  factor  of  safety 
Of  i  j-  The  height  of  each  terrace  is  to  be  10  ft.,  and  there  are  to  be  two  steps,  ab  and  cd,  each  4ft.  wide.  The 
mass  per  cubic  foot  is  y\  =  I0°  Ibs.,  and  <f>  and  h^  as  found  by  experiment  are  <p  =  31°,  h»  =  5  ft.  Find  the 
batter  for  each  terrace. 

ANS.  We  have  fi  ~—  tan  <p  —  0.6,  and  from  equation  (8),  page  468,  c  =  71.     Equation  (15),  page   470, 

becomes 
CT'      &     -<f'_ d'_ t'  A  —  ^--[284  +  907  —  2  -^189(90)'  +142)]- 

10 

For  y  =  10.  20  30 

we  have  A  =  27  159  421 


We  have  then  - 


—  x  Da'  =    27,     or     Da'  =  5.4  ft.; 


27  +  40+'—    -—  x  b'c'   =159,     or     b'c1  =6.1  ft.; 


Hence,  for  the  batter  angles, 


For  Da. 
For  be, 
For  dA, 


tan  ft  =  ~      or     ft  = 


tan  ft  =  -,     or    ft  =  31^°  ; 
=  41  $°. 


8.9 
tan/?=  x-,     or 


(9)  Find  the  batter  angle  for  a  railway  embankment  30  ft.  high,  12  ft.  top  base.  Let  ^,  =  100,  $  =  34°. 
//„  =  4  and  factor  of  safety  n  —  2.  Let  the  locomotive  weight  be  about  (>ooo  pounds  per  linear  ft.  of  trad;. 

ANS.  The  weight  of  locomotive  is  6000  pounds  on  12  square  feet,  or  500  pounds  per  sq.  ft.  This  is 
equivalent  to  a  mass  of  earth  5  feet  high. 


472  STATICS  OF  RIGID  BODIES -EARTH  EQUILIBRIUM.  [CHAP.  V. 

We  take,  then,  h  —  35  feet  in  equation  (u),  page  469,  and  have 

tan  -ft  =  ]-Tg4[o-829  +  ^0.0286]  =  0.416,     or    ft  —  45". 

The  embankment  with  this  batter  contains  47  cubic  yards  per  lineal  foot,  while  with  the  natural  slope 
of  34°  it  would  contain  62°.  There  will  then  be  a  saving  in  cost  of  construction  if  the  expense  of  protecting 
the  slope  to  preserve  the  cohesion  is  not  greater  than  the  saving  in  embankment. 

(10)  A  railway  cut  is  made  in  material  for  which  yi  =  too,  $=.34°,  At  =.5/1.  The  depth  of  cut  is 
h  =  40  ft.,  and  the  road-bed  is  '6/t.  Find  the  batter  angle  for  a  factor  of  safety  of j. 

ANS.  We  find  ft  =  47°.  The  cut  with  this  batter  is  87  cubic  yards  per  linear  foot.  For  the  natural 
slope  it  would  be  in.  There  is  then  a  saving  in  cost  if  the  expense  of  protecting  the  slope  to  preserve  the 
cohesion  is  not  greater  than  the  saving  in  excavation. 


STATICS   OF   ELASTIC   SOLIDS. 


CHAPTER   I. 

ELASTICITY  AND   STRENGTH   FOR  TENSION,   COMPRESSION  AND   SHEAR. 

Elasticity. — A  body  is  said  to  be  "elastic"  when  force  is  necessary  to  change  either 
its  volume  or  shape  and  when  the  continued  application  of  that  force  is  necessary  to  maintain 
such  change,  so  that  the  body  recovers  more  or  less  completely  its  initial  volume  and  shape 
when  the  force  is  removed. 

If  the  body  recovers  completely  its  initial  volume,  it  has  perfect  elasticity  of  volume. 
If  it  recovers  completely  its  original  shape,  it  has  perfect  elasticity  of  shape.  If  the  recovery 
is  imperfect,  the  elasticity  is  imperfect. 

All  bodies  have  some  elasticity  of  volume.  If  a  body  possesses  no  elasticity  of  shape, 
it  is  called  a  fluid,  as  air,  water,  etc.  If  a  body  possesses  elasticity  of  shape  as  well  as 
volume,  it  is  called  an  elastic  solid.  We  deal  here  only  with  elastic  solids. 

Prismatic  Body. — We  shall  treat  also  only  of  prismatic  solids.  A  prismatic  body  is  one 
whose  volume  can  be  generated  by  the  motion  of  a  plane  area,  moving  so  that  its  centre  of 
mass  describes  any  curve,  the  generating  plane  being  always  at  right  angles  to  that  curve. 

This  curve  is  the  AXIS  of  the  body,  and  the  generating  plane  at  right  angles  to  the  axis 
is  the  CROSS-SECTION.  The  area  and  form  of  this  cross-section  may  be  constant  or  variable. 

If  either  the  area  or  form  of  cross-section  varies,  the  cross-section  is  said  to  be  VARIABLE. 
If  neither  the  area  nor  form  varies,  the  cross-section  is  said  to  be  uniform. 

Stress  and  Force. — As  we  have  seen,  page  397,  we  distinguish  three  kinds  of  force  and 
stress. 

Tensile  force,  tending  to  pull  the  particles  of  a  body  apart  in  parallel  straight  lines  in 
the  direction  of  the  force,  and  tensile  stress,  resisting  such  separation. 

Compressive  force,  tending  to  push  the  particles  of  a  body  together  in  parallel  straight 
lines  in  the  direction  of  the  body,  and  compressive  stress,  resisting  such  approach. 

Shearing  force,  tending  to  make  adjacent  particles  move  past  one  another  at  right  angles 
to  the  line  joining  the  particles,  and  shearing  stress,  resisting  such  motion. 

Other  stresses  with  which  we  have  to  do  are  combinations  of  these. 

We  measure  force  and  stress,  then,  in  pounds,  and  unit  force  and  unit  stress  in  pounds 
per  square  inch. 

STRAIN — The  change  of  distance  between  two  particles  of  a  body,  in  a  direction  contrary 
to  coexisting  stress  between  those  particles,  is  called  STRAIN. 

473 


474  ST/IT1CS  OF  ELASTIC  SOLIDS.  [CHAP.  I. 

Thus  if  a  tensile  force  F  is  applied  to  a  straight  homogeneous  bar  of  uniform   cross- 
section  A   in  the  axis  of  the  bar,  and  this  force  F  is  uni- 
FIG.  i.  formly  distributed   over  the  end  area  A,  so  that  the  unit 

force  is  — ,  we  have  a   corresponding  equal   and  opposite 

unit-stress  S  at  every  point  of  the  cross-section,  and  the 
bar  is  extended  a  small  amount  A  in  a  direction  opposite  to 
coexisting  stress.     The  strain  A  between  end  particles  is 
one  of  extension,  or  tensile  strain. 
If  a  compressive  force  F  in  the  axis  is  uniformly  distributed  Fie.  2. 

over  the  end  area  A,  so  that  the  unit-force  is  — ,  we  have  a 

A  »_A_c 

corresponding  equal  and  opposite  unit-stress  S,  and  the  bar  is 
compressed  a  small  amount  A  in  a  direction  opposite  to  co- 
existing stress.  The  strain  A  between  end  particles  is  one  of 
compression,  or  compressive  strain. 

If  a  shearing  unit-force  —  is  applied,  we    have    a 
Xjil 

sponding  equal  and  opposite  unit-stress  S,  and  the  cross- 
section  slides  on  the  adjacent  one  a  small  amount  A  in  a 
direction  opposite  to  coexisting  stress.  The  strain  A  is  one 
of  shear,  or  shearing  strain. 

Strain,  then,  being  a  distance,  is  measured  in  units  of 

length,  as  feet  or  inches,  and  the  term  "  strain  "  is  general  and  signifies  always  a  change  of 
distance  between  points  of  a  body,  always  in  opposition  to  coexisting  stress  between  those 
points,  without  specifying  whether  that  change  of  distance  is  due  to  extension,  shortening, 
or  sliding,  or  whether  the  form  of  the  body  is  changed  or  not.* 

It  should  be  noted  that  when  there  is  no  coexisting  opposite  stress, 'there  is  no  strain. 
Thus  when  the  bar  in  Fig.  2  is  compressed  from  n  to  c,  let  the  force  F  be  removed.  The 
bar  would  expand  from  c  to  »,  and  during  such  expansion  we  have  unbalanced  stress  in  the 
direction  of  expansion.  Such  expansion  is  not,  then,  strain.  It  is  not  opposite  to  coexisting 
stress,  and  is  simply  displacement.  The  bar  is  not  strained  by  such  expansion,  but,  on  the 
contrary,  the  original  strain  A  diminishes  to  zero  at  the  neutral  point  ;/. 

But  after  the  end  of  the  bar  reaches  the  neutral  point  «,  if  it  still  continues  to  expand, 
as  in  Fig.  I,  the  stress  will  be  opposite  to  the  direction  of  expansion,  and  such  expansion  is 
strain.  The  bar  is  strained  by  such  expansion. 

Law  of  Elasticity. — When  a  force  F  is  applied  to  a  homogeneous  elastic  bar  of  uniform 

cross-section  A,  so  that  the  unit-force  is  —  and  acts  to  elongate,  compress,  or  shear  the  bar, 

A 

as  in  Figs.  I,  2,  3,  a  corresponding  equal  and  opposite  unit-stress  S  must  always  exist  when 
there  is  equilibrium,  and  a  corresponding  strain  A  is  always  observed  in  the  direction  of  the 
force  F,  and  therefore  opposite  to  the  coexisting  unit-stress. 

*  It  will  be  policed,  therefore,  that  strain  does  not  necessarily  imply  distortion.  In  the  two  first  cases 
above  there  is  strain  without  distortion  or  change  of  shape.  In  the  third  there  is  strain  with  distortion.  If  a 
homogeneous  sphere  is  equally  compressed  radially  in  all  directions,  it  remains  a  sphere.  It  has  strain,  but 
no  distortion.  The  word  "  distortion"  is  not,  therefore,  equivalent  to  shear,  and  in  many  cases  would  be  mis- 
applied if  used  as  equivalent.  The  term  "strain"  in  common  language  is  used  indifferently  either  for  change 
of  distance  between  points  of  a  body,  or  for  the  force  which  causes  this  change.  Our  definition  simply  restricts 
its  use  to  the  first  meaning. 


CHAP.  I.] 


SET  AND  SHOCK. 


475 


So  long  as  this  unit-stress  S  =  —  does  not  exceed  a  certain  magnitude,  all  experiments 

^1 

prove  that  the  unit-stress  6"  and  the  corresponding  strain  A  are  approximately  pioportional, 
so  that 


I    or 


=  a  constant. 


This  is  known  as  the  law  of  elasticity. 

The  point  where  this  law  begins  to  fail  is  called  the  ELASTIC  LIMIT,  and  the  correspond- 
ing unit-stress  at  the  elastic  limit  is  therefore  the  elastic  limit  unit-stress.  We  denote  it 
by  Se. 

The  law  of  elasticity  is  then  expressed  by  saying  that  for  homogeneous  bodies  and  within 
the  elastic  limit  the  strain  and  unit-stress  are  proportional. 

Beyond  the  elastic  limit  the  strain  increases  more  rapidly  than  the  unit-stress,  until 
finally  rupture  occurs. 

Set  and  Shock. — As  we  have  seen,  page  473,  if  when  the  force  F  is  removed  the  strain 
A  entirely  disappears,  the  bar  is  perfectly  elastic.  If  the  strain  A  does  not  entirely  disappear, 
the  bar  is  imperfectly  elastic,  and  the  residual  strain  is  called  the  SET. 

For  very  small  stresses  the  set,  if  any,  is  too  small  to  be  detected  by  measurement,  and 
the  bar  may  be  regarded  as  practically  perfectly  elastic.  As,  however,  no  solid  body  is 
perfectly  elastic,  the  point  at  which  set  is  first  observed  depends  upon  accuracy  of  measure- 
ment. Theoretically  there  is  a  set  for  any  stress. 

The  point  at  which  set  is  first  observed  is  often  taken  as  marking  the  limit  of  elasticity. 
But  in  view  of  the  preceding  it  should  not  be  so  taken,  and  experiments  generally  show  a 
slight  set  before  the  limit  of  elasticity  is  reached. 

A  .suddenly  applied  stress  or  shock  is  found  by  experiment  to  be  more  injurious  than  a 
steady  stress  or  a  stress  gradually  applied. 

Determination  of  Elastic  Limit  and  Ultimate  Strength. — Let  a  straight  homogeneous 
bar  of  given  material  have  a  length  /  and  uniform  cross-section  A.    Let  a  tensile" 
force  F  be  applied  in  the  axis  and  be  uniformly  distributed  over  the  end  cross- 
section  so  that  the  unit-stress  is  5  =  -r  opposite  in  direction  to  F.      Let  the 

observed  strain  of  elongation  be  A. 

Now,  according  to  the  law  of  elasticity,  provided  the  elastic  limit  is  not 
exceeded,  if  we  double  F  we  shall  observe  a  strain  2A.  If  we  apply  $F,  we 
shall  observe  a  strain  3A,  and  so  on. 

If  then  we  lay  off  the  successive  unit  stresses 

_F  _2F 

^l  ~  A1  2~  A'         "3  ~  A' 


A 


=  ~T,     etc., 


to  scale  along  a  horizontal  line  from  O,  and  lay  off  to 
scale  as  ordinates  the  corresponding  observed  strains 
A1 ,  A2  =  2 At ,  A3  =  3AP  etc.,  it  is  evident  that  within  the 

S,        S,        S,  S^  elastic  limit  we  have,  by  the  law  of  elasticity,  a  straight 

line  OL.      The  point  Z,  where  the  straight  line  begins  to  curve,  marks  the  elastic  limit,  and 
the  corresponding  unit-stress  S,  is  the  elastic  limit  unit-stress. 

As  the  straight  line  passes  into  a  curve  gradually,  it  is  evident  that  the  point  L  is  net 


476 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  I 


very  definite,  and  its  location,  even  with  very  accurate  experimental  results,  will  vary  moie 
or  less  within  certain  limits. 

When  the  elastic  limit  is  passed  the  elongation  increases  more  rapidly  than  the  unit- 
stress,  and  the  cross-section  decreases,  until  the  bar  ruptures.  The  unit-stress  at  which 
rupture  occurs  is  called  the  ULTIMATE  STRENGTH  for  tension.  We  denote  it  by  Ut. 

We  determine  the  elastic  limit  unit-stress  St  for  compression,  and  the  ultimate  strength 
for  compression  Ue ,  in  precisely  the  same  way  by  experiment.  The  force  F  is  axial  and 
uniformly  distributed  over  the  end  cross-section.  The  length  of  the  test 
specimen  should  be  small,  not  over  five  times  its  least  diameter,  in  order 
to  avoid  bending.  After  the  elastic  limit  is  reached  the  strain  increases 
more  rapidly  than  the  unit-stress,  and  the  cross-section  enlarges. 

For  shear  the  strain  A  is  as  shown  in  the  figure.  The  elastic  limit 
unit-stress  St  and  ultimate  strength  Ut  are  then  theoretically  determined  in 
the  same  way  as  for  tension  and  compression,  by  measuring  A  for 


different  values  of  — ?. 
A 

It  is,  however,  difficult  to  measure  A  for  shear,  and  hence  the  experimental  determina- 
tion of  Se  is  more  uncertain  than  for  tension  or  compression,  and  reliable  experiments  are 
wanting. 

The  values  of  Se,  Ut ,  Uc  and  Ut  vary  considerably  for  the  same  material,  according  to 
grade  and  quality.  Thus,  for  instance,  for  timber  there  are  different  kinds,  and  each  kind 
varies  according  to  soil,  climate,  season  when  cut,  method  and  duration  of  seasoning, 
direction  of  fibres  with  reference  to  stress,  etc.  So  also  iron  and  steel  vary  according  to 
quality,  process  of  manufacture,  whether  in  bars,  plates  or  wire,  etc.  We  give  in  the 
following  table  such  average  values  as  may  be  used  for  ordinary  preliminary  computations 
and  estimates.  For  actual  design,  the  necessary  values  for  the  material  used  should  be 
accurately  known  by  special  tests. 

TABLE   OF    ELASTIC    LIMIT   AND    ULTIMATE    STRENGTH — AVERAGE   VALUES. 


Material. 

Elastic  Limit  Unit-stress  in  pounds 
per  square  inch. 

Ultimate  Strength  in  pounds  per  square  inch. 

Tension. 
S, 

Compression. 
se- 

Tension. 
V* 

Compression. 

Shear. 
U,. 

Wrought  iron  
Steel  (structural)  

25000 
40000 
6000 

3000 

25OOO 
4OOOO 

55000 

lOOOOO 
20OOO 

lOOOO 

55000 
150000 
90000 

8OOO 

6OOO 
25OO 

5OOOO 
700OO 
20OOO 
<      600  longitudinal. 
)    3000  transverse. 

Timber 

3000 

Brick 

Coefficient  of  Elasticity. — In  the  experi- 
ment described  on  page  475,  we  find  for  a 
homogeneous  bar  of  given  material,  cross- 
section  A  and  original  length  /,  extended  by 
an  axial  force  F  uniformly  distributed  over  the 
end  area,  that  within  the  elastic  limit,  accord- 
ing to  the  law  of  elasticity, 


F_ 
A\ 


CHAP.  I.]  COEFFICIENT  OF  ELASTICITY.  477 

is  constant.      We  see  from  the  figure  that  this  constant  is  the  tangent  of  the  angle  which 
the  straight  line  OL  makes  with  the  vertical. 

But  if  we  had  taken  a  different  length  /x  ,  everything  else  being  the  same,  we  should 
have  for  the  same  loads  a  different  constant  or  tangent, 

F 

AX,' 

Now  if  the  bar  is  homogeneous,  the  stress  and  hence  the  strain  between  two  consecutive 
particles  in  the  line  of  strain  will  be  the  same  along  the  whole  length.  The  total  strain 
will  then  be  the  sum  of  all  the  strains  between  consecutive  particles,  or  proportional  to  the 
length. 

We  have  then  for  different  lengths 

V-  K  ::/t  :/,     or     A,  =^. 

Substituting  this,  we  have 

a 

=  a  constant. 


If  then  A,  is  known  for  any  length  /  by  experiment,  this  expression  will  give  the  constant 
or  tangent  of  the  angle  of  OL  with  the  vertical  for  any  other  length  lr 

Let  the  length  /x  be  unity.      Then  we  shall  have  for  the  constant  in  this  case 


_ 
ATC 

This  latter  constant  is  called  the  coefficient  of  elasticity,  and  we  denote  it  by  E.     We 
have  then 


Now  -T  is  the  unit-stress,  and7  is  the  strain  per  unit  of  length,  or  the  unit-strain. 
A  I 

We  can  then  define  the  coefficient  of  elasticity  as  the  ratio  of  the  unit-stress  to  the  unit- 
strain,  within  the  limit  of  elasticity. 
We  have  also,  from  (I), 


and  from  the  preceding  figure  we  see  at  once  that  we  may  also  define  E  as  that  unit-stress 
which  would  cause  an  elongation  equal  to  the  original  length  of  the  bar,  if  the  law  of 
elasticity  held  good  without  limit. 

The  value  of  E  thus  determined  by  experiments  well  within  the  elastic  li^nit  is  then  a 
measure  of  the  elasticity  of  the  body.      We  determine  the  coefficient  of  elasticity  Ec  for 


47« 


STATICS  Oh  ELASTIC  SOLIDS. 


[CHAP.  I. 


compression  in  precisely  the  same  way,  having  regard  to  the  precautions  mentioned  on  page 
476. 

For  shear  we  have  the  same  expression  for  E,  as  given  by  (I),  if  we 
put  for  /  the  distance  ab,  as  shown  in  the  figure,  and  for  A.  the  distance 
ac.  The  direct  measurement  of  \  is  difficult  and  uncertain,  but  E,  can 
be  determined  indirectly  by  experiments  on  torsion  and  bending,  as  will 
be  explained  later  (pages  511  and  555). 

The  values  of  Ett  E,  and  E,  vary  for  the  same  material  according  to 
grade  and  quality,  just  as  the  values  of  St ,  for  the  reasons  given  on  page 

476.      We  give  in    the    following    table  average   values   which   may  be   used   for  ordinary 

preliminary  computations.      For  actual  design  the  values  should   be  accurately  known  by 

special  test. 

Equation  (I)  then  holds  for  tension,  compression  or  shear,  within  the  elastic  limit,  if  we 

put  Et,  Ec ,  E,  in  place  of  E.     In  using  (I),  since  E  is  always  given  in  pounds  per  square 

inch,  we  should  take  /  and  A  in  inches  and  A  in  square  inches. 

TABLE   OF   COEFFICIENT   OF   ELASTICITY — AVERAGE   VALUES. 


Material. 

Coefficient  of  Elasticity  in  pounds  per  square  inch. 

Tension. 

*t 

Compression. 
Ec 

Shear. 
Et 

25  oooooo 
30000000 
15  oooooo 
I  500000 

25  oooooo 
30  ooo  ooo 
15  oooooo 
I  500000 
6000000 

15  oooooo 
7  oooooo 
6  ooo  ooo 
400000  longitudinal. 

Timber     . 

Brick  

As  an  example,  we  give  the  following  record  of  an  actual  series  of  experiments  made 
upon  a  wrought-iron  bar  |  inches  diameter  and  12  inches  long.     The  first  four  columns  give 


the  experimental  record. 


In  the  fifth  column  we  give  the  corresponding  values  of—. 

o 


Stress 

Elongation  A. 

T             1  C 

pounds  per 

1  olal  btress 
in  Pounds. 
F. 

square  i^ch 

Load  on 
A 

Load  off 
A 

A        / 
S^  ~E- 

A' 

in  inches. 

in  inches. 

2245 

5OOO 

O.OOI 

0.000 

o.ooo  ooo  20 

449° 

IOOOO 

O.OO4 

o.ooo 

o  .  ooo  ooo  40 

6375 

15000 

O.OC5 

o.ooo 

o.ooo  ooo  33 

8980 

20OOO 

O.OO8 

o.ooo 

o  .  ooo  ooo  40 

9878 

22000 

0.009 

o.ooo 

o.ooo  00041 

101776 

24000 

O.OIO 

.000 

o.ooo  00042 

11674 

260OO 

0.0105 

.000 

o.ooo  00040 

12572 

28000 

O.OII 

.000 

o.ooo  00039 

13470 

30000 

0.013 

.000 

o.ooo  00043 

14368 

32000 

0.014 

.000 

o.ooo  00044 

15266 

34000 

0.015 

.002 

0.00000044 

16164 

37000 

O.022 

.007 

0.000  OOQ  6l 

17062 

380OO 

0.416 

•3995 

o.ooo  ooi  09 

17960 

40OOO 

0-5445 

•523 

0.00001361 

25450 

5OOOO 

1.740 

1.707 

0.000034  80 

*?1I  *7C 

5  1600 

2.468 

''J1  /!> 

Specimen  broke  at  51600  pounds  per  square  inch. 

CHAP-  !•]  STRAIN  DUE   TO   WEIGHT.  479 

We  see  that  the  ultimate  tensile  strength  is  Ut  —  51600  pounds  per  square  inch;  also 
that  the  elastic  limit  unit  stress  Se  lies  between  34000  and  36000  pounds  per  square  inch.  If 
we  should  judge  by  the  beginning  of  noticeable  set,  the  elastic  limit  would  be  between 
32000  and  34000  pounds  per  square  inch,  and  up  to  this  point  the  elasticity  appears  to  be 
perfect.  The  elasticity,  however,  is  not  perfect,  and  more  accurate  measurement  would 
undoubtedly  show  the  set  beginning  earlier.  We  cannot,  then,  take  the  beginning  of  observed 
set  as  the  limit  of  elasticity.  Since  the  elastic  limit  lies,'  then,  between  34000  and  36000 

pounds  per  square  inch,  we  have  for  the  average  -^  up  to  this  point  o  =  O.  ooo  ooo  387  3, and 
hence  the  value  of  Et  given  by  this  experiment  is 

£t  =  —  =  0  OQO  QOO  3 87  3  =  3°  984  °°°  P°unds  Per  square  inch- 
Examples. — (i)  A  steel  rod  jo  ft.  long  and  4  square  inches  cross-section  is  subjected  to  a  tensile  force  of 
40000  pounds.     The  elongation  is  observed  to  be  — —  of  an  inch.     Find  the  coefficient  of  elasticity. 

ANS.  The  unit  stress  is  loooo  pounds  per  square  inch.  From  our  table  page  476  we  see  that  this  is  well 
within  the  elastic  limit  unit-stress  Se,  and  equation  (I)  can  therefore  be  applied  and  we  have 

Et  =  —  =  —  —  =  30000  ooo  pounds  per  square  inch. 

4X 

100 

(2)  A  rectangular  timber  strut  is  40  feet  long  and  i3\  inches  deep.  If  Ec  is  i  200  ooo  pounds  per  square 
inch,  find  the  -width  so  that  the  strain  under  a  stress  of  270000  pounds  may  be  one  inch;  all  lateral  bending 
being  prevented. 

ANS.   We  have,  from  equation  (I), 

Fl       270000  x  40  x  12 

A  =  -rrr  =  —  —  =  108  square  inches. 

-C.A          i  200  ooo  x  i 

This  gives  2500  pounds  per  square  inch,  which  we  see  from  our  table  page  476  is  within  the  elastic 
limit.  Equation  (I)  therefore  applies,  and  we  have  then  a  width  of  —,  =  8.1  inches. 

Strain  Due  to  Weight. — If  a  homogeneous  straight  prismatic  body  of  uniform  cross- 
section  A  has  a  considerable  length  /,  the  strain  due  to  its  own  weight  may  be  considerable. 

Let  <5  be  the  density  or  weight  of  a  cubic  foot,  For  any  length  x  the  weight  is  then 
6 Ax,  and  the  corresponding  strain  in  an  element  of  length  dx  is  then,  by  equation  (I), 

_  $Ax .  dx  _  d_ 

Integrating  between  the  limits  x  —  /  and  x  —  o,  we  have  for  the  entire  strain 

dl*  _  6A/Z 
A  ~  2E  ~  2AE' 
or,  since  dAl  is  the  entire  weight  W, 

Wl 


or,  as  we   see   from  (I),  the   strain  is  one-half  as  much  as  that  due  to  the  same   weight  at 
the  end. 


480  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  I. 

Example. — How  long  must  a  homogeneous  bar  of  wrought  iron  of  uniform  cross-section  A  be  in  order  that 
when  suspended  vertically  the  greatest  unit-stress  may  be  equal  to  the  elastic  limit  unit-stress,  and  what  is  the 
extension  f 

ANS.  The  weight  8AI  must  equal  S,A,  or  dl  =  S,,  or  /  =  ^.  From  our  table  page  476,  S,  =  25000  pounds 
per  square  inch;  and  since  a  bar  of  wrought  iron  one  square  inch  in  cross-section  and  3  feet  long  weighs 
jo  pounds  (page  19),  <5  =  -7  Ibs.  per  cubic  inch.  We  have  then 

2  5000  x  36 
/=     ,ox-iT-=750ofeet- 

The  extension  is,  taking  £"=25  oooooo  pounds  per  square  inch  (page  478), 

St/  25000x7500  _ 

~  2E  ~  2x25  oooooo  ~  3-75 

Stress  Due  to  Change  of  Temperature. — We  have,  from  equation  (I), 


where  A  is  the  strain  corresponding  to  the  unit-stress  5  =  -j  for   a   bar    of   length  /  and 

uniform  cross-section,  the  coefficient  of  elasticity  being  E. 

If  the  bar  is  constrained  so  that  it  cannot  change  its  length  and  then  is  exposed  to 
change  of  temperature,  there  will  be  a  unit-stress  S  equal  to  that  due  to  a  strain  equal  to  the 
change  of  length  for  the  same  unconstrained  bar  under  the  same  change  of  temperature. 

Thus  if  e  is  the  coefficient  of  linear  expansion  for  one  degree  of  temperature,  /  the 
number  of  degrees  change  of  temperature,  and  /  the  original  length,  the  change  of  length  of 
the  unconstrained  bar  would  be 

A  =  e//. 

The  coefficient  of  expansion  e  =  —r  is  then  the  strain  per  unit  of  length  per  degree. 
If  the  bar  is  constrained  so  that  its  length  cannot  change,  we  have  then  a  unit-stress 

5  =  ~  =  Eet. 

which  is  independent  of  the  length  /.     The  total  stress,  if  the  area  is  At  is  then 

AS=AEet. 

We  give  the  following  average  values  of  the  coefficient  of  linear  expansion  e  for  one 
degree  Fahrenheit : 

Wrought  iron e  =  0.0000067 

Steel e  =  0.0000065 

Cast  iron €  =  0.0000062 

Stone  and  brick e  =  0.0000050 

Example. — A  wrought-iron  rod  of  length  I  and  2  square  inches  cross-section  has  its  ends  fired  rigidly  when 
the  temperature  is  bo*  F.  Taking  the  contraction  at  0.00000671  for  one  degree,  what  tension  will  be  exerted 
when  the  temperature  is  20"  f.f 

ANS.  The  shortening  due  to  cooling' is  A  =  0.0000067  x  4°£     We  have  then,  from  equation  (I), 

EA\      E  x  2  x  0.0000067  x  4°' 
F  =  ——  =  — ~ •- —  *=  0.000536^. 


CHAP.  I.] 


WORKING  STRESS— FACTOR  OF  SAFETY. 


481 


If  E  =  30  ooo  ocx)  pounds  per  square  inch,  F  =  16080  pounds.  This  gives  the  unit-stress  8040  pounds 
per  square  inch,  which  we  see,  from  our  table  page  476,  is  well  within  the  elastic  limit,  and  equation  (i) 
therefore  applies. 

Working  Stress—  Factor  of  Safety.—  The  greatest  unit-stress  to  which  a  material  can 
safely  be  subjected  in  practice  is  called  the  WORKING  STRESS.  We  denote  it  by  Su.  The 
working  stress  should  never  equal  the  elastic  limit  unit-stress  Se  ,  since  if  it  exceeds  that 
limit  the  elasticity  is  impaired.  For  security,  then,  we  take  a  fraction  of  Se.  The  number 
/by  which  Se  is  thus  divided  is  called  the  factor  of  safety.  We  have  then  the  working  stress 

'' 


where  /  is  the  factor  of  safety. 

For  steady  stress  it  is  customary  to  take  f  —  1.5,  and  for  repeated  stress  or  suddenly 
applied  stress  /=  3. 

As  we  have  seen,  page  476,  the  exact  determination  of  Se  is  difficult.  It  is  therefore  also 
customary  to  take  Sv  a  certain  fraction  of  the  ultimate  strength  [7t,  Uc,  Us  for  tension,  com- 
pression or  shear. 

The  following  table  gives  the  factors  of  safety  usually  adopted  when  this  method  is  used: 

FACTORS    OF    SAFETY    FOR   $„  —   ^    &    E±  ^ 


Steady 
Stress 
(Buildings). 

Varying 
Repeated  Stress 
(Bridges). 

Shocks 
(Machines). 

6 

Steel  (structural).  

5 

•j 

IO 

6 

IO 

ie 

3 

Brick  

je 

25 

In  order  to  find  the  area  of  cross-section  A  for  simple  tension,  compression  or  shear,  we 
have  then  simply  to  divide  the  given  total  stress  by  the  working  stress  Sw,  We  have  then, 
when  flexure  is  not  to  be  apprehended,  for  steady  or  varying  stress  or  shocks 


A  = 


total  stress 


Sometimes  we  have  alternating  stress,  i.e.  tension  and  compression  alternating,  as  in  the 
connecting-rod  or  piston-rod  of  an  engine.      In  such  case  we  find  the  area  of  cross-section  for 
each  stress  and  take  the  greatest.      Thus,  if  flexure  is  not  to  be  apprehended, 
total  tensile  stress  total  compressive  stress 


A  = 


or 


whichever  is  the  greatest. 

When  flexure  of  long  columns  is  to  be  guarded  against  we  must  proceed  as  on  page 
Variable  Working  Stress. — The  fact  that  the  working  stress  Sw,  as  determined  in  the 
preceding  article,   is  constant  in  any  given  case,  is  by  many  engineers  considered   objec- 
tionable. 

The  total  stress  can  in  general  be  divided  into  two  portions.  The  one  portion  is  a 
steady  stress  always  existing,  such  as  that  due  to  weight  or  dead  load.  The  other  portion 
is  a  repeated  stress,  such  as  that  due  to  loads  recurring  at  intervals  of  time. 


482 


ST /tTlCS  OF  ELASTIC  SOLIDS. 


[CHAP.  2. 


Evidently  when  the  ratio  of  the  steady  stress  to  the  total  stress  is  great,  we  should  !><• 
able  to  use  a  greater  working  stress  than  when  this  ratio  is  small.  Thus  when  the  steady 
stress  is  equal  to  the  total  stress  there  is  no  repeated  stress  at  all,  and  the  working  unit- 
stress  should  have  its  greatest  value.  On  the  other  hand,  when  the  steady  stress  is  zero  we 
have  repeated  stress  only,  and  the  working  unit-stress  should  have  its  least  value. 

It  is  therefore  customary  to  take  for  the  working  unit-stress,  when  flexure  is  not  to  be 
apprehended,  for  repeated  stress  of  any  kind 


U — 


steady  stress\ 
total  stress  /' 


From  (i)  we  see  that  when  the  steady  stress  is  equal  to  the  total  stress,  that  is,  when 
there  is  no  repeated  stress,  we  have  Su  =  -^,  where    U  is  accordingly  the  ultimate  strength 

and /the  factor  of  safety,  just  as  in  the  preceding  article. 

But  when   the  steady  stress  is  zero,   we  have  only  repeated   stress,   and  (i)  gives   us 

Sv  = — -.      Hence  Up  is  the  ultimate  strength  for  repeated  stress  only,  or  the  "repetition 

strength  "  as  it  may  be  called. 

In  like  manner,  when  flexure  of  long  columns  is  not  to  be  apprehended,  we  have  for  the 
working  unit-stress  for  alternating  stress 


least  of  the  two  stresses 
greatest  of  the  two  stresse 


(2) 


From  (2)  we  see  that  when  the  least  of  the  two  stresses  is  zero,  we  have  Sw  =  — ',  as  in 
the  previous  case,  for  steady  stress  zero. 

But  when  the  two  alternating  stresses  are  equal,  we  have  Sw  =  ~ '.      Hence  U,  is  the 

ultimate  strength  for  equal  alternating  stresses,  or  the  "vibration  strength  "  as  it  may  be 
called. 

The  difficulty  and  uncertainty  of  determining  Uf  and  Uv  by  experiment,  and  the  few 
experiments  thus  far  available,  make  the  method  of  the  preceding  article  the  most  generally 
accepted.  The  present  method  is,  however,  extensively  used  with  assumed  average  values 
for  U,  Uf  and  Uv  as  given  in  the  following  table: 


w 
/ 

U-  Up 

UP 

Up-  U, 
Up 

Wood  

2 

i 

2 

± 

2 
2 

Steel  (structural)  

17800 

3 

5 
7 

15 

These  values  are  for  direct  tension  or  compression.      For  shear  we  take  four-fifths  of  Sv 
as  determined  above. 


CHAP'  *•]  WORKING  STRESS- EXAMPLES.  483 

In  order  to  determine  the  area  of  cross-section  A,  we  have,  then,  in  all  cases 

_  total  maximum  stress 
•"  —  — ^ -. 

When  flexure  of  long  columns  is  to  be  provided  against  we  must  proceed  as  on  page  569. 
Examples — (i)  A  wrought  iron  tie-rod  in  a  roof -truss  is  under  a  stress  of  20000  Ibs.     Find  its  area  of 

cross-section. 

ANS.  Taking  the  ultimate  tensile  strength  (page  476)  Ut  =  55000  pounds  per  square  inch,  and  factor  of 
safety  4  (page  481),  we  have 


e   _  Ut       55000 

••>»  —  -j-  -    — - —  =  13750  pounds  per  square  inch. 


and  hence 


1375 


Or  again,  from  (i),  page  482.  if  steady  stress  is  equal  to  total  stress,  we  have  the  same  result. 
(2)  Suppose  the  rod  is  a  bridge  member  and  the  dead-load  stress  is  5000  pounds,  the  total  stress  being  20000 
pounds  as  before. 

ANS.  We  have,  as  before,  Ut  =  55°°o.  and  (page  481)  the  factor  of  safety  6.     Hence 


W  = 

Also  by  (i),  page  482,  we  have 

£.  =  7500(1  +  -£^=9375     and     A  =  =  2.13  sq.  in. 

\        20000/  9375 

(3)  7^<?  greatest  steam-pressure  upon  the  piston  of  a  steam-engine  is  p  =  150  pounds  per  square  inch.  If 
the  area  of  the  piston  is  a  —  200  square  inches,  find  the  cross-section  of  the  steel  piston-rod,  if  lateral  bending 
is  prevented. 

ANS.  The  total  pressure  is  pa.  Taking  the  ultimate  compressive  strength  £/,=  150000  and  the  ulti- 
mate tensile  strength  Ut  =  100000  pounds  per  square  inch  (page  476),  and  the  factor  of  safety  10  (page  481), 
we  have  the  working  stress 

150000  100000 

Sw  =  —    —  =  15000     or    Sw  =  —     —  =  10000. 

IO  IO 


Taking  the  least  of  these,  we  have 

A=g  =  I5°X2°°  =  3  square  inches. 
Sw          loooo 

Also,     om  (2),  page  482,  we  have 

Sw  =  17800(1  -  ^-\=  9500. 


Hence  by  this  method 


=  150x200 
9500 


(4)  Find  the  height  to  which  a  brick  wall  of  uniform  thickness  can  be  safely  carried,  the  density  being 
d  =  125  pounds  per  cubic  foot. 

ANS.  Taking  the  ultimate  strength  Uc  =  2500  (page  476)  and  the  factor  of  safety  15  (page  481),  we  have- 
the  working  stress 

25oox  144 
Sw  =  —  -  pounds  per  square  foot. 

If  h  is  the  height,  /  the  length  and  /  the  thickness,  all  in  feet,  the  cubic  contents  is  hit  cubic  feet,  and 

£   1    9, 

the  weight  dhlt  pounds.     The  base  is  //  square  feet,  and  hence  the  pressure  on  the  base  is  —j—~  =  Sh  pounds 
per  square  foot.     Putting  this  equal  to  Sw,  we  have 

»;    or     ,.       = 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  I. 


Combined  Tension  or  Compression   and  Shear.  —  Suppose  an  element  of  a  body  of 

breadth  b,   height  h  and  width    «>,    and 


•f  S/iftJ 


"+Sado>' 


-S.dfo, 


-S(d( 


let  t  be  the  unit  stress  of  direct  tension 
and  s  the  unit  stress  of  direct  shear. 

We  have  then  the  two  equal  and 
opposite  tensile  stresses  -f-  thw  and 
—  ///«/,  and  the  two  equal  and  opposite 
shearing  stresses  -\-shw  and  —  shw. 
These  last  two  stresses  form  a  couple 
and  can  only  be  held  in  equilibrium  by 


an  equal  and  opposite  couple  -f-  sbw  and  —  sbw. 

Let  d  be  the  diagonal  and  a  the  angle  of  the  diagonal  with  the  direction  of  /.  Let  St 
be  the  unit  shear  along  the  diagonal,  and  St  the  unit  tension  normal  to  the  diagonal.  Then 
we  have  along  the  diagonal  the  components  -f  Stdw  and  —  S^w,  and  normal  to  the  diago- 
nal the  components  -|-  S,dw  and  —  S,dw. 

We  have  then  for  equilibrium 

Stdw  —  thw  cos  a  —  sbw  cos  a  -f-  shw  sin  a  =  O, 
Sdu'  —  thw  sin  a  —  sbw  sin  a  —  shw  cos  a  =  O. 


Now  we  have 


h  b 

sm  a  =  -;     and     cos  a  =  -. ; 
d  d 


hence,  dividing  these  equations  by  dwt  we  have 


S,  =  /  sin  a  cos  a -{-  s  cos2  a  —  s  sin2  a  —  —  sin  2a  -f-  s  cos  2a, 

St  =  f  sin2  a  +  2s  cos  a  sin  a  = cos  2a  +  s  sin  2a. 

dSt 


dSt 


If  we  differentiate  these  equations  and  put  the  first  differentials  -7-*  =  o  and  -;-  =  O,  we 

da  da 


have,  when  St  is  a  maximum, 


tan  2a  =  -, 


or     sm  2a  = 
and  when  St  is  a  maximum, 

tan  2a  =  — — ,     or     sin  2a  =  — 


and     cos  2  a  = 


and     cos  2  a  = 


(0 


-       (2) 


Inserting  these  values,  we  have  for  the  maximum  combined  unit  shear 


wax.  St  = 
and  for  the  maximum  combined  unit  tension 


.  St  =  -  -f  \l  /  +  - 


(3) 


(4) 


CHAP.  I.]  COMBINED   TENSION  AND  SHEAR.  485 

If  then  the  direct  tensile  and  shearing  unit  stresses  t  and  s  are  given,  we  have  from  (3) 
the  maximum  combined  unit  shear  Ss,  and  from(  i)  its  direction,  or  the  angle  a  which  it 
makes  with  the  direction  of  /.  We  also  have  from  (4)  the  maximum  combined  unit  tension 
St,  and  from  (2)  its  direction,  or  the  angle  a  which  it  makes  with  the  normal  to  t. 

For  compression  and  shear  combined  we  have  to  substitute  for  t  the  direct  compressive 
unit  stress  c,  and  then  the  same  equations  hold. 

Example.  —  A  rivet-  inch  in  diameter  is  subjected  to  a  tension  0/2000  pounds,  and  at  the  same  time  to  a  shear 

0/3000  pounds.     Find  the  combined  maximum  tensile  and  shearing  unit  stresses  and  the  angles  they  make  with 
the  axis  of  the  rivet. 

ANS.   We  have  the  area  —  ,  where  d  is  the  diameter  of  the  rivet.  Hence  the  direct  tensile  unit  stress  is 
4  >  an(j  tfre  direct  shear  is  s  —  -  -  .      .      From  (3)  we  have  Ss  =  7155  pounds  per  square  inch, 


Ttu 

making,  from  (i),  an  angle  tan  la  =  —  ,  or  a  =  9°  13'.  with  the  axis  of  the  rivet,  and  from  (4),  St  =  9420 
pounds  per  square  inch,  making  an  angle  tan  2«  =  —  -,  or  54°  23',  with  the  normal  to  the  axis,  or  35°  47' 
with  the  axis  of  the  rivet. 


CHAPTER  II. 

STRENGTH   OF   PIPES   AND   CYLINDERS.      RIVETING. 

Strength  of  Pipes  and  Cylinders.  —  A  practical  application  of  the  principles  of  the 
preceding  chapter  is  the  determination  of  the  size  of  pipes  and  cylinders,  subjected  to 
internal  pressure. 

Let  /  be  the  pressure  per  square  inch  on  the  interior  surface  of  a  pipe  or  cylinder,  due 
to  the  pressure  of  water  or  steam  or  air,  or  other  fluid.  It  is  a  well-known  principle  of 

physics  that  the  pressure  of  a  fluid  in  any 
direction  is  equal  to  its  pressure  on  a  plane 
at  right  angles  to  that  direction. 

Hence,  in  the  figure,  the  pressure  P,  say 
in  a  vertical  direction,  is  equal  to  the  pressure 
on  a  horizontal  plane  of  area  Id,  where  /  is 
the  length  and  d  the  interior  diameter.     We 
have  then 

P  —  pld. 

If  5W  is  the  working  stress  for  the  material,  and  /  is  the  thickness,  we  have  then 


Pipes  come  in  commercial  sizes,  and  the  preceding  formula  enables  us  to  select  the 
nearest  commercial  size  for  given  unit-pressure,  diameter  and  working  stress. 

If  we  consider  the  preceding  figure  as  a  closed  cylinder,  then  the  pressure  on  the  head 

Ttlf 

is/  X  ---  »  a"d  the  area  of  cross-section  is  ntd.      We  have  then 

p  X   —  =  ntdS^      or     /  =  ^-  . 
4  4-S. 

Hence  the  thickness  to  resist  longitudinal  rupture  is  twice  that  necessary  to  resist  end 
rupture.  For  water-pressure,  if  the  head  h  is  taken  in  feet,  the  pressure  in  pounds  per 
square  inch  is/  =  O.434//. 

Examples.  —  (l)   A  cast-iron  water-pipe  12  inches  diameter  and  -  inch  thick  is  under  a  head  of  300  f,'et. 
Find  the  factor  of  safety. 

ANS.   The  unit-pressure  is  300  x  0.434  =  130.2  pounds  per  square  inch.     Hence  the  working  stress  is 


Pd         130.2    X    12 

5»  =  —  =  ~       — - —  =  1230  pounds  per  square  inch. 

2    X    — 
2  X    g 


486 


CHAP.  II.] 


THEORY  AND  PRACTICE   OF  RIVETING. 


487 


sq 


From  our  table  page  476,  the  ultimate  strength  of  cast  iron  for  tension  is  20000  pounds  per  square  inch. 

The  factor  of  safety  is  then  2OOO°  __  about  jg 
1230     • 

(2)  Find  the  thickness  of  a  cast-iron  pipe  18  inches  in  diameter  under  a  head  of  water  of  300  feet,  taking 
a  factor  of  safety  of  10. 

ANS.   From  our  table  page  476,  we  have  the  ultimate  strength  for  cast  iron  in  tension  20000  pounds  per 
uare  inch.     Our  working  stress  is  then  Sw  =  2000  pounds  per  square  inch.     Hence 

/=   Pd  =  300x0.434x18  inch 

25ro  2    X    2000 

(3)  A  wrought-iron  pipe  4.3  inches  internal  diameter  weighs  12.3  pounds  per  linear  foot.     What  pressure 
can  it  carry  with  a  factor  of  safety  of  8  ? 

ANS.   A  bar  of   wrought  iron    i   square  inch  in  cross-section  and  3  feet  long  weighs  10  pounds.     Hence 

the  area  of  cross-section  of  the  pipe  is  12.5  x  --  =  3.75  square  inches.     The  thickness  is  then  /  =  —  7-^  =  - 

10  -zitr       4 


inch.     We  have,  from  our  table 


page 


481,  the  working  stress  Sw  =  ^= — .     Hence 


,       2tS,u       2  x  55000^  . 

^  =  — T-  =  TTT-. =  763  pounds  per  square  inch. 

Theory  and  Practice  of  Riveting. — Another  important  practical  application  of  preceding 
principles  is  the  determination  of  the  size  and  number  of  the  rivets  with  which  plates  are 
fastened  together. 

KINDS  OF  RIVETED  JOINTS. — We  may  distinguish  the  following  FIG.  i. 

joints : 

ist.  Simple  "Lap-joint,  Single -riveted. — Fig.  I  shows  this 
joint  front  and  side.  The  two  plates  overlie  each  other  by  an  amount 
equal  to  the  "  lap, "  and  are  united  by  a  single  row  of  rivets.  The 
distance  p  from  centre  to  centre  of  a  rivet  is  called  the  pitch.  We 
denote  the  diameter  of  rivet  by  d,  and  the  thickness  of  plate  by  /. 

2d.  "  Lap  "Joint,  Double-riveted. — This  joint  is  similar  to  the 
preceding,  except  that  two  rows  of  rivets  are  used.  In  both  cases 
the  rivets  are  in  single  shear. 

In  all  cases  where  more  than  one  row  of  rivets  is  used  the  rivets  are  "  staggered,"  or  so 
spaced  that  those  in  one  row  come  midway  between  those  in  the  next,  as  shown  in  Fig.  2. 

Lap  joints  are  used  in  tension  only. 


o  „  o 


FIG.  3. 


FIG.  2. 


O       O       O 

ooo     o 


o    o     o 


o    o     o 


3d.  "Butt"  Joint,  Single-riveted,  Two  Cover-plates. — Here  the  two  plates  are  set  end  to 
end,  making  a  "  butt  "joint,  and  a  pair  of  "cover-plates"  are  placed  on  the  back  and  front  and 
riveted  through  by  a  single  row  of  rivets  on  each  side  of  the  joint  (Fig.  3).  The  plates  in 
such  a  joint  are  in  general  not  allowed  to  actually  touch,  and  the  entire  stress,  whether 
tensile  or  compressive,  is  therefore  transmitted  by  the  rivets.  The  thickness  of  the  cover- 


488 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  II. 


FIG.  4. 


o     o    o 
o    o    o    o 


o     o    o    o 

000 


plates  should  not  be  less  than  half  the  thickness  of  the  plates  joined,  except  when  this  rule 
would  give  a  thickness  less  than  £  inch.  Owing  to  deterioration  of  the  metal  by  the 
action  of  the  weather,  no  plate  is  used  in  construction  less  than  ±  inch  in  thickness.  Hence  if 

the  plates  joined  arc  less  than  £  inch,  the  cover-plates 
should  be  ^  inch. 

4th.  "Butt"  Joint,  One  C<rver  -plate.  Single- 
riveted.  —  This  is  the  same  as  the  preceding,  except 
that  one  cover-plate  only  is  used,  of  the  same  thick- 
ness as  the  plates  themselves. 

5th.   Double-riveted  "Butt"  Joint,     Two  Cover- 
plates.  —  This  joint  is  the  same  as  case  3,  except  that 
we  have  two  rows  of  rivets  on  each  side  of  the  joint. 
The  thickness  of  the  cover-plates  is  determined 
by  the  same  considerations  as  in  case  3. 

6th.  "Butt"  Joint,  One  Cover-plate,  Double- 
riveted.  —  This  is  the  same  as  the  preceding  case, 
except  that  there  is  only  one  cover-plate  of  the  same 
thickness  as  the  plates  themselves. 

7th.  Chain  Riveting.  —  When  we  have  more  than  two  rows  of  rivets  on  each  side  of  a 
butt  joint  the  system  is  called  chain  riveting.  Such  a  disposition  becomes  necessary  when 
the  requisite  number  of  rivets  is  so  great  that  they  cannot  be  disposed  in  two  rows 
without  unduly  weakening  the  plates. 

SIZE  AND  NUMBER.  —  In  a  riveted  joint  the  resistance  of  the  rivets  should  equal  the 
strength  of  the  plates  joined.  A  rivet  may  fail  by  shearing  across  or  by  being  crushed. 
The  plate  may  fail  by  rupture  between  the  rivets  or  by  tearing  through  of  the  rivets  at  the 
edge  of  the  plate.  The  rivets  should  be  so  proportioned  and  spaced  that  the  strength  for 
any  method  of  failure  may  be  equal  and  the  plates  weakened  as  little  as  possible. 

Notation.  —  Let  STObe  the  working  unit  stress  of  the  plates,  either  compression  or  tension, 
Sw  the  working  unit  stress  for  compression,  Sws  the  working  stress  for  shear,  /  the  thickness 
of  the  plates,  d  the  diameter  of  rivet,  p  the  pitch  of  rivets  in  a  row,  or  the  distance  from 
centre  to  centre  in  a  row,  and  «  the  number  of  rivets. 


Diameter  of  Rivets.  —  Then  the  area  of  a  rivet  is  --  =  O./854*/2. 

4 

ance  of  a  rivet  is  o.7854^2Sw,  and  the  total  shearing  resistance  of  n  rivets  is 
The  bearing  surface  of  a  rivet   is  dt,  of  »  rivets   ndl,  and  the  resistance  to  crushing 
For  equal  strength  of  crushing  and  shearing  we  have  for  single  shear,  or  lap  joint, 


o. 


a,J  =  ndtSm  ,     or     d  =  Q  • 


For  double  shear,  or  butt  joint  with  two  cover-plates,  we  have 
<*Sws=ndfSw,     or     '=^3- 


The  shearing  resist- 


0) 


(2) 


For  threefold  shear  we  have  3  X  0.7854  in  place  of  0.7854  in  (i),  and  so  on. 
It  is  customary  to  take  Sw  =  12500  Ibs.  per  square  inch  and  S.^  =  7500  Ibs.  per  square 
inch  for  wrought-iron  rivets  in  single  shear. 
We  have  then 

d =  2.\2t  for  single  shear,  ) 

r (3) 

d=  I.o6/for  double  shear.  ) 


CHAP.  II.]  THEORY  AND  PRACTICE  OF  RIVETING.  4^0 

Practical  Value  of  d. — Owing  to  risk  of  injury  to  the  material  in  punching,  the  diameter 
of  rivet  must  always  be  at  least  a$  large  as  the  thickness  of  the  thickest  plate  through  winch  it 
passes,  and  the  diameter  as  given  by  (i),  (2)  or  (3)  must  be  chosen  with  reference  to  this 
restriction.  The  least  allowable  thickness  of  a  plate  is  Jinch.  We  should  have  then  as  a 
lower  limit  for  double  shear,  d  =  \  inch.  But  rivets  as  small  as  this  are  rarely  used.  Usually 
•£  inch  is  the  least  diameter  allowable.  A  common  practical  rule  is 


where  d  is  the  diameter  of  rivet,  and  t  the  thickness  of  the  plate  in  inches.  When  this 
rule  gives  d  greater  than  (i),  (2)  or  (3),  we  use  it;  otherwise  we  use  (i),  (2)  or  (3),  unless 
considerations  of  pitch,  as  given  in  what  follows,  prevent. 

Pitch  of  Rivets. — The  area  of  plate  between  two   rivets  is  (p  —d}t;    and   if  Sw  is  the 
working  unit  stress  of  tension  or  compression  for  the  plates,  and  Sws  the  working  unit  stress 
for  shear,  we  have  for  equal  strength : 
for  single  shear  or  lap  joint 

(p  -  d}tSw  =  -— SWJ,     or    /  =  d(i  +  0.7854-^)  ; 

for  double  shear  or  butt  joint 

(p  -  d}tSm  =  ^j-5«,     or    p  =  d(\  +  1.5708-^). 

Since  Sws  and  Sw  are  nearly  equal,  we  have  practically,  if  A  is  the  area  of  cross-section 
of  a  rivet, 

for  single  shear 

I  d\  A    ' 

/=  d(i+o.7*S4j)  =^  +  7'> 

for  double  shear 

I  d\  2.A 

p  -  d\\  +  1.57087]  =  ^+ -y- 

The   plate   section  is  reduced  by  punching  from  //  between  two  rivets  to  (p  -  d)t,  so 
that  in  the  case  of  a  tension  joint  the  strength  is  reduced  in  the  ratio 


or 


We  see  at  once  that  for  a  given  thickness  /  a  large  rivet  gives  a  large  pitch  and  less 
reduction  in  strength  than  a  small  rivet.     Small  rivets  allow  a  less  pitch  at  a  sacrifice 
strength       But  the  less  the  pitch  the  tighter  the  joint.     When  strength  rather  than  tightness 
is  desired,  as  in  bridges  and  parts  of  buildings  and  machines,  we  should  then  use  a  large 
When  tightness  is  essential,  as  in  steam-boilers,  we  should  use  a  small  rivet  witl 


rivet. 


sacrifice  of  strength. 

Practical  Restrictions.-^'^  to  risk  of  injury  to  the  material  in  punching  and  liability 
to  tear  out,  rivets  are  not  allowed  a  pitch  of  less  than  3  diameters,  or  if  this  distance 


490 


ST/tTICS  OF  ELASTIC  SOLIDS. 


[CHAP.  II. 


than  3  inches,  as  it  usually  is,  less  than  j  inches.  Rivets  should  not  be  spaced  farther  apart 
than  6  inches  in  any  case,  or,  when  the  plate  is  in  compression,  16  times  the  thickness  of  the 
thinnest  outside  plate.  This  last  is  to  guard  against  buckling  of  the  outside  plate  between 
rivets.  With  these  restrictions  we  may  apply  (5). 

Number  of  Rivets. — Guided  by  the  preceding  restrictions  and  rules,  we  can  select  in  any 
case  a  suitable  size  of  rivet.  This  done,  we  can  easily  determine  the  number  required. 

A  rivet'  is  considered  as  failing  either  by  shearing  across  or  by  crushing.  In  any  case, 
then,  the  diameter  being  chosen,  we  must  take  such  a  number  as  shall  give  security  against 
these  two  methods  of  failure,  choosing  the  greater  number.  In  general  the  number  to  resist 
crushing  will  be  more  than  enough  to  resist  shear.  Still  we  should  try  for  both.  The  bear- 
ing area  of  a  rivet  is  the  projection  of  the  hole  upon  the  "diameter,  or  dt. 

The  allowable  compressive  stress  is  about  12500  Ibs.  per  square  inch.  The  allowable 
shear  is  taken  at  7500  Ibs.  per  square  inch  for  single  shear. 

In  the  following  table  we  have  given  the  safe  shearing  and  bearing  resistance  for  rivets 
of  different  sizes  and  for  different  thicknesses  of  plate.  Having  chosen,  then,  the  size  of 
rivet, -an  inspection  of  the  table  will  give  its  resistance.  The  stress  to  be  resisted  being 
known,  the  number  to  resist  this  stress  either  by  bearing  or  shearing  is  easily  determined. 
The  greatest  of  these  two  numbers  is  taken,  with  enough  over  in  any  case  to  complete  the 
row  or  rows.  As  most  practical  cases  are  double  shear,  the  greatest  number  will  usually  be 
determined  by  the  bearing  resistance. 

Distance  from  End  to  Edge. — The  distance  between  the  end  and  edge  of  any  plate  and 
the  centre  of  rivet-hole,  or  between  rows,  is  fixed  by  practice  at  never  less  than  i£  inches, 
and  when  practicable  it  should  be  at  least  2  diameters  for  rivets  over  f  inch  diameter. 

Joints  in  Compression — The  size  and  number  of  rivets  are  determined  for  joints  in  com- 
pression precisely  as  for  joints  in  tension,  because  the  joints  are  not  considered  as  in  contact, 
and  hence  the  rivets  must  transmit  the  stress  in  either  case. 

RIVET  TABLE. 
SHEARING  AND  BEARING  RESISTANCE  OF  RIVETS. 


Diameter  of 
Rivcl  in  inches. 

Area 
of 
Rivet 
in 
square 
inches. 

Single 
Shear  at 
7500  Ibs. 
per 
square 
inch. 

Hearing  Resistance  in  pounds  for  Different  Thicknesses  of  Plate  at  17500  Ibs.  per  square 
inch  =  12500  x  at. 

Frac- 
tion. 

Deci- 
mal. 

i" 

iV 

t" 

TV" 

i" 

TV 

r 

H" 

1" 

1  3" 
T« 

1" 

t 

0-375 

0.1IO4 

828 

1170 

1465 

1760 

V 

0-4375 

0.1503 

II3O 

1370 

1710 

2050 

2390 

* 

0.5            0.1963 

14/0 

1560 

1950 

2340 

2730 

3125 

rV 

0.5685 

0.2485 

1860 

1760 

220O 

2640 

3080 

3520 

3955 

1 

0.625 

0.3068 

2300 

1950 

2440 

2930 

3420 

3900 

4300 

488o 

H 

o  6875 

0.3712 

2780 

2I5O 

2680 

3220 

3760 

4290 

4830 

5370 

5908 

I 

0.75 

0.4418 

33io 

2340 

2930 

3520 

4100 

4690 

527" 

5%o 

6440 

7030 

it 

0.8125 

0.5185 

3890 

2540 

3170 

3800 

4440 

5080 

57»o 

6350 

€980 

762O 

8250 

I 

0.875 

o  6013 

45io 

2730 

3420 

4100 

4780 

5470 

6150 

68*0 

7520 

8200 

8890 

9570 

il 

0.9375 

0.6903 

5180 

2930 

3660 

4390 

5130 

5860 

6590 

7320 

8050 

8790 

9520 

10250 

i 

I 

0.7854 

5890 

3125 

3900 

4690 

5470 

6250 

7^30 

7810 

8590 

9370 

10160 

10940 

IA 

1.062? 

0.8866 

6650 

3320 

4150 

4980 

5810 

6640 

74/0 

8300 

9130 

9900 

10790 

11620 

n 

1.125 

9.0940 

7460 

3520 

4390 

5270 

6150 

7030 

7910 

8790 

9667 

10550 

11420 

12300 

IA 

1.1875 

1.1075 

8310 

3710 

4640 

5570 

7420 

8350 

9280 

I02fX> 

HI30 

12060 

12990 

CHAP.  II.]  WETTING— EXAMPLES.  49 I 

Examples. — (i)  A  boiler  is  to  be  made  of  ivr ought-iron  plates  f  inch  thick,  united  by  single  lap-joints. 
Find  the  size  and  pitch  of  rivets.  If  the  boiler  is  30  inches  in  diameter  and  carries  a  pressure  of  100  Ibs.  per 
square  inch  above  the  atmosphere,  find  the  factor  of  safety,  taking  the  ultimate  strength  at  55000  Ids.  per 
square  inch. 

ANS.  From  (4),  page  489,  we  have  £-in.  rivets.  But  from  (3),  page  488,  we  have  f-in.  This  size  would 
be  chosen  for  ordinary  construction  work.  In  this  case  we  wish  a  tight  joint,  and  therefore  use  a  small 
rivet  at  sacrifice  of  strength.  Let  us  take,  then,  f-in.  rivets.  Then  from  (5),  page  489,  we  find  the  pitch  f  in. 
But  this  violates  the  practical  restriction  that  rivets  should  not  have  a  less  pitch  than  three  diameters.  We 
take  the  pitch,  then,  2  inches.  The  pressure  on  a  length  equal  to  the  pitch  is  30  x  2  x  100  =  6000  Ibs.  If  S  is 

the  unit  stress,  the  resisting  stress  is  5(2  —  |-]/  =  p5.     Hence  S  =  — =  11640  Ibs.  per  square  inch. 

\         °  /        °4  .33 

The  factor  of  safety  is  then  about  5.  If  this  is  considered  too  small,  we  should  use  a  less  pitch  or  a  larger 
rivet.  A  larger  rivet  would  not  be  tight  enough.  For  a  less  pitch  the  holes  must  be  drilled  and  not 
punched. 

(2)  Required  to  unite  two  \-inch  plates  by  a  butt  joint  with  two  cover-plates  ;  the  stress  to  be  transmitted 
being  40000  Ibs.  and  the  unit  working  stress  10000  Ibs.  per  square  inch. 

ANS.  The  area  of  the  plates  must  then  be  4  square  inches  net  if  the  joint  is  in  tension, gross  if  in  com- 
pression. The  cover-plates  can  be  each  \  inch  thick.  Our  rule  (4),  page  489,  gives  for  diameter  of  rivet 
d=  \l  inch.  This  is  greater  than  given  by  (3).  page  488,  therefore  we  take  it.  From  our  table  page  490  we 
have  for  the  resistance  to  shear  of  a  ff-inch  rivet  3890  Ibs.  The  rivets  are  in  double  shear  in  a  butt  joint, 

hence  we  require  ~ —  =  about  5  rivets.     The  bearing  resistance  from  our  table  is  5080  Ibs.     We  require, 

40000 
then,  for  bearing  — ^ —  =  about  8  rivets.     This,  then,  is  the  number  we  should  use. 

For  the  pitch  we  have  from  (5),  page  489,  2.887  inches.  This  is  less  than  3  inches.  We  therefore  take 
the  pitch  3  inches.  We  must  have  at  least  ij  inches  for  distance  from  end  and  ed<re  (page  490). 

If  the  plates  are  8|-  inches  wide,  we  must  then  have  three  rows  of  rivets,  three  in  the  first  and  last  and 
two  in  the  middle  on  each  side  of  the  joint.  The  cover-plates  must  then  be  10  inches  long.  The  student 
can  now  sketch  the  cover-plates  with  the  rivet-holes  properly  spaced. 

(3)  A  plate  girder  is  17  feet  long  and  27  inches  deep.      The  uniformly  distributed  load  is  55000  pounds. 
The  thickness  of  the  web  is  \  inch  and  of  the  flange  angle-irons  T98  inch.     Find  the  size,  number  and  spacing  of 
the  rivets  to  unite  the  web  and  flanges. 

ANS.  Our  rule  (4),  page  489,  gives  for  diameter  of  rivet  d=\  inch.  This  is  greater  than  the  size  given 
by  (3),  page  488.  We  take  the  rivets,  then,  £  inch  diameter. 

If  we  neglect  the  web,  the  stress  S  of  compression  in  the  upper  flange  or  of  tension  in  the  lower,  at  any 
point  distant  x  feet  from  the  end,  is  given  by 


Sd-^x  +  ^  =  o,     or     S  =  ™(l-x), 

where  /  is  the  length  in  feet,  d  the  depth  in  feet  and  TV  the  uniform 
load  per  foot  of  length. 

Inserting  /  =  17,  d=~—,  w  —  ----—, 

s  =  55PPQT/    _£ 
4-5     \        17. 

If  we  take  x  =  o  2.5  ft.  5  ft.  8.5  feet, 

we  have  the  stress  at  these  points         S  =  o  26062  43 137  5r944  pounds. 

We  have  then  for  the  fin-t  space  of  2.5  feet  the  horizontal  stress  26062  pounds  or  13  tons  to  be  taken  by 
the  rivets  in  that  space  ;  in  the  second  space  of  2.5  feet,  43137  —  26062  =  17075  pounds  or  8.5  tons  ;  and  in 
the  third  space  of  3.5  feet  we  have  51944  —  43137  =  8807  pounds  or  4.4  tons  to  be  taken  by  the  rivets  in 
that  space. 

For  the  shear  at  any  point  distant  x  feet  from  the  end  we  have 

Shear  = wx  = 


If  we  take  x  =  o  2.5ft.  5ft.  8.5  feet, 

we  have  the  shear  at  these  points         27500  19400  11300  o  pounds. 


492  ST/1T1CS  OF  ELASTIC  SOLIDS.  [CHAP.  II. 

We  have  then  for  the  first  space  of  2.5  feet  the  shear  27500  —  19400  =  8100  pounds  or  4  tons  to  be  taken 
by  the  rivets  in  this  space.  In  the  second  space  of  2.5  feet  we  have  19400  —  11300  =  8100  pounds  or  4 
tons  ;  and  in  the  third  space  of  3.5  feet  we  have  11300  pounds  or  5.65  tons. 

Hence  the  combined  shear  (page  484)  in  the  first  space  of  2.5  feet  is 


/ 

4/ 


—  =  7.63  tons  =  15260  pounds. 


In  the  second  space  of  2.5  feet 

|/4*  +  —  =  5.9    tons  =  11800  pounds. 
In  the  third  space  of  3.5  feet 


/  44 

\  5-65'  +  ~ —  =       6  tons  =  12000  pounds. 

The  bearing  resistance  of  a  seven-eighths  rivet  is,  from  our  table  page  490,  2730  pounds.  We  require 
then,  for  bearing,  in  the  first  space  of  2.5  feet,— — -  =  6  rivets  ;  in  the  next  2.5  feet,  — =  5  rivets;  in  the 

third  space  of  3.5  feet, =  5  rivets. 

We  must  not  pitch  the  rivets  less  than  3  inches  nor  more  than  6  inches  (page  490).  A  pitch  of  4  inches 
for  the  first  2.5  feet,  then  5  inches  for  the  next  2.5  feet,  and  then  6  inches  to  the  middle  will  therefore  give 
more  rivets  than  are  necessary. 


CHAPTER  III. 


FIG.  i. 


FIG.  2. 


STRENGTH   OF   BEAMS.    TORSION. 

Flexure  or  Bending  Stress — When  a  body  is  in  equilibrium  under  equal  and  opposite 
couples  in  the  same  plane,  the  stress  is  one  of  bending  or  pure  flexure  in  that  plane.  Thus 
if  a  beam,  Fig.  i,  supported  at  A  and  B,  is  loaded  at 
the  ends  a  and  b  with  equal  loads  F,  the  upward  pressure 
or  reaction  at  the  supports  A  and  B  will  be  F,  and  if  the 
distances  aA,  bB  are  equal,  the  portion  of  the  beam  be- 
tween A  and  B  is  in  equilibrium  under  opposite  and  equal 
couples  in  the  same  plane,  and  is  therefore  under  pure 
flexure  or  bending  stress,  and  the  beam  bends  as  shown 
in  Fig.  2. 

Beyond  the  supports  A  and  B  we  have  bending 
stress  and  shear  combined ;  between  the  supports  A  and 
B  we  have  pure  bending  stress,  or  flexure,  only. 

As  already  stated,  we  consider  only  prismatic  bodies 
(page  473).      We  consider  all  such  bodies  as  composed  at  any  cross-section  ab,  Fig.  2,  of  an 
indefinitely  large  number  of  parallel  filaments  or  fibres  of  indefinitely  small  crossrsection. 

Assumptions  upon  which  the  Theory  of  Flexure  is  Based. — We  assume  in  all  that 
follows  in  our  discussion  of  beams,  first,  that  the  coefficient  of  elasticity  E  is  constant  ; 
second,  that  any  section  which  is  plane  before  bending  remains  plane  after  bending; 
third,  that  the  deflection  is  very  small  compared  to  the  length;  fourth,  that  the  elastic 
limit  is  not  exceeded.  * 

Upon  these  assumptions  the  theory  of  flexure  rests.  The  comparison  of  its  results  with 
experiment  shows  them  to  be  correct  so  long  as  the  elastic  limit  is  not  exceeded. 

The  reader  should  therefore  never  apply  the  theoretical  formulas  to  cases  where  the 
elastic  limit  is  exceeded. 

Neutral  Axis  of  Cross-section  and  Body. — When  a  body  is  bent  it  is  evident,  as  shown 
pIG    j  pIG    j.  in  Fig.  I,  that   for  any  cross-sec- 

tion ab  the  outer  layer  of  fibres 
on  the  convex  side  at  a  is  ex- 
tended and  the  outer  layer  on  the 
concave  side  at  b  is  compressed. 
Between  these  two  layers  there 
must  then  exist  a  layer  cc,  Fig.  2, 
which  is  neither  extended  nor 
compressed.  This  layer  is  the  neutral  axis  of  the  cross-section.  It  is  at  right  angles  to  the 
plane  of  bending  through  some  point  C,  Fig.  i. 

For  any  layer  of  fibres  above  or  below  cc,  if  we  assume  that  any  plane  cross-section 
before  bending  remains  plane  after  bending,  the  strain  will  be  proportional  to  the  distance 

4Q3 


«-0/a----£  

r  ^^ 

_.  0 

-!-__<, 

—  *^_  , 

c 

6 

494 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  III. 


from  cc,  and,  by  the  law  of  elasticity  (page  475),  if  the  elastic  limit  is  not  exceeded,  the 
stress  is  proportional  to  the  strain. 

Let  Sf  be  the  unit  stress  in  the  most  remote  fibre  at  a  distance  v  from  cc.  If  a  is  the 
cross-section  of  this  fibre,  Sfa  is  the  stress.  Let  Sa  be  the  stress  in  any  other  fibre  at  the 
distance  v'  from  cc. 

Then  if  the  elastic  limit  is  not  exceeded  and  if  any  cross-section,  plane  before  bending, 
remains  plane  after,  we  have 


Sa  :  Sfa  :  :  v' :  v,     or    Sa  •=  —Sfa. 


(0 


Now  for  equilibrium  the  algebraic  sum  of  all  the  parallel  fibre  stresses  must  be  zero,  or 

-f          =  o. 


But  if  2av'  =  o,  the  fibre  layer  cc  must  pass  through  the  centre  of  mass  C  of  the  cross- 
section,  and  this  layer  is  not  strained  by  bending. 

Hence  if  the  clastic  limit  is  not  exceeded  and  if  any  cross-section,  plane  before  bending, 
remains  plane  after,  the  neutral  axis  for  any  cross-section  passes  through  the  centre  of  mass  of 
that  cross-section  at  right  angles  to  the  plane  of  bending. 

If  then  we  draw  a  line  AC  through  the  centre  of  mass  of  every  cross-section,  this  line 
is  not  strained  by  bending  and  is  called  the  neutral  axis  of  the  body. 

Hence  if  the  elastic  limit  is  not  exceeded  and  if  any  cross-section,  plane  before  bending, 
remains  plane  after,  the  neutral  axis  of  the  body  passes  through  the  centres  of  mass  of  all  t  lit- 
er oss-  sect  ions  . 

Bending  Moment.  —  The  resultant  moment  relative  to  the  neutral  axis  cc  of  any  cross- 
section,  of  all  the  external  forces  on  the  right  or  left  of  that  cross-section,  causes  bending. 

For  reasons  to  be  given  later  (page  497),  we  always  take  the  resultant  moment  on  the 
left  of  the  cross-section,  and  call  this  the  BINDING  MOMENT  for  that  cross-section.  The 
resultant  moment  on  the  left  of  any  cross-section  is  the  algebraic  sum  of  the  moments  of  all 
the  external  forces  on  the  left. 

We  define,  then,  the  bending  moment  for  any  cross-section  as  the  algebraic  sum  of  the 
moments  of  all  the  external  forces  on  the  left  of  that  cross-section,  and  denote  it  by  M* 

If  all  the  external  forces  are  co-planar,  we  take  the  plane  of  XY  as  the  common  plane, 

and  the  'horizontal  line  through  the  centre  of  mass  C 
of  the  cross-section  as  the  axis  of  X. 

Hence,  as  shown  in  Fig.  I,  rotation  counter- 
clockwise from  X  to  F  is  positive,  upward  forces 
and  forces  towards  the  right  are  positive,  and  dis- 
tances on  the  right  of  C  and  above  C-Jfare  positive. 

Thus  if  we  have  any  number  of  vertical  forces, 
Pv  Pv  P3,  etc.,  on  the  left  of  C  with  the  lever-arms 
jfp  xv  x3,  etc.,  the  bending  moment  at  C  is 


FIG.  i. 


where  every  term  in  the  algebraic  sum  is  to  be  taken 
with  its  proper  sign.  In  Fig.  I,  for  instance,  we 
should  have,  taking  the  forces  as  given, 


CHAP.  III.] 


BENDING  MOMENT. 


495 


FIG.  2. 
— f-— ' 


FIG.  3. 


where  the  numerical  values  of  the  forces  and  lever-arms  are  to  be 
inserted  without  sign. 

CASE  i. — Thus  for  a  beam  of  length  /,  Fig.  2,  fixed  horizon-    \ 
tally  at  the  right  end,  with  a  load  P  at  the  left  end,   the  bending 
moment  at  any  point  C  distant  x  from  the  left  end  is 

Mx=  +  Px,      .     .     .     '.     .     .     .     (2) 

when  the  numerical  values  of  P  and  x  are  to  be  inserted  without 
sign. 

The  bending  moment  is  a  maximum  and  equal  to  M2  =  -)-  PI 
at  the  fixed  end,  and  the  bending  moment  at  any  point  is  given  to 
scale  by  the  ordinate  to  a  straight  line,  as  shown  in  Fig.  3. 

FIG.  4.  If  the  beam  is  fixed  at  the  left  end,  as  shown  in  Fig.  4,  the 

moment  on  the  right  is  —  Px;  and  since  for  equilibrium  the  moment 
on  the  left  must  be  equal  and  opposite,  we  still  have  for  moment 
on  left,  as  before, 

M,  =  +  Px, 

where  P  and  x  are  to  be  taken  without  sign. 
CASE  2. — If  the  beam  is  covered  with  a  uniformly  distributed  load  of  w  per  unit  of  length, 
Fig.  5,  the  load  on  the  left  of  C  is  wx.  and  this  load  can  be  con-  FIG.  5. 

sidered  as  acting  at  its  centre.      The  bending  moment  is  then 

t-2 

• (3) 


WX" 


FIG.  6. 


where  the  numerical  values  of  w  and  x  are  to  be  inserted  without 

sign.      Here,  again,  the  bending  moment  is  a  maximum  at  the  fixed  -wx 

end  and  equal  to  M2  =  -\ •,  and  the  bending  moment  at  any  point  is  given  to  scale  by  the 

ordinates  to  a  parabola  whose  vertex  is  at  the  left  end,  as  shown 
in  Fig.  6. 

If  Fig.   5  were  reversed,  we  should  have,  as  in   Fig.  4,   the 

tux* 
moment  on  the  right  —     — .    Hence,  the  moment  on  the  left  being 

equal  and  opposite,  we  should  still  have  for  the  moment  on*  the  left 


FIG.  7. 


CASE  3. Let  a  beam  of  length  /,  Fig.  7,  rest  horizontally  on  two  supports  and  have  .. 

load  P  at  a  distance  kl  from  the  left  end  and   (i  —  k}l  from  the  right  end,  where  k  is  any 

given  fraction.      Thus  if  the  load  is  at  the  centre,  k  =  -  and 

kl  =  -/      If  the  load  is  at  -/  from  the  left  end,  k  =.  -,  etc. 
2   '  4  4 

We  have  then  the  left  reaction  Rl  given  by 


-, \ 


_  Rj  _|_  />(!  _  k]l  =  o,      or     R,  =  (i  -  k)P. 

The  bending  moment  for  any  point  C^  between  the  left 
cud  and  the  load  is  then 


h 


i-*)z—  ^ 


M  = 


=  -  (i  -  k}Px. 


(4) 


STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  III. 

For  any  point  <7a  between  the  load  and  the  right  end  the  bending  moment  is 

x).       .     ....     .     (5) 


In  both  cases  the  bending  moment  is  a  maximum  at  the  load  and  equal  to  —  k(i  —  k)Pl, 
and  the  bending  moment  at  any  point  is  given  to  scale  by  the  ordinates  to  two  straight  lines 
as  shown  in  Fig.  8. 

I  r  pi 

If  the  load  is  at  the  centre,  k  =  -,  and  we  have  R,  —  -P,  M  '=  --  . 

2     '  4 

CASE  4.  —  If  the  beam  has  a  uniformly  distributed  load  of  w  per  unit  of  length,  Fig.  9,  the 


FIG.  o. 


wl 

reaction  at  the  left  end  Rl  =  — ,  and  the    bending    mo- 
ment at  any  point  C  is 

•wlx  .   wx* 

Mm  —  — . 

2  2 

The  bending  moment  is  a    maximum  at  the  centre 

wl* 
and  equal  to ^-,  and  the  bending  moment  at  any  point 


is  given  to  scale  by  the  ordinates  to  a  parabola 
whose  vertex  is  at  the  centre  as  shown  in  Fig.  10. 
(See  page  420.) 

Resisting  Moment. — The  resultant  moment  of  all  the  fibre  stresses  at  any  cross-section, 
relative  to  the  neutral  axis  of  that  cross-section,  resists  bending.  It  must  therefore  be 
equal  and  opposite  to  the  bend- 
ing moment.  We  call  it  there- 
fore the  resisting  moment.  Since 
we  have  taken  for  the  bending 
moment  the  resultant  moment  of 
all  external  forces  on  the  left,  the 

resisting  moment  must  be  taken     (  C 

for  all  fibre  stresses  on  the  right 
of  the  cross-section. 

Let  5"  be  the  unit  stress  in  any 
fibre  of  the  cross-section,  and  a 
the  area  of  cross-section  of  the  fibre.  The  stress  in  the  fibre  is  then  Sa.  If  v'  is  the  dis- 
tance of  the  fibre  from  the  neutral  axis  cc  of  the  cross-section,  we  have  for  the  moment  of 
the  fibre  stress 

-  Sai/. 

The  resisting  moment  for  the  cross-section  is  then 

-  2Sav'. 

But  from  equation  (i),   page  494,  within  the  limit  of  elasticity,  if  any  cross-section 
plane  upon  bending  remains  plane  after, 


CHAP.  III.]  RESISTING  MOMENT.— EXAMPLES.  497 

where  Sf  is  the  unit  stress  in  the  most  remote  fibre  at  a  distance  v.     Hence  we  have  for  the 
resisting  moment 


But  2av'*  is  the  moment  of  inertia  /  of  the  reaction  relative  to  its  neutral  axis  cc. 
Hence  the  resisting  moment  is 

Sfl 

resisting  moment  = *— . 

Now  if  M  is  the  resultant  moment  of  all  the  external  forces  either  on  right  or-left  of 
section  relative  to  the  neutral  axis  of  the  section,  we  have  in  general  for  equilibrium,  within 
the  limit  of  elasticity,  if  any  cross-section  plane  before  bending  remains  plane  after, 

M  *  •  -£—  =  o.     or    — —  =  M. •     (II) 

v  v 

If  in  equation  (II)  we  always  take  M  on  the  left  of  the  section,  it  is  evident  that  Sf  will 
be  positive  when  the  stress  is  tension  and  negative  when  the  stress  is  compression.  It  is 
for  this  reason  that  we  always  take  the  bending  moment  M  on  the  left  of  the  section. 

In  applying  (II),  since  Sf  is  always  given  in  pounds  per  square  inch,  all  distances  should 
be  taken  in  inches.  In  finding  M  the  change  of  lever-arm  due  to  bending  is  neglected. 
That  is,  tJie  deflection  is  assumed  as  very  small  compared  to  the  length. 

(    By  inserting  for  M  its  value  as  found  on  page  495,  we  can  find  in  any  case  the  load 
which  will  give  any  desired  outer  fibre  stress  Sf  within  the  limit  of  elasticity. 

For  values  of  /  for  different  cross-sections  see  page  42.  We  give  here  some  of  the 
most  common. 

I  d 

Rectangular  cross-section /  =  — bd*t          v  =  — . 

Circular  "  /  =  — ,  v  =  r. 

4 

Triangular  "  /  =  —  v  =  —A,          -h. 

36  33 

Examples. — (i)  A  timber  beam  of  length  I  =  10  feet  and  constant  rectangular  cross-section  of  breadth 
b  =  4  inches  and  depth  d  =  6  inches,  rests  horizontally  upon  two  supports  and  sustains  a  load  of  P  =  3000 
pounds  at  the  centre.  Find'the  maximum  fibre  stresses.  P 

P                                                                                   \   ' 
ANS.  We  have  R\  =  — .     The  maximum  bending  moment  is  at        L . 2 

the  centre  and  given  by  (page  496)  L p 


2  4  £ 

The  maximum  fibre  stress  Sf,  if  the  elastic  limit  is  not  exceeded,  is  then  given,  from  (II),  by 

c  _  Mv 
•V-     /   • 


Now  /  =  —  bd*\  and  since  v  =  -\  —  for  the  top  fibre,  we  have  for  the  top  fibre 

•\Pl  T,  x  2000  x  1  20 

Sf  =  ~  M  =  ~      2  x  4  x  36~  =  -  25 

The  minus  sign  shows  that  the  top  fibre  is  in  compression 


•\Pl  T,  x  2000  x  1  20 

Sf  =  ~  M  =  ~      2  x  4  x  36~  =  -  25°°  Pounds  per  square  inch. 


49^  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  TIL 

For  the  bottom  fibre  we  have  v  = ,  and   hence  S/  =  +  2500  pounds  per  square  inch  tension.     We 

see  from  our  table  page  476  that  this  is  within  the  elastic  limit. 

(2)  Let  the  cross-section  be  triangular  with  the  point  up,  the  base  of  the  triangle  being  S  inches  and  the 
height  to  inches.     Find  the  maximum  fibre  stresses. 

ANS.  We  have  as  before  M  —  — — .     But  /  =  — g,  and  for  the  top  fibre  v  =  +  -h.     Hence  for  the  top 
fibre 

6/Y»  6  X  2000  X    120 

6/ = j-rr  = = —          —  =  —  looo  pounds  per  sq.  in. 

bhl  8  x  loo 

For  the  bottom  fibre,  since  v  = h, 

sf  =  +  ^Ta  =  +  9°°  pounds  per  sq.  in. 

The  maximum  fibre  stresses  are  then  1800  pounds  per  square  inch  compression  in  the  top  fibre  and  900 
pounds  per  square  inch  tension  in  the  bottom  fibre,  at  the  centre  of  the  span. 
We  see  from  our  table  page  476  that  the  elastic  limit  is  not  exceeded. 

(3)  Required  the  depth  of  a  rectangular  beam  supported  at  the  ends  and  carrying  a  load  P  at  the  middle 

in  order  that  the  elongation  of  the  lou>est  fibre  shall  equal of  its  original  length. 

ANS.  We  have  from  (i),  page  477,  the  elongation  A  given  by 

Sfl         I  E 

A  =  -FT  =    — .          Hence  Ss  =  — 
E        1400  1400 

Also  from  (II),  page  497, 

s  _Mv E_ 

Now  from  (4),  page  495,  J/  =      — .    Also  in  the  present  case  v  =  - -,  I  =~M*.     Hence 

« nr  rr 

or     d  =' 


(4)  A  cast-iron  rectangular  beam  rests  horizontally  upon  supports  12  feet  apart,  and  carries  a  load  of 
2000  pounds  at  the  centre.     If  the  breadth  is  one  half  the  depth,  find  the  area  of  cross-section,  so  that  the  unit 
stress  stress  in  the  loiuer  fibre  may  nowhere  exceed  4000  pounds  per  square  inch. 

ANS.  The  given  unit  stress  4000  is  less  than  the  elastic  limit  (page  476),  so  (II)  applies.     The  greatest 
will  be  at  the  centre,  where  the  bending  moment  is  greatest. 
We  have  then  for  tension  in  lower  fibre 

Mv 
Sf  =  4000  =  — . 

In  the  present  case,  from  (4),  page  495,  M  = ,v  =  —  -,/  =  _.     Hence 

r     ^  =  3x2000x13x1^     or     </  = 
4000 

(5)  A  wrought-iron  plate  girder  27  \  inches  deep  centre  to  centre  of  the  flanges  rests  horizontally  upon 
supports  20  J  eft  apart.     Its  bottom  flange  is  a  =  48  square  inches  area  of  cross-section.     Neglecting  the  web, 
find  the  load  at  the  centre  which  would  cause  a  unit  stress  of  15000  pounds  per  square  inch  in  the  bottom  flange. 

ANS.  The  unit  stress  1 5000  is  less  than  the  elastic  limit  (page  476),  so  (II)  applies.   We  have  then  at  centre 

*,=  , 5000-*?. 


CHAP.  III.]  DESIGNING  AND  STRENGTH  OF  BEAMS.  499 

In  the  present  case,  from  (4),  page  495.  M  =  -  ^,  v  =  -  ^and,  neglecting  the  -web,  I  =  ad\     Hence 

8^=I5coo,     or    P  =  8^—f/^  =  5o7692  pounds. 

(6)  A  cast-iron  plate  girder,  fixed  horizontally  at  one  end  and  free  at  the  other,  8  feet  long  and  12  inches 
deep  centre  to  centre  of  flanges,  causes  a  uniformly  distributed  load  of  16000  pounds.  Find  the  area  of  the  top 
flange,  neglecting  the  web,  so  that  the  unit-stress  shall  not  exceed  3000  pounds  per  square  inch. 

ANS.  From  (3),  page  495,  the  maximum  moment  is  M  =  ~.  From  (II),  Sf  =  3000  =  — .  In  the  present 
case  v  =  +  - ,  neglecting  the  web  7  =  ad"1,  and  iv  =  i6^?  Hence 

wt*  _  <.  _  w?_  _  16000  x  8  x  12  _ 

*ad  ~  ~  v*Sf  ~  4  x  12  x  3000  =       ^  square  inches. 

Designing  and  Strength  of  Beams— Crippling  Load—  Coefficient  of  Rupture.— Let  A 

be  the  area  of  cross-section  of  a  beam  at  anypoint,  V  the  shear  at  that  point  or  algebraic 
sum  of  all  the  vertical  external  forces  between  the  point  and  left  end  (see  page  398),  and 
Sws  the  working  unit  stress  for  shear. 

Then  for  safety,  as  regards  shear,  we  must  have  at  every  point 


From  (II)  we  have 


-=¥• 


where  Sf  is  the  unit  stress  within  the  elastic  limit  in  the  most  remote  fibre  of  any  cross-sec- 
tion at  a  distance  v  from  the  neutral  axis  of  that  cross-section,  /is  the  moment  of  inertia 
of  the  cross-section  relative  to  that  neutral  axis,  M  is  the  bending  moment  as  defined  and 
found  on  page  495,  that  is  the  algebraic  sum  of  the  moments  of  all  the  external  forces  on 
the  left  of  the  section. 

If  in  (2)  we  replace  Sf  by  the  elastic  limit  unit  stress  St,  and  M  by  the  maximum  bend- 
ing moment,  we  have 

max.JJ/=-<-.   ...........     (3) 

Equation  (3),  if  we  put  for  maximum  M  in  any  case  its  value  as  found  page  495,  will 
give  the  load  which  will  strain  the  beam  to  its  elastic  limit.  We  call  this  load  the  CRIPPLING 
LOAD,  since  it  cannot  be  exceeded  without  overstraining.  The  working  load  may  be  taken 
a  suitable  fraction  of  this. 

But,  as  we  have  seen  (page  4/6),  Se  is  difficult  of  exact  determination  by  experiment. 

The  customary  method  of  estimating  the  strength  of  a  beam,  therefore,  is  to  put  equa- 
tion (3)  in  the  form 

max.  M=™,    .......     ....     (4) 

where  R  is  determined  by  direct  experiments  carried  to  the  point  of  rupture.  The  value  of 
R  thus  found  by  experiment  is  called  the  COEFFICIENT  OF  RUPTURE. 

This  use  of  (3)  is  of  course  applying  it  beyond  the  elastic  limit,  and  (4)  is  then  a  purely 
empirical  formula  whose  form  only  is  dictated  by  theory. 


Soo  St/tTlCS  OF  ELASTIC  SOLIDS.  [CHAP.  HI. 

When  satisfactory  experimental  values  of  R  are  not  at  hand,  the  best  we  can  do  is  to 
replaced  by  the  ultimate  tensile  strength  -Ut  or  the  ultimate  compressive  strength  Ue,  using 
whichever  one  gives  the  least  value  for  the  breaking  load.  For  the  safe  load  we  then  take  a 
suitable  factor  of  safety  (page  481). 

We  give  in  the  following  table  average  values  of  R  for  different  materials. 

R. 
Pounds  per  square  inch. 

Steel  (structural)  .................................  120000 

Wrought  iron  ...................................  55000 

Cast  iron  .......................................  35000 

Timber  .......................................  '.  9000 

Stone  ..........................................  2000 

If  then  we  put  in  (4)  for  max.  M  its  value  as  given  page  495,  we  have  the  breaking 
load  in  the  various  cases  there  given  : 

CASE   i.  —  For  beam  fixed  horizontally  at  one  end  with  load  P  at  free  end 


CASE.  2.  —  For  same  beam  uniformly  loaded 


CASE  3.  —  For  beam  resting  horizontally  on  two  supports  with  load  at  a  distance  kl  from 
left  end  and  (i  —  £)/  from  right  end,  where  k  is  any  given  fraction  of  /, 


CASE  4.  —  For  the  same  beam  uniformly  loaded 

'=**=*-%•  ...........  co 

If  in  these  equations  R  is  not  known,  we  can  replace  it  by  Ut  or  Uc,  whichever  gives  the 
least  value  for  P. 

If  we  replace  R  by  S«,we  obtain  the  crippling  load. 

If  we  replace  R  by  Sr  we  obtain  the  load  which  will  cause  any  desired  maximum  outer 
fibre  stress  Sf  within  the  elastic  limit. 

Examples.—  (i)  A  timber  beam  of  length  I  =  10  feet  and  uniform  rectangular  cross-section  of  breadth 
h  =  4  inches  and  depth  d  =  6  inches,  rests  horizontally  upon  two  supports.  Find  the  cripplittg  and  breaking 
load  at  the  centre,  and  the  working  load. 

i  d  I 

ANS.  From  (7)  we  have  for  breaking  load,  since  /  =  —  bd*,  v=  —,  k  =  — 


and  for  crippling  load 

If  we  take  S<  =  3000  pounds  per  square  inch  as  given  by  our  table  page  476,  we  have  the  crippling  load 


CHAP.  III.]  EXAMPLES.  SOT 

If  we  take  R  =  9000  pounds  per  square  inch  as  given  by  our  table  page  500,  we  have  the  breaking  'oad 

P  —  7200  pounds. 

If  we  take  a  factor  of  safety  of  8  (page^Si),  we  have  for  the  working  load 

P  =  900  pounds. 

If  we  take  Uc  =  8000,  as  given  by  our  table  page  476,  in  place  of  R,  we  should  have  for  the  breaking  load 
7 '  =  6400  pounds,  and  a  working  load  of  800  pounds. 

(2)  Let  the  same  beam  have  a  uniform  triangular  cross-section  with  the  point  up,  the  base  b  =  8  inches  and 
the  height  h  =  10  inches. 

ANS.  We  have  from  (7)  as  before,  since  7=  — ,  v  =  -//,  k  =  -  for  breaking  load, 

36  3  2 


and  for  crippling  load 


_  ±RI  __  Rbh* 
~    vl   ~    61  ' 


T>      _- 

'  "     61  ' 


Taking  S'  =  3000  as  before,  we  have  for  the  cfippling  load 

„        3000  x  8  x  loo 
P.  =  -  ^—  -  =  3333*  pounds. 

Taking  R  =  9000  as  before,  we  have  the  breaking  load 

P  =  loooo  pounds, 

or  for  a  factor  of  safety  of  8  (page  481)  a  working  load  of  1250  pounds. 

If  R  were  unknown,  we  should  have,  taking  the  ultimate  strength  for  tension  and  compression  Ut  and 

Uc,  in  the  place  of  R,  from  our  table  page  476,  and  v  —  -h  for  the  lower  or  tension  fibre  and  -h  for  the  upper 

3  3 

or  compression  fibre, 

_  \UtI      12  x   loooo/ 

"      =    ~~ 


4*7,7       6  x  8ooo7        8000  x  8  x   100 

or  P  =  -  --  =  --  -.-.  -  =  --  ?  -  =  88884  pounds. 

•2.  hi  6  x  120 

—hi 
3 

The  second  gives  P  the  least  value  and  should  be  taken. 

(3)  A  beam  of  yellow  pine,  14  inches  wide,  15  inches  deep,  resting  horizontally  upon  supports  10  feet  6 
inches  apart,  broke  under  a  uniformly  distributed  load  of  12  1940  pounds,  rupture  beginning  in  the  lowest  fibre 
at  the  centre.     Find  the  coefficient  of  rupture. 

ANS.  We  have,  from  (8), 

Pvl       3/Y      3  x   121940  x   126 

R  =  -%-=•  =  -2;-=  =  -  —          -  --  =  3658.2  pounds  per  sq.  in. 
87       4W.     4  x   14  x   15  x  15 

(4)  A  wroitght-iron  beam  4  inches  deep  and  i\  inches  wide,  fixed  horizontally  at  one  end  and  loaded  with 
7000  pounds  at  the  free  end,  ruptured  at  a  point  2  feet  8  inches  from  the  load.      Find  the   coefficient   of 
rupture  R. 

ANS.  From  (5), 

Pvl      6PI      6  x  7000  x  32 
R  =  —  =_—=:  -  L  2-  =  56000  pounds  per  sq.  in. 

|  x  4  x  4 

(5)  A  ivr  ought-iron  beam  2  inches  wide  and  4  inches  deep  rests  horizontally  upon  supports  12  feet  apart. 
Find  the  uniformly  distributed  load  it  will  carry  in  addition  to  its  own  weight  for  a  factor  of  safety  of  4, 

ANS.  From  our  table  page  500,  R  =  55000  pounds  per  square  inch. 


502 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  III. 


A  bar  of  iron  3  feet  long  and  one  square  inch  in  cross-section  weighs  10  pounds.     The  weight  of  the 
beam  is  then  320  pounds. 

We  have  then,  from  (8),  for  the  breaking  load 


Hence  for  a  factor  of  4 


P  +  32°  = 


vl 


8AY      Rbd*       55000  x  2  x  16 


3x  .44 


=  4°74  P°" 


and  P  =  3754  pounds. 

(6)  Find  the  length  of  a  beam  of  ash  6  inches  square  which  would  break  of  its  ovvn  weight  when  supported 
horizontally  at  the  ends,  the  weight  of  the  timber  being  jo  pounds  per  cubic  foot  and  R  =  7000  pounds  per 
square  inch. 

ANS.  The  weight  per  cubic  inch  is  -|jg;  and  since  the  cross-section  is  36  square  inches,  the  weight  per 

inch  in  length  is  «/  =  3° '  *g3    =  g-.     We  have  then,  from  (8), 


Hence 


V 


32  x  7000  x  6  x  36 


1796  inches  =  149!  feet. 


(7)  A  wrought-iron  beam  12  ft.  long,  2  inches  wide  and  4  inches  deep  is  supported  at  the  ends.  The  beam 
weighs  \  pound  per  cubic  inch.  Taking  R  at  54000  pounds  per  square  inch,  find  the  uniform  load  it  can  carry 
to  rupture. 

ANS.  Without  the  weight  of  beam,  16000  pounds. 
Besides  the  weight  of  beam,  15712  pounds. 

Comparative  Strength. — We  see  at  once  from  (5),  (6),  (7)  and  (8),  page  500,  that,  taking 
the  semi-beam  of  Case  I  as  unity,  the  same  beam  with  unifoim  load  will  carry  twice  as  much, 
the  beam  on  two  supports  with  load  in  centre  will  carry  four  times  as  much,  and  the  same 
beam  uniformly  loaded  eight  times  as  much. 

Examples. — (i)  A  round  and  a  square  beam  are  equal  in  length  and  have  the  same  loading.  Find  the 
ratio  of  the  diameter  to  side  of  square  so  that  the  two  may  be  of  equal  strength,  R  being  the  same  for  both. 

ANS.  Since  P,  R,  I  are  the  same  in  both  cases,  we  see  from  our  equations  page  500  that  —  must  be 

Diameter         \/  2 
equal  in  both  cases.    Hence  — gr^ —  =f  2|/ 

(2)  Compare  the  relative  strengths  of  a  cylindrical  beam  and  the  strongest  rectangular  beam  that  can  be 
cut  from  it. 

ANS.  Let  D  be  the  diameter  of  the  cylinder,  and  b  and  d  the  breadth  and  depth  of  the  rectangle.     Tlie 

/       bd* 
strength  of  the  rectangle  is  proportional  to;-  =  —/-,  or  to  btP.     But  we  have 


Hence 


or    t    =  j    — 


=  W  -  b*. 


If  we  differentiate  for  l>  and   put  the  differential  coefficient  equal  to  zero,  we 
have  for  maximum  strength 

D  /~2 

D*  —  3<J»  =  o,     or    b  =  — -,  and  hence  ,/  =  D  {/  - 

^3  K  3 


CHAP.  III.]  BEAMS  OF  UNIFORM  STRENGTH.  503 

For  the  strongest  rectangular  beam  we  have  then 

-  =  — 
v  ~    6 
For  the  cylinder  we  have 

7  =  7r03 

v  ~    32  " 
We  have  then 

Strength  of  cylindrical       9*1/3 
Strongest  rectangular   ~      32 

(3)  Compare  the  relative  strengths  of  a  square  beam  to  that  of  the  inscribed  cylinder. 

ANS.   *  =  1.7. 
3* 

(4)  Compare  the  strength  of  a  square  fyeam  with  its  sides  vertical  to  that  of  the  same  beam  -with  a  diagonal 
vertical. 

ANS.        S'de  vertical       =  ^  = 
Diagonal  vertical 

Beams  of  Uniform  Strength. — If  the  unit  stress  Sf  in  the  outer  fibre  is  the  same  at 
every  cross-section,  the  beam  is  of  uniform  strength.  We  have  then,  from  (II),  for  the 
condition  for  uniform  strength 

Mv 

Sf  =  —j~  =  a  constant  quantity. 

CASE  I. — For  beam  fixed  horizontally  at  one  end  with  load  P  at  the  free  end,  we  have 
then,  since  M  =  Px,  for  Sf  at  any  cross-section  distant  x  from  the  free  end 

Pvx 

Sf—  — Y~  —  constant. 

At  the  end  cross-section  let  x  =  /,  v  =  v^  and  /  =  7r     Then,  since  5  is  constant,  we  have 


A        I 

If,  for  instance,  the  beam  is  rectangular,  we  have  the  breadth  and  height  at  the  fixed 
end  bl  and  /tl,  and  at  any  point  b  and  h.     Hence  /  =  — bhz,   7X  =  — bji*  and  v  =  -,  VY  =  — •  . 


12  12 

We  have  then  for  uniform  strength,  from  (i), 

.£.          7 

^//a  ~~~  b.h?' 


If  the  height  is  constant,  h  =  hl ,  and  we  have  for  the  breadth  at  any  point  distant  x  from 
the  free  end 

b=*jx (3) 

The  breadth  then  varies  as  the  ordinates  to  a  straight  line,  from  ^  at  the  fixed  end  to 
zero,  theoretically,  at  the  free  end.  Practically  the  breadth  cannot  be  zero  at  the  free  end, 
but  must  have  a  value  bQ  such  that  the  area  A  =  bji^  at  the  free  end  may  resist  the  shear 
safely. 

We  have  then  from  (i),  page  499, 

**,  =  —  -—       or     b  P 


.•504 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  III. 


Substituting  this  value  of  £„  for  b  in  (3),  we  find  that  the  cross-section  must  be  constant 

p 
and  equal  to  bji^  =  —  -  for  a  distance  x^  from   the 

->«// 
free  end  equal  to 

PI 


For  any  value  of  x  greater  than  XQ  the  breadth 
is  given  by  (3). 
If  the  breadth  is  constant,  we  have,  in  (2),  b  =  bl  and  hence 


(4) 


The  height  then  varies  as  the  ordinates  to  a  parabola  from  bl  at  the  fixed  end  to  zero, 
theoretically,  at  the  free  end.      Here,  again,  we  must  have  the  height  at  the  free  end  equal  to 


Substituting  this  for  h   in  (4),   we  find  that 
the  cross-section   must  be  constant  and  equal  to 

p 
^j//0  =  r; —  for  a  distance  x§  from  the  free  and  equal  to 


For  any  value  of  x  greater  than  x^  the  height  is  given  by  (4). 

If  both  b  and  //  vary,  but  the  cross-section  at  every  point  is  rectangular,  we  have 

bl :  //j : :  b :  //,      or     b  =  £-,          h  —    '  . 


Substituting  these  in  (2),  we  have 


,=*, 


(5) 


The  height  and  breadth  vary,  then,  as  the  ordinates 
to  a  cubic  parabola  from  //,  and  £,  at  the  fixed  end  to 
zero,  theoretically,  at  the  free  end.  The  area  at  -any 
point  is  then 


A  = 


The  area  at  the  free  end  should  be,  then, 
P 


i  ' 

The  cross-section  should  therefore  be  constant  and  equal  to  £</*„  =  v, —  for  a  distance  .r0 
from  the  free  end,  given  by 

'•  =  k?S    \/ktS    • 


CHAP.  III.] 


BEAMS   OF  UNIFORM  STRENGTH. 


5°5 


In  a  similar  way  we  can  find  the  shape  for   uniform   strength  for  any  other  form  of 
cross-section  by  substituting  in  (i)  the  values  of  /,  flt  v  and  vlf      t 

CASE  2. — For  beam  fixed  horizontally  at  the  end  and  loaded  uniformly,  we  have,  since 


wvx* 


vx* 


For  rectangular  cross-section  we  have  I  =  — bh*,  v  =  -.  L=  — b.h*    v,  =— ,  and  hence 

12  2  12  *  2 

3*  P 


For  constant  height  h  =  hv  and 


(3) 


The    breadth    then    varies    as    the   ordinates    to    a 
parabola  from  bv  to  zero,  theoretically,  at  the   free  end, 
and  to  provide  against  strain  the  cross-section  must  be          x0 
constant  for  the  distance  x0  from  the  free  end  equal  to 


For  any  value  of  x  greater  than  this  the  breadth  is 
given  by  (3).      For  constant  breadth  b  —  b^  in  (2),  and 


(4) 


To  provide  against  shear,  we  have  for  the  height  hQ  at  the  distance 


Substituting  in  (4),  we  find  that  in  order  to  resist  shear  the  end  cross-section 

wl 


If  then  the  end  cross-section  is  safe  for  shear, 
every  cross-section  is  safe,  and  for  any  value  of  x  the 
height  is  given  by  (4).  The  height  then  varies  as  the  ordinates  to  a  straight  line  from  //,  at 
the  fixed  end  to  zero. 

If  both  b  and  h  vary,  we  have  for  rectangular  cross-section 


506 


STATICS   OF  ELASTIC  SOLIDS. 


[CHAP.  III. 


Substituting  in  (2),  we  have 


(5) 


To  provide  against  shear  we  must  have 


,       or 


Hence  from  (5)  the  cross-section  must  be    constant  and 
equal  to  b0hQ  =— : — ?  for  a  distance  x9  from  the  free    end 


given  by 


For  any  value  of  x  greater  than  JTO  the  height  and  breadth  are  given  by  (5). 

In  a  similar  way  we  can  find  the  shape  for  uniform  strength  for  any  other  form  of 
cross-section  by  substituting  in  (i)  the  values  of  /,  r,  Iv  and  ?•,. 

Theory  of  Pins  and  Eyebars. — A  direct  application  of  our  principles  is  to  the  design- 
ing of  pins  and  eyebars.  A  pin  is  a  round  beam  subjected  to  bending  and  shear.  It  is 
also  subjected  to  crushing.  Hence  its  bearing  resistance  should  equal  the  greatest  pressure 
upon  it  due  to  any  plate  through  which  it  passes. 

BEARING. — If  </is  the  diameter  of  pin,  /  the  thickness  of  any  plate  through  which  it 
passes,  then  lit  is  the  bearing  area.  Let  Smc  be  the  working  unit-stress  for  compression, 
then  dtSw  is  the  bearing  resistance  of  the  pin.  This  should  equal  the  stress  transmitted  by 
the  plate,  or 

dtSwe  =  stress. 

We  may  take  Swf  at  6.25  tons  per  square  inch  for  wrought-iron  pins.  The  stress  trans- 
mitted is  always  known.  For  a  transmitted  stress  of  one  ton  the  required  bearing  area 
is  then 


and  hence  we  have 


lineal  bearing  on  pin  per  ton  of  stress  = 


,. 


(0 


From  (i),  having  given  the  diameter  d,  we  can  find  the  corresponding  lineal  bearing  or 
thickness  of  plate  for  every  ton  of  transmitted  stress.  We  have  only  to  multiply  this  by 
the  number  of  tons  transmitted  stress  in  any  case  to  find  the  requisite  thickness  of  the  plate. 

DIAMETER  OF  PiN. — Let  /  be  the  thickness  of  plate  or  eyebar,  and  //  its  depth,  then 
///  is  the  area  of  cross-section  of  plate  or  eyebar.  If  £„,,  is  the  working  unit  stress  for 
tension,  then  thSwt  is  the  transmitted  stress.  Now  if  d  is  the  diameter  of  the  pin,  and  the 
thickness  of  the  eyebar  head  is  equal  to  the  thickness  of  the  bar,  we  have  td  for  the  bearing 
area  of  pin,  and  tdSwc  for  its  bearing  resistance.  We  must  have,  then,  for  equal  strength 


tdSul,  =  thS,t  ,      or     d  = 


" 


CHAP.  III.]  MAXIMUM  BENDING  MOMENT. 

We  can  take  the  ratio  -^        *-.      Hence  the  least  diameter  of  pin  is 


5°7 


The  diameter  of  pin  may  need  to  be  greater  than  this,  but  it  cannot  be  less,  unless  the 
thickness  of  eyebar  head  is  made  greater  than  the  thickness  of  the  bar  itself. 

When  this  is  the  case,  if  ^  is  the  thickness  of  the  bar  and  /  the  thickness  of  the  head, 
we  have  for  the  least  diameter  of  pin 


or         =-- 


and  for  the  thickness  of  head 


Ad  ' 


(3) 

(4) 


For  a  beam  subjected  to  flexure  we  have  from  (II),  page  497, 


„*•    V 

max.  M =  -£-, 

where  r  is  the  radius  of  the  pin,  and  Syis  the  unit-stress  in  the  outer  fibre.      Now  /=  — . 
Hence 


max.  M  = 


(5) 


where  max.  Mis  the  maximum  bending  moment.  The  usual  value  taken  for  S/  is  15000 
pounds,  per  square  inch  for  iron  and  20000  pounds  per  square  inch  for  steel. 

We  have  then  in  any  case  to  find  the  maximum  bending  moment,  and  then  from  (5) 
we  can  find  d. 

Maximum  Bending  Moment.  —  In  general  for  any  pin  we  must  resolve  the  stress  in 
every  bar  through  which  the  pin  passes  into  its  vertical  and  horizontal  components.  The 
stress  in  each  bar  is  considered  as  acting  along  the  centre  line  or  axis,  and  hence  the  point 
of  application  of  each  vertical  and  horizontal  component  is  at  the  centre  of  the  bearing  of 
the  corresponding  bar. 

Let  Mh  be  the  maximum  bending  moment  of  all  the  horizontal  and  Mv  of  all  the  verti- 
cal forces.  Then  the  resultant  maximum  bending  moment  is  p,  R,  F5 


M        —  4/M,*  4-  M* 

^max.  ~~        Mk       \     •*"»  • 

From  (5)  we  then  find  the  diameter  d  of  the  pin. 

Let  the  parallel  horizontal  or  vertical  components  on  one 
side  of  the  centre  of  pin  be  T7,  ,  F2,  F3,  F4,  etc.,  the  odd 
indices  F1,  F3.  etc.,  acting  in  one  direction,  and  the  even 
indices  F2,  F±,  etc.,  acting  in  the  other.  Let  /t  be  the  dis- 
tance between  centres  of  bearing  Fl  and  F2,  /„  the  distance 
between  F2  and  F3  ,  etc.  We  can  now  easily  find  the  maxi- 
mum moment  by  trial. 

Thus  the  moment  at  F2  is  FJr     Add  to  this  (F1  —  F2)/2 

and  we  have  the  moment  at  Fy      Add  again  (Fl  —F^-\-F3)l3  and  we  have  the  moment  at 
and  so  on.      The  greatest  of  all  these  is  the  moment  required. 


508  STATICS  OF  ELASTIC  SOLIDS.  [CHAI  .  III. 

Since  all  the  forces  Flt  Ft,  Fs,  etc.,  on  one  side  are  equal  to  all  on  the  other,  F2,  F4, 
Ft,  etc.,  they  reduce  to  a  couple  on  each  side  of  centre  of  the  pin,  and  hence  the  moment 
at  any  point  /'beyond  the  last  force,  as  F6,  is  constant.  We  have  then  only  to  find  the 
greatest  moment  Mk  or  Mv  by  trial  as  directed. 

PRACTICAL  SIZES  FOR  PINS.  —  Pins  are  furnished  in  sizes  differing  by  £  inch,  and  all 
sizes  are  an  even  number  of  sixteenths.  A  pin  must  always  be  ordered  at  least  one  six- 
teenth larger  than  the  hole  it  is  to  fit,  in  order  that  it  may  be  turned  down  to  fit.  We  must 
then  add  -jV  inch  to  the  calculated  size,  and  if  this  gives  an  even  number  of  sixteenths  it  can 
be  ordered  ;  if  not,  add  -^\  more. 

Thus  if  the  size  of  a  pin  is  4f  inches  by  calculation,  it  should  be  ordered  at  least  4T\; 
but  since  only  even  sixteenths  are  furnished,  we  should  order  4^  and  turn  down  to  fit 
the  hole. 

Examples.  —  (i)  A  pin  3  inches  diameter  passes  through  the.  web  of  a  channel  bar  three  fifths  of  an  inch 
thick.  The  transmitted  stress  is  55500  Ids.  Find  the  thickness  of  re-enforcing  plate  necessary  to  give  sufficient 
bearing  on  the  pin. 

ANS.  The  thickness  for  each  ton  (page  506)  is 

-?  -  -T-  =  ^  -  =  0.0533  inches. 
6.25^       6.25  x  3 

For  55500  Ibs.  =  27.75  tons  we  should  have  a  thickness  of  0.0533  x  27-75  =  I-4&  inches. 

The  channel  web  is  only  —  =  0.6  inch  thick.     In  order  to  have  the  proper  thickness  for  safe  bearing  on 

the  pin.  we  must  then  increase  the  thickness  by  1.48  —  0.6  =  0.88  inch.    Two  re-enforcing  plates  on  each  side 
of  the  web,  each  0.44  inch  thick,  or  about  ^  inch  each,  will  then  give  the  required  thickness. 

(2)  If  the  depth  of  an  eyebar  is  fo  inches,  find  i  he  least  diameter  of  pin  which  can  be  used  without  having 
the  thickness  of  the  head  greater  than  that  of  the  bar. 

ANS.  (Page  507.)    d=i\  inches. 

(3)  A  bar  8  in.  by  \  in.  has  a  pin  4$  inches  diameter  passing  through  it.     Find  the  thickness  of  bar  head. 
ANS.  The  least  diameter  without  having  the  head  thicker  than  bar  is  6  inches.     As  the  pin  is  less  than 

this,  the  head  must  be  thicker  than  the  bar  and  equal  to 


4  x 


=  i  A  inches. 


(4)  In  a  Panel  of  a  bridge  truss  we  have  at  each  end  of  the  pin  two  eyebar  s  on  one  side,  4  in.  by  r-j*,  in.  , 
and  on  the  other  side  one  eyebar  4  in.  by  /1\  in.  Also  one  tie  on  each  side  of  centre  sf  pin  /,*,  in.  thick.  The  tie 
is  packed  close  to  the  vertical  post,  which  consists  of  two  channels  of  \  in.  thickness.  The  burs  are  packed  snug. 
The  vertical  compression  in  the  half  post  is  40000  Ibs.  The  working  unit-stress  of  the  bars  is  10000  Ibs.  per 
square  inch.  Find  the  size  of  pin  required. 

ANS.  We  have  here  on  one  side  acting  horizontally 

F\  =  F»  =  4  x  i*ff  x  10000  =  47500  Ibs., 
and  on  the  other  side 

Ft  =  4  x  i,V  x  loooo  =  4/500  Ibs. 
The  horizontal  component  of  the  tie-stress  is 

Ft  =  2  x  47500—  57  500  =  57  500  Ibs. 


CHAP.  III.]  TORSION.  509 

The  distances  are 

*S)  =  i  ft  inches, 


/,  =     (i  A  +  iT9g)  x       =  2^  inches. 

2  o 

We  have  then  at  jpa  the  moment  FJi  =  4750x3  x  iTBf  =  62344  inch-lbs., 
at  F3  we  have  62344  +  (F,  —  F*)l*  =  49219  inch-lbs., 
at  Ft  we  have  492I9  +  (^i  —  F*  +  ^»X«  =  133594  inch-lbs. 
The  maximum  horizontal  bending  moment  is  then 

Mh  =  133594  inch-lbs.  =66.797  inch-tons. 
The  vertical  compression  in  post  is  40000  Ibs.     Its  lever-arm  is 


Hence 

Mv  =  40000  x  i^  =  48750  inch-lbs.  =  24*375  inch-tons. 

The  resultant  maximum  bending  moment  is  then 


+  MJ  =    4/66.S2  +  24.4*  =  71.11  inch-tons  =  142220  inch-lbs. 
We   have  then   for  size  of   pin   about  4$  inches  diameter,  or  4$  commercial  size.      The  least  allowable 
diameter  is  —h  —  3  inches.     Hence  the  bearing  is  abundant. 

Torsion. — Torsion  occurs  when  the  external  forces  acting  upon  a  body  tend  to  twist  it, 
so  that  each  section  turns  on  the  next  adjacent  section  aBout  a  common  axis  at  right  angles 
to  the  plane  of  section. 

Let  a  horizontal  shaft  of  length  /  be  fixed 
at  one  end,  and  let  a  force  couple  -f-  F,  —  F  act 
at  the  free  end  whose  moment  about  the  axis 
AC'isFp. 

The  shaft  will  be  twisted  about  the  axis  AC 
so  that  any  radial  line,  as  aC,  moves  to  bC 
through  the  angle  aCb=  6. 

If  the  elastic  limit  is  not  exceeded,  any  longitudinal  plane  aBAC  before  twisting 
remains  plane  after,  as  bBAC,  and  when  the  couple  -f-  F,  — F  is  removed  the  line  bC 
returns  to  its  original  position  aC.  Also,  the  angle  aCb  is  proportional  to  F  and  to  the 
distance  AC  =  /of  the  cross-section  from  the  fixed  end.  Thus  if  0  is  the  angle  aCb  at  the 

distance  /  from  the  fixed  end,  the  angle  a^Cvbv  at  the  distance  x  from  the  fixed  end  is  j6.     If 

the  elastic  limit  is  exceeded,  this  proportionality  does  not  hold,  the  line  bC  does  not  return 
to  its  original  position  when  the  couple  -f-  F,  — F  is  removed,  and  if  the  twist  is  great 
enough  we  have  rupture. 

NEUTRAL  Axis. — Consider  the  shaft  to  be  made  up  of  an  indefinitely  great  number  of 
parallel  fibres.  Since,  within  the  elastic  limit,  stress  is  proportional  to  strain,  as  one  cross- 
section  of  the  shaft  turns  about  the  axis  and  slides  upon  the  adjacent  cross-section,  the 
strain  and  therefore  the  shearing  stress  on  each  fibre  of  a  cross-section  is  proportional  to  its 
distance  from  the  axis  AC.  For  the  fibre  at  the  axis  AC  there  is  then  no  shearing  stress.  The 
axis  AC  is  then  the  NEUTRAL  AXIS. 


5*°  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  III. 

POSITION  OF  THE  NEUTRAL  Axis.  —  Let  a  be  the  cross-section  of  any  fibre,  and  S,  the 
unit  shearing  stress  within  the  elastic  limit  for  that  fibre  in  any  cross-section  most  remote 
from  the  neutral  axis  at  the  distance  v.  Then  the  shearing  stress  for  the  most  remote  fibre 
in  any  cross-section  at  the  distance  v  is  S,a,  and  for  any  other  fibre  in  that  cross-section,  at 

•*•  T 

—  *•  the  distance  r,  it  is  -S/i.     The  sum  of  all  the  fibre  stresses  of  any  section  in  any 

£ 

straight  line  perpendicular  to  the  axis  is  then  —2ra. 

v 

But  the  sum  of  the  external  forces  -f  F,  —  F  is  zero,  hence  for  equilibrium 
we  must  have  2ar  =  o. 

Therefore  the  neutral  axis  AC  must  pass  through  the  centre  of  mass  of  the 
cross-sections. 

TWISTING  AND  RESISTING  MOMENT.  —  All  the  external  forces  acting  upon  the  shaft 
reduce  to  a  couple  -j-  F,  —  F,  as  shown  in  the  figure,  whose  moment  Fp  with  reference  to 
the  neutral  axis  is  the  TWISTING  MOMENT  Mt.  This  moment  is  the  same  at  every  point  of 
the  neutral  axis  AC,  and  therefore  tends  to  make  each  cross-section  turn  on  its  adjacent 
cross-section  nearest  the  fixed  end,  about  the  axis  AC,  so  that  there  must  be  for  equilibrium 
between  every  two  cross-sections  an  equal  and  opposite  RESISTING  MOMENT  due  to  the 
shearing  stress  between  these  two  cross-sections. 

Since  for  any  cross-section  the  shearing  stress  for  any  fibre  at  a  distance  r   from   the 

neutral  axis  is  -S.a,  the  moment  of  that  stress  about  the  neutral  axis  is  —  ar*,  and  the  sum 
v  v 

of  the  moments   of  all   the  stresses  for  any  cross-section  about  the  axis,  or  the  resisting 

moment,  is  then  —2ar*. 

v 

For  equilibrium  this  is  balanced  by  the  twisting  moment  Mt. 

But  2ar*  is  \.\\e  polar  moment  of  inertia  It  of  the  cross-section  with  reference  to  the  axis 
through  the  centre  of  mass  (page  34). 

We  have  then  for  equilibrium,  without  reference  to  direction  of  rotation, 


where  S,  is  the  unit  shearing  stress  within  the  limit  of  elasticity  in  the  most  remote  fibre  of 
any  cross-section  at  the  distance  v  from  the  neutral  axis,  /,  is  the  polar  moment  of  inertia  of 
the  cross-section  with  reference  to  that  axis,  which  always  passes  through  its  centre  of  mass, 
and  Mt  is  the  twisting  moment. 

The  reader  will  note  the  similarity  of  this  equation  with  that  for  flexure,  page  497. 

From  (i)  we  can  find  Mt  for  any  given  St  when  7,  and  v  are  known  and  the  elastic  lim.'t 
is  not  exceeded,  or,  inversely,  for  given  Mt  and  S,  we  can  find  the  dimensions. 

Coefficient  of  Rupture  for  Torsion.  —  Equation  (i)  holds  within  the  elastic  limit,  or  for 
all  values  of  S,  less  than  St.  If  we  consider  it  as  holding  without  limit,  and  find  the  value 
of  S,  in  this  case  from  experiments  carried  to  the  point  of  rupture,  we  call  the  value  of  S,  thus 
obtained  the  coefficient  of  rupture,  and  denote  it  by  X.  We  have  then  for  rupture 


<„ 


CHAP.  III.]        COEFFICIENT  OF  ELASTICITY  FOR  SHE4R  DETERMINED  BY   TORSION.  $H 

From  equation  (2),  then,  if  R  is  known,  we  can  compute  the  couple  ±  /''which,  acting 
with  the  lever-arm  /,  will  rupture  a  given  shaft,  or  the  dimensions  of  a  shaft  to  resist  rupture 
with  any  desired  factor  of  safety.  If  R  is  in  pounds  per  square  inch,  P  should  be  taken  in 
pounds,  /  in  inches,  and  all  dimensions  in  inches.  The  distance  v  in  inches  is  the  distance 
to  the  remotest  fibre  of  the  cross-section. 

We  have,  page  38,  for 


circular  cross-section  of  radius  r /, .  =  — ,     v  =  r; 


rectangle  with  sides  b  and  d /,  = 1 ,     v  =  —  \  &  4-  d*  ; 


d*  d 

square  cross-section /,  =  -—-,     v  =  — 7=-. 

o  1/2 

We  give  here  also  average  values  of  the  coefficient  of  rupture  for  different  materials: 

x. 
Pounds  per  square  inch. 

Steel  (structural) 75000 

Wrought  iron 50000 

Cast  iron 2  5000 

Timber 2000 

Coefficient  of  Elasticity  for  Shear  determined  by  Torsion. — Let  the  length  of  shaft  be  /, 
and  let  the  angle  of  torsion  be  0  in  radians,  and  the  twisting  moment  be  Mt.  Then,  within 
the  limit  of  elasticity,  the  strain  of  the  outer  fibre  for  the  end  cross-section  is  v6,  and  the 

strain  per  unit  of  length  is  -j- .  The  shearing  unit  stress  of  the  outer  fibre  of  the  end  cross- 
section  is  Ss.  Then  from  page  477,  since  the  coefficient  of  elasticity  is  the  ratio  of  the  unit 
stress  to  the  unit  strain, 


where  v  is  the  distance  of  the  outer  fibre  of  the  end  cross-section  from  the  neutral  axis,  and 
0  is  taken  in  radians. 

If  we  substitute  for  Ss  its  value  from  (i),  we  have 

i  '  *=i£ •'•  I  -  •'•  •  •  (3) 

from  which  Et  can  be  computed  if  the  other  quantities  are  known  and  the  elastic  limit  is  not 
exceeded.      The  angle  0  is  taken  in  radians. 
Inversely  we  have 


From  (4)  we  can  find  Mt  for  any  given  0  in  radians,  when  £.,  /,  and  /  are  given  and  the 
elastic  limit  is  not  exceeded. 


512  ST/4T1CS  OF  ELASTIC  SOLIDS.  [CHAP.  III. 

Work  of  Torsion.  —  If  8  is  the  angle  of  torsion/in  radians  for  any  cross-section,  the  strain 
of  any  fibre  in  that  cross-section  at  a  distance  r  from  the  neutral  axis  is  r&,  and  the  stress 

for  that  fibre  is  —S,a.    The  work  is  then  one  half  the  product  of  the   stress  and  strain  (page 


The  work  of  all  the  fibres  is  then  —'Sar*,  or,  since  2ar*  =  /,,  we  have,  from  (4)  and  (i), 


w_  OS/.  _  MtB  __       ,     _ 

~~^~  :~-T         2l     ~          ......... 


for  the  work 


where  6  is  taken  in  radians. 

Transmission  of  Power  by  Shafts. — Work  is  the  product  of  a  force  by  the  distance 
through  which  it  acts,  and  is  measured  therefore  in  foot-pounds.  Power  is  time-rate  of  work, 
and  is  measured  in  foot-pounds  per  minute  or  per  second.  A  horse-power  is  33000  ft.- 
pounds  per  minute  or  550  ft. -pounds  per  second.  If  a  shaft  makes  n  revolutions  per  minute 
and  the  twisting  force  is  F  with  a  lever-arm  /,  then  2np  X  n  is  the  distance  and  2nnpF  is  the 
work  per  minute,  and  the  horse-power,  if  p  is  taken  in  inches,  is 

H  P    =       27rnFP 

33000  X  12' 

But  Fp  =  Mt  =  ~     Hence 

H  P   =     "xnSJ*  ff- 

1980007;'  I  ) 

where  n  is  the  number  of  revolutions  per  minute,  7,  and  v  are  to  be  taken  in  inches  and  S,  in 
pounds  per  square  inch. 

Combined  Flexure  and  Torsion — We  have  for  bending,  from  equation  (II),  page  497, 

Mv 


and  for  torsion,  from  equation  (i),  page  510, 

&'-M*. 

Hence   for  the   combined   unit  stresses  of  shear  and  compression  or  tension  we  have, 
from  equations  (i)  and  (2),  page  484, 

S=\A'  +  ^. .....     (7) 

S,orS,=%^+A/S,'  +  Sl (8) 


CHAP.  III.]  EXAMPLES. 


513 


Examples.—  (i)  An  iron  shaft  3  feet  long  and  2  inches  diameter  is  twisted  through  an  angle  of  7  degrees 
by  a  couple  of  ±  5000  pounds  with  a  leverage  of  6  inches,  and  on  the  removal  of  the  couple  returns  to  its  origi- 
nal position.  Find  the  value  of  Es  for  shear. 

ANS.  From  (3),  page  511,  we  have,  since  /=  5  x  12  =  60  inches,  r  =  i  inch,  Mt  =  5000  x  6  =  30000  inch- 

itr*       7t    ..       -j-jf 
pounds,  Iz  —  —  =  -,   0  =  '—-  radians, 

„          60  X   3OOOO  X   2   X    I  80 


=  9  390  ooo  pounds  per  square  inch. 


(2)    What  is  the  couple  which,  acting  with  a  lever-arm  of  12  inches,  will  rupture  a  steel  shaft  1.4  inches 
diameter,  the  coefficient  of  rupture  by  torsion  being  75000  pounds  per  square  inch  ? 

ANS.  From  (2),  page  510,  since/  =  \-z,R  —  75000,  v  —  0.7,  /,  =  *  *  *'4  , 

64 

,          '    ,    or     7^=1683  pounds. 


(3)  A  circular  shaft  twisted  by  a  couple  of  90  pounds  with  a  lever-arm  of  27  inches  has  a  unit  shearing 
stress  of  2000  pounds  per  square  inch.     If  the  same  shaft  is  twisted  by  a  couple  of  40  pounds  with  a  lever-arm 
of  57  inches,  what  is  the  unit  shearing  stress  f 

ANS.    From   (i),   page  510,  since  Ss  =  2000,  Mt  =  90  x  27,  we  have  —  =  4P  =  9°  X  27.     Hence  for 

V  i,  2OOO 

Mt  =  40  x  57  we  have 

_  40  x  57  x  2000 
•*»  —        g0  x  27 —    =  J°77  pounds  per  square  inch. 

(4)  An  iron  shaft  5  feet  long  and  2  inches  diameter  is  twisted  by  a  couple  of  5000  pounds  with  a  leverage 
of  6  inches.     If  Es  is  9  390  ooo  pounds  per  square  inch,  find  the  angle  of  twist. 

5000  x  6  x  60                                  5000  x  6  x  60  x  1 80 
ANS.  From  (4),  page  511,  9  =  939OOOO  x  -a.  radians>    or    9  =    —       QQQQ  x     t =  7  degrees. 

(5)  Compare  the  strength  of  a  square  shaft  with  that  of  a  circular  shaft  of  equal  area. 

ANS.  From  (2),  page  510,  we  see  that  the  strength  is  proportional  to—.      For  a  square  shaft  —  = =. 

v  v        34/2 

For  a  circular  shaft  —  =  — .     The  ratio  is  then         3  .     For  equal  areas  we  have  icr*  =  d*.     Hence  the 

ratio  is   V2* . 

(6)  A  circular  shaft  2  feet  long -is  twisted  through  an  angle  of  7  degrees  by  a  couple  of  200  pounds  with  a 
lever-arm  of  6  inches.     Find  the  angle  for  the  same  shaft  4  feet  long  when  twisted  by  a  couple  of  500  pounds 
with  a  lever-arm  of  18  inches. 

ANS.    105  degrees. 

(7)  Find  the  combined  unit  stresses  for  a  wrought-iron  shaft  3  inches  diameter  and  12  feet  long,  resting  on 
bearings  at  each  end,  which  transmits  4°  horse-power  while  making  120  revolutions  per  minute,  upon  which  a 
load  of  800  pounds  is  brought  by  a  belt  and  pulley  at  the  centre. 

ANS.  The  unit  stress  for  flexure  is 

Mr      Plr       PI        800  x   12  x  12  x  8 
Sf  =  —7-  =  — -r  =  — 3  =  —  —  =  10800  pounds  per  square  inch. 

1  4-*  1tr  it    X.    2y 

The  unit  stress  for  torsion  is,  from  (6),  page  512, 

198000  x  40  x  r       198000  x  40  x  2  x  8 

,SS  =  —  -1— a =  — — 5 — =  4000  pounds  per  square  inch. 

it  x   I20/,  TT"  x  120  x  27 

The  combined  stresses,  then,  from  (7)  and  (8),  are,  for  shear,  S,  =  6700  pounds  per  square  inch,  and  for 
tension  or  compression,  St  or  Sc  =  12100  pounds  per  square  inch. 


514  STATICS  OF  ELASTIC  SOLIDS.  [CHAi-.   III. 

(8)  A  vertical  shaft  weighing  with  its  loads  6000  pounds  is  subjected  to  a  twisting  moment  bv  >i  f.'ne 
couple  of  joo  pounds  acting  with  a  leverage  of  4  feet.  If  the  shaft  is  of  wrought  iron  4  feet  long  and  2  inches 
diameter,  find  its  unit  stress,  provided  flexure  is  prevented. 

ANS.  The  unit  stress  of  compression  is       ,   =  1910  pounds  per  square  inch.     The  unit  shearing  stress 

for  torsion  is 

_       Mflf       300  x  48  x  2 

S$  =  — j-  •  =9172  pounds  per  square  inch. 

From  equations  (i)  and  (2),  page  484,  for  combined  compression  and  shear,  we  have  for  the  shearing 
unit  stress 


S,  =  4/9172*  +  —    —  =  9215  pounds  per  square  inch, 
and  for  the  compressive  unit-stress 


Se  =—  -   +  -1/9172*  +  -    —  =  10170  pounds  per  square  inch. 

(9)  Find  the  diameter  of  a  short  vertical  steel  shaft  to  carry  a  load  of  6000  pounds  when  twisted  by  a  force 
of  300  pounds  with  a  leverage  of  4  feet,  taking  unit  stress  for  shear  at  7000  and  for  compression  at  10000 
pounds  per  square  inch. 

ANS.  About  2.5  inches. 


CHAPTER   IV. 

WORK  OF  STRAINING.    DEFLECTION  OF  FRAMED  STRUCTURES.    PRINCIPLE  OF  LEAST 
WORK.     REDUNDANT   MEMBERS.     BEAMS   FIXED   HORIZONTALLY  AT   ENDS. 

Work  of  Straining.  —  If  the  force  F  is  gradually  applied,  increasing  from  zero  up  to  Ft 

F 
the  average  force  is  —  ;   and  if  A,  is  the  strain,  the  work  done  by  I7  is 

p 
work  =  —  A. 

Since  the  stress  is  equal  and  opposite  to  F,  the  work  of  straining,  within  the  elastic 
limit,  is  one  half  the  product  of  the  stress  and  strain. 
Now,  from  (I), 

Fl 


Hence  we  have  for  the  work  done  by  the  force  or  against  the  stress,  within  the  elastic 
limit, 

FH 


Since  E  is  always  taken  in  pounds  per  square  inch,  if  we  take  A  in  square  inches,  .Fin 
pounds  and  /  in  feet,  the  result  will  be  foot-pounds. 

Work  and  Coefficient  of  Resilience.  —  If  the  unit  force  —  is  equal  to  the  elastic  limit  unit 

A 

stress  Se,  so  that 


we  have,  from  (II),  for  the  work  done  in  straining  up  to  the  limit  of  elasticity 

S?AI 

work  = 

If  we  regard  the  body  as  practically  perfectly  elastic  up  to  the  elastic  limit,  this  is  the 
work  which  the  body  can  do  in  returning  to  its  original  dimensions  when  the  force  is 
removed.  It  is  therefore  called  the  WORK  OF  RESILIENCE. 

Since,  for  uniform  A,  Al  is  equal  to  the  volume  V,  we  can  write 

S?  S? 

work  of  resilience  =  —  ^  .  Al  =  —^  .  V. 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  IV. 


s? 

The  coefficient  —jr,  or  the  work  per  unit  of  volume,  is  called  the  coefficient  of  resilience. 

The  work  of  resilience  measures  the  ability  of  the  body  to  withstand  shock  or  suddenly 
applied  force  due  to  the  impact  of  another  body.  To  bring  such  a  body  to  rest  requires 
work.  If  this  work  is  not  greater  than  the  work  of  resilience,  the  elastic  limit  is  not 
exceeded. 

From  our  tables,  pages  476  and  478,  for  average  values  of  S,  and  £,  we  can  compute  the 
following  average  values  of  the  coefficient  of  resilience  : 

Coefficient  of  Resilience. 
Wrought  iron  ....................  12.5  inch-pounds  per  cubic  foot 

Steel  (structural)  .................  26.6          « 

Cast  iron  .......................    1.2          "  "        "        " 

Timber  ......  ...................   3  "  "        "        " 

Deflection  of  a  Framed  Structure.  —  Equations  (I)  and  (III)  find  direct  application  in 
the  computation  of  the  deflection  of  a  framed  structure. 

Thus  let  S  be  the  stress  in  any  member  due  to  the  actual  loading,  and  /  and  a  the  length 
and  area  of  cross-section  of  the  member.  Then,  from  (I),  the  strain  of  the  member  is 


Now  let  s  be  the  stress  in  the   same  member  due  to  any  arbitrary  load  /,  supposed  to 
rest  at  the  point  where  the  deflection  is  required.      The  work  due  to  this  load,/  is 


s\ 
2 


Sst 
zaE' 


The  total  work  in  all  the  members  due  to  this  load  /  is  then 

Ssl 


Now  if  A  is  the  deflection  at  the  point  where/  acts,  the  work  done  by/  is.—  .      Hence 
we  have 


1 

-j~ 
pE 


Sst 

- 

a 


Example.—  Suppose  a  truss  (see  figure)  composed  of  two  inclined  rafters  of  length  bo  inches,  two  vertical 

ties  of  length  48  inches,  an  upper  chord  of  length  60 
inches  and  a  lower  tie  of  length  /  J-?  inches,  the  two 
end  panels  36  inches  and  the  centre  60  inches.  Lft 
there  be  a  diagonal  strut  cf  whose  length  is  76.84 
inches.  Suppose  a  load  of  j  tons  atf  and  10  tons  at  e. 
Required  the  deflection  at  e,  the  areas  of  cross-section 
being  known. 

ANS.  Let  the  coefficient  of  elasticity  E=  12500 
tons  per  square  inch,  and  the  areas  of  cross-section 
of  the  members  as  given  in  the  following  table. 


f 
5  tons 


10  tons 


CHAP.  IV.] 


REMARKS   ON   THE  PRECEDING   EXAMPLE. 


517 


Member. 
af,  

Length 
/ 
in  inches. 

60 

in  tons 
per  sq.  in. 

I25OO 
12500 
I25OO 
12500 
12500 
12500 
12500 
12500 
12500 

5 
in  tons. 

—     7-9545 

—     4-7727 
-  10.7954 

+     6.4772 
+     6.4772 
+    4-7727 
+     6.3636 
-f-  IO.OOOO 

+ 

+ 

+ 

-f 

+ 

n  tons. 
3.4091 
2.0454 
9.0909 

5-4545 
5-4545 
2-0454 

2.7272 

IO.OOOO 

Cross-section 
in  sq.  ins. 

1.85 
I.OO 

1.85 

1-5 
1-5 
i-5 
2.0 
2.0 

/ 
3 

9 

+  14-6582 
+    9.7621 
+  53.0436 
+  23.5532 
+  23.5532 
+    6.5080 

+    8.6777 
+  50.0000 
+  12.7067 

A 

in  inches. 

\    0.0372 
!•    0.0199 

J 

V  0.0308 

be 

60 

cd            

60 

de 

16 

ef 

60 

31250 
3 

9 

l6 

bf.  .  . 

48 

31250 
6 

48 

15265 
6 

,f 

76  84 

15265 
76.84 

125000 

A  =    0.0879  i°-  at  e 

We  take  for  the  value  of/  the  load  of  10  tons  at  e  and  find  the  stresses  s  in  every  member  due  to  this 
single  load.  We  also  find  the  stresses  S  in  every  member  due  to  the  actual  loading.  In  the  product  Ss  these 
stresses  must  be  taken  with  their  proper  sign.  Thus  if  .yis  compression  or  minus  and  S'ls  also  compression  or 
minus,  the  product  Ss  is  positive.  If  one  is  tension  or  positive  and  the  other  compression  or  negative,  the 
product  is  negative.  If  the  signs  of  S  and  s  are  carefully  observed,  the  signs  of  the  products  Ss  will  thus  take 
care  of  themselves. 

If  we  take  E  in  pounds  or  tons  per  square  inch,  S,  s  and  p  must  be  taken  in  pounds  or  tons,  and  /  in 
inches  and  a  in  square  inches. 

We  have  taken  /  at  e  equal  to  10  tons,  or  the  load  actually  acting  there.  But  if  there  were  no  load 
acting  there,  we  could  still  assume/  =  10  tons  or  i  ton  or  any  convenient  amount,  and  proceed  as  before. 

The  stresses  S  due  to  actual  loading  are,  strictly  speaking,  affected  by  the  change  of  shape.  This  can, 
however,  be  disregarded  without  perceptible  error,  as  the  deflection  in  all  practical  cases  is  very  small  com- 
pared to  the  span. 

Remarks  on  the  Preceding  Example. — In  our  example  we  assume  E  as  constant  for  all 
-members.  We  also  assume  that  every  member  has  its  exact  length  and  area  of  cross-section, 
that  all  pins  fit  perfectly  tight,  and  all  adjustable  members,  if  any,  are  accurately  adjusted. 

A  truss  after  erection  may  then  be  tested  by  calculating  the  deflection  at  the  centre  for 
a  given  loading  and  comparing  with  the  actual  observed  deflection  due  to  this  loading. 

A  good  agreement  is  thus  a  test  of  the  close  fit  of  all  pins,  of  the  proper  adjustment  of 
all  adjustable  members,  of  the  agreement  of  the  lengths  and  cross-sections  of  members  with 
those  called  for  by  the  design,  of  the  constant  value  of  E  and  its  proper  assumption  as  to 
magnitude,  and  finally  of  the  fact  that  the  elastic  limit  is  not  exceeded. 

It  is  evident,  however,  that  when  so  many  conditions  must  concur,  a  discrepancy  between 
the  observed  and  the  calculated  deflection  has  little  practical  significance.  The  last-mentioned 
fact,  that  the  elastic  limit  is  not  exceeded,  is  the  most  important,  and  this  is  proved,  not  by 
any  close  agreement  between  actual  and  calculated  deflections,  but  by  observing  whether  the 
deflection  is  constant  under  repeated  applications  of  the  same  loading  after  the  structure 
has  attained  its  permanent  set. 

Computations  of  deflection  are  then  of  little  value  as  a  means  of  testing  framed  structures, 
and  the  calculated  result  cannot  be  expected  to  agree  very  closely  with  the  actual  deflection. 

Principle  of  Least  Work. — We  have  seen,  page  311,  that  for  stable  equilibrium  of  a 
material  system  the  conditions  of  static  equilibrium  must  be  fulfilled,  and,  also  that  the 


STATICS   OF  ELASTIC  SOLIDS. 


[CHAP.  IV. 


potential  energy  is  a  minimum.  When  an  elastic  body  is  strained  by  external  forces  within 
the  elastic  limit,  the  work  done  by  the  external  forces  is  the  work  the  body  can  do  when 
released,  or  the  potential  energy  is  equal  to  the  work  of  the  external  forces. 

Hence  for  stable  equilibrium  of  an  elastic  body  the  work  of  the  external  forces  must  be 
the  least  possible  consistent  with  the  conditions  of  static  equilibrium.  This  is  called  the  prin- 
ciple of  least  work. 

As  an  illustration  of  the  application  of  this  principle,   suppose  a  rectangular  table  of 
L  length  L  and  breadth  b  to  have  four  legs,  all  of  equal  length  / 

and  uniform  cross-section  A,  one  at  each  corner. 

Let  a  load  Prest  on  the  table,  and  let  x  and  y  be  the  co- 
ordinates of  its  point  of  application. 

Let  Pl ,  P2,  P3,  Pt  be  the  loads  carried  by  the  legs.     We 
have  for  the  conditions  of  static  equilibrium 


P3L  +  P2L  -  Px  =  o, 
P,b  +  PJ>  -  Py  =  o. 


From  these  equations  we  obtain 


where  P^  ,  P2  and  P3  are  given  in  terms  of  P4.  But  P^  is  unknown.  We  have  four  unknown 
quantities  and  only  three  equations  of  condition.  We  need  another  equation  of  condition. 
This  is  furnished  by  the  principle  of  least  work. 

Thus  from  equation  (III)  we  have  for  the  work  of  compressing  the  legs,  assuming  the 
floor  and  table-top  to  be  rigid, 


=  [/>«  +  P*  +  P*  +  P*]. 

If  in  this  we  substitute  the  values  of  Pl  ,  P2,  P3,  we  have 

-  zT-  '4  -  zk  + 


We  thus  have  the  work  given  in  terms  of  Pt.  Now  Pt  must  have  such  a  value  that  the 
work  shall  be  a  minimum.  Therefore  putting  the  differential  of  the  work  relative  to  P4 
equal  to  zero,  we  have 


CHAP.  IV.]  PRINCIPLE  OF  LEAST  WORK.  519 

Hence 

P       Px       Py  P      Px       py 

^  =  4~2-Z+2*      and     P*=-4+2Z+2-f 

P-^_^0.^ 

2  ~   4          2b  ~^~  2U 

=  $P_r^_Py 
1       4      2£      2b' 

If  P  is  at  the  centre,  x  =  — ,  y  =  -  and 


P  —  P  =  P  —  P  —  -P 
f\  —  -*a  ~  *  a  —  *4.      v*« 

4 

If  P  is  at  the  middle  of  a  side  /,  x  —  —  ,  y  =  o  and 


If  P  is  over  leg  three,  x  =  Z,  ^  =  <£  and 

?!=--,  />,  =  +-,  />=+^/>,  />     -+£  ' 

4  ^4  ^4  4 

The  minus  sign  for  /^  shows  that  the  stress  is  reversed  and  leg  one  must  be  fastened 
down.  If  not,  it  is  lifted  off  the  floor  and  we  have  the  table  supported  on  three  legs  only. 

Example  —  Take  t/ie  same  table  with  a  fifth  leg  at  the  centre,  and  show  that  the  load  carried  by  this  leg  is 
always  -/'  no  matter  where  the  load  P  may  be  placed. 

Remarks  on  the  Preceding.  —  The  preceding  problem  of  the  four-leg  table  illustrates 
the  principle  of  least  work.  It  also  illustrates  much  more  —  it  furnishes  an  example  of  theory 
misapplied.  The  theory  is  sound,  and  the  results  are  therefore  correct  provided  the 
assumptions  are  realized  in  fact.  But  these  assumptions  are  not  realized  by  any  actual 
table.  For  instance,  we  have  assumed  the  floor  and  table-top  rigid,  every  leg  of  precisely 
equal  length  and  the  same  uniform  cross-section.  Such  a  table  is  an  ideal  which  has  no 
physical  existence.  The  theory  is  then  misapplied,  since  the  assumptions  do  not  correspond 
to  fact.  A  very  small  discrepancy  in  the  length  of  the  legs  alone  would  entirely  change 
the  results. 

The  reader,  then,  must  regard  the  problem  simply  as  an  illustration  of  the  principle  of 
least  work,  and  should  note  that  even  sound  principles  need  care  in  application,  and  that  for 
proper  application  the  assumptions  should  accord  with  reality,  otherwise  the  results  are 
worthless. 

Thus  in  the  present  case  the  legs  should  be  designed  for  three  only.  Then  if  any 
others  are  desired,  they  can  be  added  of  the  same  dimensions.  This  practical  solution  is  not 
only  simpler,  but  it  is  actually  more  accurate  and  scientific. 


FIG.  i. 
c 


B 


520  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  IV. 

I 

Redundant  Members. — The  principle  of  least  work  is  also  illustrated  by  the  calculation  of 
the  stresses  in  a  framed  structure  with  redundant  members.  The 
method  of  procedure  is  the  same  as  for  the  table  with  four  legs. 
Thus  take  the  simple  truss  shown  in  Fig.  I,  resting  on 
supports  at  a  and  b  with  a  load  P  at  the  centre.  Let  the 
inclined  braces  make  the  angle  6  =  45°  with  the  vertical,  so 
that  the  lengths  of  all  horizontal  and  vertical  members  are 
equal  to  the  height  h  of  the  truss.  The  length  of  the 
inclined  members  is  then  h  ^2,  and  tan  0  =  I,  sec  d  =  ^2. 
Let  the  stress  and  area  of  cross-section  of  aA,  aC,  ac,  Ac,  AC,  Cc  be  sl  and  al ,  st  and 

"*>S*    and    «•'    CtC"aS   lndi-  FIG.,.  FIG.  3. 

cated  in  Fig.  i.     This  truss  A 

consists  of  two  statically  de- 
terminate trusses,  as  shown  in 
Fig.  2  and  Fig.  3,  super- 
posed upon  each  other. 

Let  the  truss  of  Fig.  2  a 
carry  a  certain  fraction  <pP  of 
the  load,  and  the  truss  of  Fig. 
3  carry  the  rest,  or  (i  —  0)/>. 


CHOP 


Then  we  have  the  stresses  (page  409) 


=  -*¥ 


2  2    '  2 

=  -(i-0) /'tan  6  =  -  (i  -  0)/>, 


s6  =  />0. 


The  value  of  0  must  be  that  which  makes  the  work  a  minimum.  We  have  for  the 
work,  from  equation  (II),  since  the  stresses  and  areas  in  the  left  portion  of  Fig.  i  are  the 
same  as  the  right, 


or,  substituting  the  values  of  the  stresses  as  found, 


work  = 


, 


,_ 


2^  L          2«,  a2  za3  a4  a, 

If  we  differentiate  with  reference  to  0  and  put  the  value  of  -         —  =  o,  we  have  for 
the  value  of  0  which  makes  the  work  a  minimum 


2JK2_,    I 

~^~       as 


21/24 
a.         a. 


CHAP.  IV.]  NQ  ECONOMY  DUE    TO  REDUNDANT  MEMBERS.  521 

We  can  therefore  find  slt  s2  ,  etc.,  by  inserting  this  value  of  <£  in  the  equations  for  the 
stresses  already  given. 

It  will  be  noted  that  the  cross-section  alt  a2,  etc.,  must  be  known  for  each  member  in 
advance. 

If  the  cross-sections  are  all  equal,  we  have 


Evidently  the  same  remarks  apply  here  as  in  the  case  of  the  table  with  four  legs. 
Every  member  must  be  of  absolutely  true  length  and  of  the  exact  cross-section  assigned. 
Any  variation  from  ideal  conditions  invalidates  the  result.  As  such  ideal  conditions  do  not 
and  cannot  exist,  the  actual  stresses  in  any  given  case  will  not  agree  with  the  computed 
stresses. 

We  see,  then,  that  the  use  of  redundant  members  in  a  structure  not  only  makes  the 
calculation  of  stresses  very  involved  and  laborious,  but  also  that  the  results  obtained  hold 
only  for  an  ideal  structure  under  ideal  conditions  and  are  by  no  means  the  actual  stresses. 

No  Economy  due  to  Redundant  Members.  —  It  remains  to  inquire  whether  there  can  be 
any  compensating  advantages  in  economy  in  the  use  of  redundant  members  to  offset  the 
objections  already  noted. 

This  inquiry  is  directly  answered  by  the  preceding  article.  We  see  that  Fig.  I  is 
composed  of  two  statically  determinate  trusses,  Fig.  2  and  Fig.  3,  superposed  upon  each 
other.  Each  of  these  carries  its  own  proportion  of  the  loading.  It  is  also  evident  that  one 
of  these  trusses  is  more  economical  than  the  other.  The  combined  truss  of  Fig.  I  must 
therefore  have  an  economy  intermediate  between  the  two,  and  therefore  must  necessarily  be 
less  economical  than  one  of  the  two. 

The  same  holds  for  any  structure  with  redundant  members.  We  may  consider  it  as 
formed  by  the  superposition  cf  a  series  of  statically  determinate  trusses.  The  economy  of 
the  combination  must  be  less  than  the  economy  of  some  one  of  this  series.  Hence  any 
structure  with  redundant  members  is  less  economical  than  some  statically  determinate  struc- 
ture included  in  the  redundant  structure. 

There  is,  then,  no  gain  of  economy  by  the  use  of  redundant  members. 

Work  of  Bending.  —  As  before,  let  5^  be  the  unit  stress  in  the  most  remote  fibre  of  any 
cross-section  at  a  distance  v  from  the  neutral  axis  of  that  cross-section  ;  then  the  unit  stress 
5  in  any  fibre  at  a  distance  v'  is,  from  equation  (i),  page  494, 

S=^-S/.     hence     5/  =  7-,S  .........     (i) 

Also,  if  is  the  bending  moment,  or  moment  of  all  the  external  forces  on  the  left  of 
the  cross-section  (page  494),  we  have,  from  (II),  page  497, 


v 

Inserting  the  value  of  S^from  (i),  we  have  for  the  unit  stress 


5«2  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  IV. 

If  ds  is  the  distance  measured  along  the  neutral  axis  between  two  consecutive  cross- 
sections,  and  a  the  area  of  the  cross-section  of  the  fibre,  we  have,  from  (I),  page  477,  putting 
Sa  tor  F,  a  lor  A,  and  ds  for  /,  for  the  strain  of  any  fibre 

SJs 

*~    £' 


or,  inserting  the  value  of  5"  from  (2), 

_  Mv'ds 
A~     El   ' 


(3) 


Now  the  work  on  the  fibre  is  half  the  product  of  the  stress  and  strain  (page  515),  or, 
from  (2)  and  (3), 


A. 

*"  '   2 

and  since  2ai/*  =  /,  the  work  on  all  the  fibres  of  the  cross-section  is 


2EI' 
For  the  total  work,  then,  for  all  the  cross-sections  we  have 


"•=/ 


(IV) 


Equation  (IV)  is  general  whatever  the  shape  of  the  beam  before  flexure.      If  the  beam 
is  straight  before  flexure,  we  have  dx  in  place  of  dst  and  /  in  place  of  s.      Hence 


w=  /    -^ 


Beams  Fixed  Horizontally  at  the  Ends. — By  means  of  equation  (IV)  and  the  principle 
of  least  work,  we  can  solve  all  cases  of  beams  fixed  horizontally  at  the  ends. 

CASE   i.  BEAM  OF  UNIFORM  CROSS-SECTION  FIXED   HORIZONTALLY  AT  ONE  END 
AND   SUPPORTED  AT  THE  OTHER — CONCENTRATED   LOAD — Reaction. — Let   Rl   be  the 

reaction  at  the  supported  end  A   on   left,  and  let  the  load 
R,  |R*  P  be  at  a  distance  kl  from  the  supported^  end  and   (i—  k)l 

4 f  —          — >|  ||  from  the  fixed  end  B.      Then  for  any  point  of  the  neutral 

axis  distant  x  from  the  supported  end  we  have  the  bending 
moment, 

when  x  <  kl,          Mx  —  —  R^x ; 
R(  when  x  >  kl,          Mx  =   -  R,x  +  P(x  —  kl}. 

The  beam  may  be  fixed  at  the  end  B  either  by  letting 


it  into  a  wall  or  by  continuing  it  over  the  support  at  B 
and  applying  a  load  P .  In  cither  case  the  tangent  to  the 
deflected  beam  must  be  Jiorizontal  at  the  fixed  end  />',  and 

the  value  of  R^  must  be  that  which  makes  the  work  of  bending  from  A  to  B  a  minimum. 

The  work  on  the  remainder  of  the  beam,  if  any,  can  be  neglected. 


CHAP-   IV]  BEAMS  FIXED  HORIZONTALLY  AT   THE  ENDS. 

From  (IV),  then,  we  have  for  the  work  of  bending 


523 


If  we  differentiate   with   respect   to  Rl  and   put      iwor  ')  _  Q>we  havg  for  the  value  of 

dRl 
which  makes  the  work  of  bending  from  A  to  B  a  minimum,  since  E  and  /are  constant, 


=  o. 


Performing  the  integrations,  we  obtain 


If  the  load  is  at  the  centre,  we  have  k  =  —  and  R,  =  —P. 

2  16 

Moment.  —  The  moment  M2  at  the  fixed  end  is 


and  is  positive.      Hence  RJ  is  less  than  P(i  —  k)L      The  moment  at  the  load  is  —  RJsl  and' 
is  negative.      If  we  subtract  this  from  M2,  we  have 


which  is  positive  since  Pis  greater  than  Rr  The 
moment  at  the  fixed  end  is  therefore  the  greatest, 
and  the  moment  at  any  point  is  given  to  scale  by 
the  ordinates  to  two  straight  lines  as  shown  in 
the  figure. 

Point  of  Inflection.  —  We  see  that  at  the  point  C  in  the  figure  the  bending  moment  is 
zero.  This  point,  as  we  shall  see  (page  547),  is  the  point  at  which  the  curvature  of  the 
deflected  beam  changes  from  concave  to  convex.  We  can  easily  find  its  position.  Thus, 
from  the  figure,  if  xl  is  the  distance  of  C  from  the  supported  end  A,  we  have 


or     x,  = 


Substituting  the  values  of  M2  and  R^  already  found, 

2/ 


If  the  load  is  at  the  centre,  k  =  —  and  x^=.  —  /. 

Breaking  Load.  —  Since,  as  we  have  proved,  the  bending  moment  M2  at  the  fixed  end  is 
greatest,  we  have,  from  equation  (4),  page  499, 


524  ST/1T1CS  OF  ELASTIC  SOLIDS.  [CHAP.  IV. 

or,  inserting  the  value  of  Mt,  we  have  for  the  breaking  weight 

2RI 


P  = 


vlk(\  - 


For  the  load  at  the  centre  k  ==  -  and  P  =  -  -,  or  -  as  much  as  for  the  same  beam 

2  ->>vl          3 

supported  at  both  ends  (page  500). 

The  moment  M2is  a  maximum  for  k  =  A  /-  =  0.5774,  that  is  when  the  load  -Pis  distant 

O.5774/  from  the  supported  end. 

For  this  position  we  have  the  least  breaking  load 


vl 

CASE  2.  BEAM  OF  UNIFORM  CROSS-SECTION  FIXED  HORIZONTALLY  AT  ONE  END 
AND  SUPPORTED  AT  THE  OTHER  —  UNIFORM  LOAD.  —  Reaction.  —  In  the  preceding  case  we 
have  found  for  concentrated  load 


If  for  P  we  put  wdx,  and  for  k  we  put  -j,  we  have  in  the  present  case 


Performing  the  integration,  we  have 

*,=!«/. 

Moment.  —  The  moment  M2  at  the  fixed  end  is  then 


wl*  wl* 

"~~          '' 


The  moment  at  any  point  distant  x  from  the  supported  end  is  then 


This  is  the  equation  of  a  parabola  whose  vertex 

is  at  -/  from  the  supported  end  as  shown  in  the  figure 
o 


the  moment  at  this  point  being  —  z—  5. 

1  2o 

Point  of  Inflection.  —  Here  again  the  moment  is  zero  and  we  have  a  point  of  inflection  6 
at  a  distance  x1  from  the  supported  end  given  by 


CHAP.  IV.] 


BEAMS  FIXED  HORIZONTALLY  AT    THE  ENDS. 


525 


Breaking  Load. — Since  the  moment  M2  at  the  fixed  end  is  the  greatest,  we  have,  from 
equation  (4),  page  499, 


RI 

=>     °r 


wl*      RI 


Hence  the  breaking  load  is 


8RI 


or -as  much  as  for  the  same  load  at  the  centre  and  just  the  same  as  for  the  same  beam 

supported  at  the  ends  (page  500). 

CASE  3.  BEAM  OF  UNIFORM  CROSS-SECTION  FIXED  HORIZONTALLY  AT  BOTH  ENDS — 
CONCENTRATED  LOAD. — In  this  case  we  have  the 
bending  moment  for  * 


x<kl, 


M,, 


and  for 


kly 


Mx=  M  — 


+  P(x  — 


AA 


The  beam  may  be  fixed  at  the  ends  either  by 
letting  it  into  a  wall  at  each  end,  or  by  having  it 
project  beyond  the  supports  and  applying  loads  Pr t 
P"  at  the  ends.  In  either  case  the  tangent  to  the 
deflected  beam  must  be  horizontal  at  the  ends,  and 
the  values  of  R^_  and  Ml  must  be  such  as  make  the 
work  of  bending  from  A  to  B  a  minimum.  The  work  on  the  remainder  of  the  beam,  if  any, 
can  be  neglected. 

From  (IV),  then,  we  have  for  the  work  of  bending 

/dx          Cl  dx 

\M,  —  R,x~]z  —=-=  4-   /    [M.  —  R,x-}-  P(x  —  £/)]2  — r— . 
2EI       J^  2EI 

Differentiating  with  respect  to  R^  and  Ml  and  putting  — -^ —   =  o  and 

1 

we  have  for  the  values  of  Rl  and  Ml  which  make  the  work  a  minimum 

/*/  /v 

(M.  —  R^x\dx  +  /    P(x  —  kl}dx  =  o, 
Jki 

r  r 

I   (-M,x  +  R,x*}dx  +  \    -P(x- 
J0  Jkt 


dR 


=  °' 


kl}x  dx  =  o. 


Perorfming  the  integrations,  we  obtain 


-  3^  +  2^7-^(2  -  3^+  /P)  =  o. 


526  STATICS  OF  ELASTIC  SOLIDS. 

From  these  two  equations  we  have 

Je,  =  P(i  -  k)\i  +  2k),         MI  =  Plk(i  -  Kf. 


[CHAP.  1\-. 


If  the  load  is  at  the  centre,  k  —  —  and 


The  reaction  at  the  right  end  is 

R^=P-Rl  =  Pk(i  -  k)(i  +  2k\ 

and  the  moment  M2  on  the  left  of  the  right  end  is 

MI  =  Ml  -./?,/  +  Pl(i-  k)  =  PlP(i  -  k). 


The  moment  at  any  point  is  given  to  scale  by  the 
M2  ordinates  to  two  straight  lines,  as  shown  in  the  figure,  the 
3  moment  at  the  load  being 


Ml  -  RJk  =  - 


^P  The  greatest  moment  will  then  beat  the  nearest  fixed 

end  to  the  load. 

Points  of  Inflection. — The  moment  is  zero,  and  we  have  a  point  of  inflection  at  Cl  and 
C3.     If  •*!  is  the  distance  of  Cl  from  the  end  A,  we  have  at  once,  from  the  figure, 


kl 


For  the  distance  xz  of  C2  from  B  we  have 


or     *„  = 


3  -  2 


If  the  load  is  at  the  centre,  k  =  -and  xl  =  x2  =  -. 

Breaking  Load. — The  greatest  moment  is  at  the  end  nearest  to  the  load.      Let  this  be 
the  left  end.     Then,  from  equation  (4),  page  499, 


,=,      or     P 


RI 


vlk(i  - 


T  ft  A*  T 

For  the  load  at  the  centre  k  =  -  and  P  =   — j-,  or  twice  as  much  as  the  same  beam 
supported  at  the  ends.     The  moment  M^  is  a  maximum  for  k  =  — .      That  is,   the  greatest 


CHAP.  IV.]  BEAMS  FIXED  HORIZONTALLY  AT  THE  ENDS.  527 

moment  at  the  nearest  end  occurs  when  the  load  is  distant  -/  from  that  end.      The  value  of 

3 

this  greatest  moment  is  Ml  =  —  .      Hence  the  least  breaking  weight  is  given  by 

4«_*£  p_ 

27  ~   v  '  ~  4»/    ' 

27 
or  —?  as  great  as  for  the  same  beam  supported  at  the  ends. 

CASE  4.  BEAM  OF  UNIFORM  CROSS-SECTIO,N  FIXED  HORIZONTALLY  AT  BOTH  ENDS 
—  UNIFORM  LOAD.  —  Reaction.  —  In  the  preceding  case  we  have  found  for  concentrated  load 


If.  for  P  we  put  wdxy  and  for  k  we  put    ,  we  have  in  the  present  case 


Performing  the  integration,  we  have 

_«,/ 

^  =  T- 

Moment.  —  We  have  found  in  the  preceding  case 
MI  =  Plk(i  -  k)\ 

If  for  P  and  k  we  put  wax  and  j,  we  have 


C        (X        2X* 

Mi=     w/b  ~  ~P 

J0        * W  '-> 

Performing  the  integration,  we  have 


The  moment  Mt  on  the  left  of  the  right  end  is  the  same. 
The  moment  at  any  point  distant  x  from  the  left  end  is 


.. 
Mx  =  M,  -  R,x  +  —  =  -^-^V~ 


wx*      wl*      vox 
—  =  -^-^ 

This  is  the  equation  of  a  parabola  whose  vertex  is 


at  '/  as  shown  in  the  figure,  the  moment  at  this  point  being  -  ~.     The  moment  at  the  fixed 
,?nds  is  the  £reatest. 


528  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  IV. 

Breaking  Load.  —  Since  the  moment  is  greatest  at  the  fixed  end,  we  have,  from  equation 
(4),  page  499, 

wl*      RI  \2RI 

—  =  —  ,    or     wl  —  --  . 

12  V  V 

Bending  and  Tension  Combined.  —  A  beam  may  sometimes  be  subjected  to  bending  and 
at  the  same  time  to  tension.  Thus,  for  instance,  a  lower  chord  panel  of  a  bridge  truss  may 
be  in  tension  and  at  the  same  time  sustain  loads  applied  by  means  of  cross-ties  between  the 
panel  points. 

In  such  a  case  let  M  be  the  maximum   bending  moment,  y  the  deflection,  Tthe  tensile 

stress  and  A  the  area  of  cross-section,  so  that  —  is  the  tensile  unit  stress.     Then  we  have  the 

A 

maximum  moment 


and  from  equation  (II),  page  497,  the  unit  stress  in  the  outer  tensile  fibre  is 

T      (M+  Ty]v 
A^          I 

If  Sw  is  the  working  unit  stress  adopted,  we  have  then 

T  ,  (M+Ty)v 
•^  -  ^  -t  f 

The  neutral  axis  is  now  no  longer  at  the  centre  of  mass  of  the  cross-section,  and  a  strict 
solution  leads  to  results  of  great  complexity.  If,  however,  we  disregard  the  small  deflection 
y,  we  have  in  all  practical  cases 


Sv  =  -:  -\ — p ,     hence     M  = 


If  we  put  for  /its  value  Ax2,  where  K  is  the  radius  of  gyration   of  the  cross-section 
(page  32),  we  have 


From  (i)  we  can  find  in  any  case  the  load  for  a  given  Sw,  T  and  cross-section  A,  and 
from  (2)  the  cross-section  for  a  given  load. 

Bending  and  Compression  Combined.  —  This  case  is  just  the  same  as  the  preceding, 
except  that  we  must  put  the  compressive  stress  C  in  place  of  T  and  take  Sw  the  working 
stress  for  compression.  If  flexure  is  to  be  apprehended,  we  must  take  Su,  as  given  on 
page  569. 

Examples.—  (i)  Find  the  area  of  cross-section  for  a  square  beam  of  ra  ft.  span  -which  sustains  a  load  of 
joo  pounds  at  the  centre  and  has  at  the  same  time  a  direct  longitudinal  tension  of  2000  pounds,  the  working 
stress  being  taken  at  1000  pounds  per  square  inch.  . 

ANS.  We  have  Sw  =  1000,  T=  2000,  v  =  —  ,  K*  =  —  ,  A  =  d*,  M  =  I5ox6x  12.     Hence,  from  (2), 
A=d*  =  ¥^  +  2,     or    d  =  4.  1  8  inches. 


CHAP.  IV.]  EXAMPLES.  529 

(2)  'Find  the  area  of  cross-section  for  a  square  beam  of  12  ft.  span  which  sustains  a  load  of  50  pounds  per 
foot  uniformly  distributed  and  has  at  the  same  time  a  direct  longitudinal  tension  of  2000  pounds,  the  working 
unit  stress  being  taken  at  1000  pounds  per  square  inch. 

ANS.  We  have  5W  =  1000,  T  =  2000,  v  =  d-,  *  =  £.  A  =  d\  M  =  ^  =  5<>x  12x12x12  x»     Hence> 
from  (2), 

A  =  </*  =  324+2,    or    d—  4.18  inches. 
5* 

(3)  A  rectangular  iron  beam  12  feet  long  and  2  inches  wide  has  a  longitudinal  tension  of  20000  pounds 
and  supports  a  load  of  5000  pounds  at  the  centre.     Find  the  depth  in  order  that  the  unit  stress  shall  not  exceed 
10000  pounds  per  square  inch. 

ANS.  We  have  Sw  =  10000,  T  =  20000,  v  =  ^,  *•'  =  ^,  A  =  zd,  M  =  5°^°_>l6x^     Hence,  from  (2), 
^4  =  2^=^  +  2,    or    d  =  7.86  inches. 


CHAPTER   V. 

DEFLECTION   OF   BEAMS. 

Deflection  of  a  Beam. — Let  ds  be  the  distance  measured  along  the  neutral  axis  between 
any  two  consecutive  cross-sections,  and  v'  the  distance  of  any  fibre  from  the  neutral  axis ; 
then  the  strain  A  of  the  fibre  is,  from  equation  (3),  page  522, 

Mv'ds 


where  M  is  the  bending  moment  of  all  the^external  forces  on  the  left. 

Take  the  origin  at  the  left  end  A,  and  let  x,  y  be  the 
co-ordinates  of  any  point  Pr  At  this  point  let  a  force  F 
be  supposed  to  act,  and  let  its  moment  relative  to  any 
point  Pz  on  the  right  of  Pl  given  by  the  co-ordinates  ~x,  ~y 
be  m. 

Then  from  equation  (2),   page  521,  the  stress  due  to 
I  this  force  is 

mad 


The  work  of  this  force  on  any  fibre  of  the  cross-section  at  P2  is  then  half  the  product  of 
the  stress  and  strain  (page  515),  or 

mMa&ds 

On  all  the  fibres  of  the  cross-section,  since  2aiP  =  o,  we  have  then 

mMds 
work  =      Ej  , 

and  for  all  the  cross-sections  between  the  right  end  B  and  />,,  if  AB  =  s  and  APl  =  slt 

'mMds 


work 


=f 

*/  tt 


This  equation  is  general  whatever  the  shape  of  the  beam,  whether  straight  or  curved,  or 
the  direction  of  F. 

Let  F  be  vertical.     Then  its  moment  m  at  Pt  is 

m  =  F(x  —  x), 

530 


CHAP.  V.]  DEFLECTION  OF  A  BEAM.  S31 

and,  from  (i),  its  work  is 

CsF(x  -  x)Mds 
work  =     /    -^ =-t . 

./,,          2EI 

But  if  Ay  is  the  vertical  deflection  at  Plt  the  work  of  F  is  ^,  and  this  must  be  equal 

and  opposite  in  sign  to  the  work  done  against  the  fibre  stresses,  or 


FA^  _     C*F(x-x)Mds 
2     ~V,,  2EI 

Hence  the  vertical  deflection  at  any  point  Pl  is  given  by 
C*M(x  —  x)ds        C'Mxds 

A>=J,-'-ET-  -XTF 

Again,  let  Fbe  horizontal.     Then  its  moment  m  at  P^  is 


andj  from  (i),  its  work  is 

work 


-£: 


If  Ax  is  the  horizontal  deflection  at  Plt  we  have  then 

F(y-y)Mds 
2EI 


FA,  _  r* 

2  / 

J  *i 


Hence  the  horizontal  deflection  at  any  point  Pl  is  given  by 

Mds 


C'Myds  ,     '    r* 

*=-     -ir+y 

j  j,          j  ^  , 


rr 


Equations  (V)  and  (VI)  are  general  whatever  the  shape  of  the  beam,  whether  straight  or 

curved. 

For  a  straight  beam  y  =  o  and  the  horizontal  deflection  Ax  =  o.  The  vertical  deflection 
is  given  by  (V)  if  we  put  ds  =  dx,  sl  =  x  and  s  =  /.  Hence  for  a  straight  beam,  calling  the 
vertical  deflection  y,  we  have 

,=  r%£—*  f^f-- .'...'...  (vn) 

If  we  differentiate  this,  we  obtain 

MX  dx      MX  dx  Cl  M^x 


or 

vjftfr 


dy  __    C 

~fc~         I     El 

.     j  * 

If  we  differentiate  again,  we  have 


(VIII) 


53* 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  V. 


Equation  (VIII)  is  the  differential  equation  of  the  curve  of  deflection  for  a  stra ight  beam 
By  integrating  it  we  obtain  (2),  which  gives  the  inclination  at  any  point.  Another  integra- 
tion gives  (VII),  the  deflection  y  at  any  point. 

When  the  deflection  is  very  small  compared  to  the  length,  we  have,  from  the  Calculus, 

d*y       i 
~dx*~~~p' 

where  p  is  the  radius  of  curvature.     Hence  we  can  write 

El 


=  -  M 


(IX) 


Equations  (VIII)  and  (IX)  can  also  be  directly  deduced  as  follows: 


Let  ad,  aldl  and  a262  be  three  cross-sections  all  parallel 
before  bending,  the  distance  between  the  first  two  being  x, 
and  the  last  two  being  consecutive  at  the  distance  dx. 

After  bending  let  these  cross-sections  be  still  plane,  and 
intersect  at  some  point  C.  Let  ccfa  be  the  neutral  axis,  so 
that  cC  —  c^C  —  ctC  =  p.  Let  the  angle  aCa^  =  a  be  very 
small.  Then  the  angle  a^Ca2  =  da,  and  the  distance  ccl  =  s 
will  be  equal  to  x,  and  c^ct  =  ds,  to  dx,  approximately. 

Through  cl  draw  a'b'  parallel  to  a2bz.  Then  the  angle 
a^CjOt'  =  da. 

For  any  fibre  at  a  distance  v  from  the  neutral  axis,  then, 
the  strain  is 

A  =  vda. 

Since  the  original  length  is  dx,  if  the  area  of  cross-section 


is  a,  we  have  from    I),  page  477,  the  stress  of  the  fibre 

Eavda 


Ea\ 
dx 


dx 


The  moment  relative  to  cl  is  then 


EaiPda 
dx 


For  all  the  fibres  of  the  cross-section,  then,  since  2av*  =  /,  we  have  the  moment 

Elda 

dx   ' 

This  moment  must  be  equal  and  opposite  to  the  bending  moment  M.     Hence 


. 
dx 

Now  for  very  small  deflection  a  =  ~  and  hence  da—  ~.     Substituting,  we  have 

dx  dx 


which  is  equation  (VIII). 


CHAP.  V.]      BEAMS  FIXED  HORIZONTALLY  AT  ONE  END  AND  LOADED  AT   THE  OTHER. 
Again,  from  similar  triangles,  we  have 

vda  :  v  :  :  as  :  p, 

or,  since  ds  =  dx9 


533 


da 

dx 


Substituting  this,  we  have 


which  is  equation  (IX). 

Beam  Fixed  Horizontally  at  One  End  and  Loaded  at  the  Other.  _(*)  UNIFORM  CROSS- 
SECTION. — For  uniform  cross-section  /  is  constant.      Take  the  ori- 
gin at  the  free  end.     Then  for  any  point  of  the  neutral   axis  at  a 
distance  x  the  bending  moment  is 


We  have  then,  from  (VIII), 


Integrating, 


where  Cl  is  the  constant  of  integration. 

Since   the  beam   is   fixed  horizontally  at  the  right  end,  the  tangent  at  the  right  end  to 

dy  PP 

the  curve  of  deflection  is   horizontal,  and   hence  — -  =  o  when  x  =  /.      Hence  Cl  =  +  — . 

We  have  then 


Integrating  again, 


PPx 

— 


where  C2  is  the  constant  of  integration. 

Since   the   deflection   at  the  fixed   end  is  zero,  we  have  y  =  o  when  x  =  /,  and  hence 


We  have  then 


PPx      Px* 

-          6~ 


This  equation  gives  the  deflection  at  any  point.      The  deflection  is  evidently  greatest  at 
the  free  end.      Making,  then,  x  =  o,  we  have  the  maximum  deflection 

'      '       ^-ffr       '    ;    :    :   • 

The  minus  sign  shows  that  the  deflection  is  downwards. 


534  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.    V. 

If  the  cross-section  is  rectangular,  /  =  —  bd*  and  we  have 

4/Y« 


(b)  UNIFORM  STRENGTH.  —  For  uniform  strength  the  cross-section  varies  and  hence  / 
is  variable.      We  have  then  in  general  for  rectangular  cross-section  and  uniform  strength 

d*y  _  _Px_  i2Px 

d**~        El  Ebd*' 

where  b  and  d  are  variable. 

Constant  Depth.  —  If  the  depth  is  constant  and  always  equal  to  dl  ,  then,  as  we  have 

seen  (page  503),  b  =  jbl  ,  where  bl  is  the  breadth  at  the  fixed  end.      Hence  for  rectangular 

cross-section 

d*  12.PI 


As  we  have  seen  (page  532),  -^  =  —  i  where  p  is  the  radius  of  curvature.       Hence  for 

constant  depth  and  uniform  strength  the  radius  of  curvature  p  is  constant  and  the  curve  of 
deflection  is  a  circle. 

Integrating,  since  ^-  =  o  for  x  =  /, 

dy  \2Plx         I2/V2 

Tz-    ~'Eb^' 

Integrating  again,  since  y  =  o  for  x  —  /, 


_  . 

Ebji?        EbJ* 

For  the  deflection  J  at  the  free  end  x  =  o  and 

6/V8 


or  —  as  much  as  for  same  beam  of  uniform  cross-section. 

Constant  Breadth. — For  constant  breadth  bl  we  have  (page  504)  d*  =  jd*.      Hence 

d*y  _         I2/Y4/7 

Integrating,  since  -r-  =  o  for  x  =  /, 

~dx 
Integrating  again,  since  7=0  for  x  =  /, 

y=  — 


CHAP.  V.]      BEAMS  FIXED  HORIZONTALLY  AT  ONE  END  AND  LOADED  AT   THE   OTHER.  535 

For  the  deflection  A  at  the  free  end  x  =  o  and 


or  twice  as  much  as  for  same  beam  of  uniform  cross-section. 

For  similar  cross-sections  we  have  (page  504)  d*  =  *d*.      Hence  bd*  =    ^\**  and 


Integrating,  since  ~  =  o  for  x  =  /, 


Integrating  again,  since  ^  =  o  for  x  =  /, 

_  S*^1** 

-pV^  + 

For  the  deflection  J  at  the  free  end  *  =  o  and 

A-  , 

or  —  as  much  as  for  same  beam  of  uniform  cross-section. 

The  maximum  deflections  are  then  as  -,  2  and  -,  or  as  15,  20  and  18. 
If  we  call  the  volume  of  the  beam  of  constant  cross-section  V,  then  in  the  first  case  the 
volume  is  —Vt  in  the  second  case  —V,  in  the  third  case  —V,  or  the  volumes  are  as  15,  20  and 

1  8.      The  maximum  deflections,  then,  for  a  beam  of  uniform  strength  in  three  cases  are  as 
the  volumes. 

Beam  Fixed  Horizontally  at  One  End  and  Uniformly  Loaded.  —  (a)  UNIFORM  CROSS- 
SECTION.  —  If  w  is  the  load  per  unit  of  length,  we  have  for  any  point  of  the  neutral  axis  at  a 
distance  x  from  the  free  end,  the  bending  moment 

.   wx* 
J/  =  +—  . 

Hence,  from  (VIII), 

A 


dy 

Integrating,  since  —  =  o  for  x  =  /, 


dy  wx*      wts 

EIdx=     -~6+~^' 


S36  ST/tTlCS  OF  ELASTIC  SOLIDS.  [CHAP.  V. 

Integrating  again,  since  y  =  o  for  x  —  /, 

wx*      wl*x      wl* 

"iTH  ~6      r- 

The  deflection  A  at  the  end  is  for  x  =  o 


or  only  «  as  much  as  for  an  equal  load  at  the  end. 
o 

(£)    UNIFORM    STRENGTH.  —  We    have    then    in    general    for    uniform    strength    and 
rectangular  cross-section 


Constant  Depth.  —  For  constant  depth  and  uniform  strength,  as  we  have  seen  (page  505), 

X* 

b  —  j^bl  and  d  —  dr     Hence 

<Py_ 
dx* 

d*y       i 
As  we  have  seen  (page  532),  -j-^  =  -,  where  p  is  the  radius  of  curvature.      Hence   for 

constant  depth  and  uniform  strength  the  radius  of  curvature  p  is  constant  and  the  curve  of 
deflection  is  a  circle. 

Integrating,  since  -j-  =  O  for  x  =  /, 

dy  _        6wPx       6wl* 
dx  ~        ~~~ 


Integrating  again,  since  y  =  o  for  x  =  /, 


_ 

~  ~Eb~d?       ~Eb~d* 

The  deflection  J  at  the  end  is  then  for  x  =  o 


or  twice  as  much  as  for  a  beam  of  constant  cross-section. 

Constant  Breadth  —  If  the  breadth  is  constant,  we  have  (page  505)  d  —    d^  and./^  —  b}: 
hence 


CHAP-  V' 


APPLICATION   TO  METAL  SPRINGS. 


537 


Integrating  as  before, 


Integrating  again 


dy_ 
dx 


'• 


—  log 

' 


For  x  =  o  the  deflection  A  at  the  free  end  is 


.  _ 

~ 


or  eight  times  as  much  as  for  a  beam  of  constant  cross-section. 

For  similar  cross-sections  we  have  (page  505)  d*=  jg*t  b  =  dl  l/~.     Hence 


dx* 


Integrating,  since  -=-  =  o  for  x  =  /, 

•  dy_  _ 

' 


Integrating  again,  since  y  =  o  for  x  =  /, 


For  x  =  o  the  deflection  A  at  the  free  end 


A  =  — 


or  three  times  as  much  as  for  beam  of  constant  cross-section. 

Application  to  Metal  Springs. — The  most  common  examples  of  bodies   of  uniform 
strength  are  metal  springs,  such  as  dynamometer-springs  and  wagon-springs. 

In  the  figure  we  have  a  spring  dynamometer  made  of  two  parabolic  springs  AA  and 
BB,  united  at  their  ends  A  by  the  links  AB.     Such 
a  dynamometer  measures  the  force  P  applied  at  D    A( 
by  the   deflection  indicated  by  the   pointer  at  Z, 
which  is  equal  to  the  sum  of  the  deflection  of  the 
two  springs. 

Since  the  springs  are  of  uniform  breadth,  and 
the   depth   varies   as   the   ordinates  to  a  parabola, 

p 
they   are  (page    504)   beams  of  uniform   strength,   fixed  at  the  end  with  a  load  —  at   the 

free  end. 


538  ST/tTlCS  OF  ELASTIC  SOLIDS.  [CHAP.  V. 

For  such  beams  we  have  found  (page  535)  the  deflection 

8/V» 


p 
In  the  present  case,  then,  if  we  take  the  length  /=  A  A,  we  have  to  insert  —  for  P  and 

-  for  /,  and  we  have  for  the  deflection  of  each  spring  under  a  load  P 


The  total  deflection,  then,  measured  by  the  pointer  at  Z  is  twice  this,  or 


where  £,  and  dv  are  the  breadth  and  depth  at  the  centre  C  and  D.      Hence 


where  A  is  measured  by  the  pointer  at  Z. 

Example,  —  Let  a  steel-spring  dynamometer   such  as  described  have  a  length  7=3  ft.,  a  breadth  bi  = 
2  inches  and  depth  d\  =  I  inch.  • 

For  steel  the  limit  of  elasticity  Se  is  about  40000  pounds  per  square  inch  (page  476). 
We  have  then,  from  equation  (II),  page  497,  for  the  crippling  load 


PI 

In  the  present  case  M  =  —  .     Hence  the  crippling  load  is 

•  4SeS       4  x  40000  x  2  x  2 

f  =  ±-L-=*—  —  =  1480  pounds. 

vl  36  x  1  2 

The  capacity  of  the  instrument  is  about  1400  pounds,  and  /'/  should  not  be  used  to  measure  greater  forces. 
If  we  take  E  =  30000000  (page  478),  we  have  for  load  1400  pounds  the  deflection 

A  =    ™L-  =  HOC  x  36x36x36  =  ,  og  inches 
b\d\E          2x30000000 

The  graduation  should  not  extend,  then,  over  I  inch. 

Finally,  by  direct  experiment,  suppose  we  find  that  with  a  load  of  loco  pounds  the  deflection  J  observed 
is  0.8  inch.     Then  the  coefficient 

hdSE       P      looo 

-7i-=j  =  -^r  =  I25°- 

Hence  for  this  instrument  we  have  the  equation 

P  =  I250J, 

where  d  is  the  reading  of  the  pointer. 

If  instead  of  piarabolic  springs  we  have  springs  of  uniform  strength,  of  constant  depth, 
and  theiefore  '^.ngular  shape  on  top  (page  504),  we  have  found  for  such  beams  (page  534) 
the  ^tfleaio^ 

6/V3 


CHAP.  V.]  APPLICATION  TO  METAL  SPRINGS. 


539 


P  I 

In  the  present  case,  then,  putting  —    for  P  and  -  for  /,  we  have  for  the  deflection  of 


each  spring 


The  total  deflection  measured  by  the  pointer  would  be,  then, 


and  we  have 


or  one  third  greater  than  in  the  preceding  case. 

Wagon-springs  are  usually  formed  of  a  number  of  springs  laid  one  over  another. 

If  we  have  thus  a  compound  spring  composed  of  n  springs  of  constant  rectangular  cross- 
section,  laid  one  upon  another,  we  have  when  the  breadth,  depth  and  length  for  each  spring 
is  d,  d  and  /,  and  the  load  at  the  end  of  the  entire  spring  is  P,  from  page  534,  the  deflection 


nEbd*' 
For  the  crippling  load  we  have  (page  499) 

PI      SJ  „       nSebdz 

—  —  — '         or     ^  —      <zi    • 
n         v  61 

Hence,  also, 

25,/2  A      2SJ 

or      —  = 


The  ratio  -  measures  the  flexibility  of  the  spring. 


If  the  spring  is  composed  of  n  springs  of  uniform  strength  and  constant  depth,  and 
therefore  of  triangular  shape  on  top  (page  504),  as  shown  in  the 
figure,  the  deflection,  as  given  page  534,  is 


while  the  crippling  load  P  is  unchanged  and  given  by 

*SM? 
61 

Setz  A       Sel 

*  =  ±i;   or  T  =  M' 

or  the  flexibility  for  the  same  strength  is  |  as  much  as  for  uniform  cross-section.  Hence 
beams  of  uniform  strength  are  preferable  for  springs,  since  the  object  is  to  get  with  maximum 
strength  maximum  flexibility. 


540 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  V. 


If  the  unit  stress  in  the  outer  fibre  is  S/t  we  have  in  this  case  for  the  total  end  cross- 
section 


Hence  the  amount  of  material  is 


+  P 


+P 


A  common  form  of  spring  consists,  as  shown  in  the  figure  of  n  flat  superposed  bars,  each 
of  uniform  depth  d  and  breadth  b  and  different  lengths,  triangular  at  the  ends,  so  placed  that 

the  point  of  each  triangle  just  reaches 
the  base  line  of  the  one  below,  the  last 
bar,  AJ3,  being  triangular  only. 
J 


The  portion  AAl  is  a  beam  of  uni- 
form strength  (page  503)  which  bends 
in  a  circle  (page  534).  If  its  length  is 

-,  the  radius  of  curvature  is  (page  534) 
n 


-P 


•rP         -P 


P  = 


nEbd* 

121P' 


Every  other  triangular  portion,  A^AZ,  A2A3,  bends  in  a  circle  with  the  same  radius  of 
curvature  and  is  of  uniform  strength,  since  AA^  in  bending  exerts  a  pressure  -f-  P  at  Al ,  AVAZ 
the  same  pressure  at  A2,  etc. 

PI 

As  to   the  rectangular  portions,   we  have   from  Av  to  B  a  constant  moment  M  =  — '• 

n 

due  to  the  couple  -j-  P,  —P.      Hence  -=-  is  constant  for  each  rectangular  portion,  which  is 

therefore  also  of  uniform  strength  and  bends  in  a  circle  of  the  same  radius  of  curvature  as  the 
triangular  portions.     The  deflection  A  of  the  entire  spring  is  then 

6/Y8 


The  crippling  load  is,  as  before, 


Hence 


•V' 


A 
or 


The  flexibility  is  then  just  the  same  as  in  the  preceding  case  of  a  number  of  superposed 
bars  of  the  same  length  and  uniform  depth,  triangular  on  top. 


CHAP'  V'J  APPLICATION   TO  METAL  SPRINGS.  541 


The  end  cross-section  of  each  bar  in  the  present  case  is 


The  amount  of  material  in  the  first  bar  is  then 

6*V-/_ 

~*      ~; 


for  the  next  bar, 

-^f-4-/  - 

nS,d[_2n  ~*      ~  *  J  ' 

for  the  next, 

^[L  +  t    3/l. 

*  Viz*  ^      ~  £J  ' 
and  so  on  up  to  the  last,  which  is 

—("-  +  /       -1 
.       «VL2»  »  J" 

We  have  then  the  total  amount  of  material, 


But  I  -f  2  -f-  .  .  .  n  =  —  --  -.      Hence  the  total  amount  of  material  is 


v 

The  amount  of  material  is  then  just  the  same  as  in  the  preceding  case  of  a  number  of 
superposed  bars  of  the  same  length  and  uniform  depth,  triangular  on  top.  There  is,  then, 
no  gain  either  in  material  or  flexibility. 

It  is  not  necessary  to  make  the  ends  of  the  bars  triangular.  We  can  have  any  other 
form,  provided  the  radius  of  curvature  is  constant. 

Thus  in  the  general  expression 

EI_  _  El  _  Ebd* 

—  —      »  P—  -fif  —  I2px> 

if    we    make    the    breadth    constant    and    equal    to    ^,    and    the    depth    variable,    so    that 
d  =  di\/  —T-  "P  to  the  next  bar  and  then  uniform,  p  will  be  constant  still. 

Such  a  spring  is  shown  in  the  figure.      The      A 
crippling  load  and  flexibility  are  as  before,  the 
length  /  being  measured  from  the  ends  BD,  BD, 
and  not  from  the  centre. 


542  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  V. 

Beam  Supported  at  Both  Ends    Uniform  Cross-section — Concentrated  Load. — Let  the 

load  P  be  distant  from  the  left  end  a  distance  /•/,  where  k  is  any  given  fraction.      Then  the 

R/  •      Ri     distance  from  the  right  end  is  (i  —  >&)/,  and  the  reaction 

at  the  left  end  is  P(  i  —  K). 

Take  the  origin  at  the  left  end.     Then  for  any  value 
of  x  less  than  kl  we  have 

when  x<kl        M  =  -  I\\  -  K)x\ 
and  for  x  greater  than  kl  we  have 

when  x>kl         M  =   -  P(\  —  k)x  +  P(x  -  kl). 
Hence  from  (VIII)  we  have 

when  x<kl        Ef^  =  P(i  -  k)x ; 

when  x>kl         ^fj~5  =  ^l  ~  ®)x  ~~  P^x  ~  &fy 
Integrating,  we  have 

dy      P(i  —  k}x*  _  dy       P(i  —  k\x*      Px* 

EI-r  —  —  pC,      and       EI-j-  =  — h  Pklx 

dx  2  dx  2  2 

dy 

For  x  =  kl  these  two  values  of  -7-  are  equal,  and  hence 

dx 


2i  2      ' 

Inserting  this  value  of  C2  and  integrating  again,  we  have 

k**      Px*       Pklx* 


Ely  =       —-  -  +  Cvx  +  C,     and     Ely  = 


In  the  first  of  these  equations,  when  x  =  o,  y  =  o,  and  hence  C3  =  o.     For  x  =  kl  these 


two  equations  are  equal,  hence  £"4  =  — -= — .      For  x  =  o,  y  in  the  second  equation  is  zero, 
hence 

-  k)(2  -  k) 


and  therefore 

Pl*k(2  +  £*) 
C*~  ~6~ 

Substituting  these  constants,  we  have 

when  x<kl         y=-  ^"/^W-  PP-  ^); 


when^r>^/         y=-'     ^       ~ '  (2lx  -  &P  -  x*). 
brLl 


CHAP.  V.]  BEAM  SUPPORTED  AT  BOTH  ENDS.  543 

If  we  make  x  —  kl,  we  have  th6  deflection  at  the  load 


If  we  insert  the  values  of  C^  and  C2  in  the  equations  for  ~  and  put  -^  =  o,  we  have  for 
the  value  of  x  which  makes  the  deflection  a  maximum 


when  x<kl 


when  x>kl         x=l  — 


When  x  =  kl  in  these  equations  the  maximum  deflection  will  be  at  the  load  and  will  be 
the  greatest  possible.  Placing,  therefore,  x  =  kl,  we  obtain  from  both  these  equations  the 
condition 


That  is,  the  greatest  maximum  deflection  is  at  the  load  when  the  load  is  at  the  centre. 
For  any  other  position  of  the  load  the  maximum  deflection  is  on  the  right  of  the  load  when 

/£<  —  ,  and  on  the  left  of  the  load  when  k>  —  .       That  is,   the  maximum  deflection  is  always 

between  the  load  P  and  the  farthest  end. 

Inserting,  then,  these  values  of  x  in  the  values  for  y  when  x>kl,  or  when  x<kl,  we  have 
the  maximum  deflection  in  either  case, 

—  k}(2  — 


If  the  load  is  at  the  centre,  we  have  k  —  -,  and  the  equation  of  the  curve  of  deflection  is 


'=  - 


and  the  maximum  deflection  in  this  case  is 


A  -  — 


or  only  —  as  much  as  for  a  beam  of  same  length  fixed  at  one  end  and  loaded  at  the  other. 


544 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  V. 


Beam  Supported  at  Both  Ends— Uniform  Cross-section— Uniform  Load. For  a  load 

w  per  unit  of  length  the  bending  moment    at  any  point  of  the 


J  neutral  axis  distant  x  from  the  left  end  i 


Hence,  from  (VIII), 


cPy  wlx       wx* 

£L  I    ""7 Q*   —    •   """"    ' 

dx*  2  2 


dy  I 

Integrating,  since ^  =ofor;r  =  — ,  we  have 

PJ  dy  _  wlx*       wx*  __  wl* 
dx  ~      4  6  24  ' 

Integrating  again,  since  for  x  =  o,  y  =  o,  we  have 

_  Wlx*          WX*         wl*X 

Ely  =  ~rr~   ~  ~^~A         rr~ 


The  maximum  deflection  A  is  at  the  centre,  and  hence 


Beam  Supported  at  One  End  and  Fixed  Horizontally  at  the  Other— Uniform  Cross- 
section — Uniform  Load. — In  this  case  we  have  for  the  moment  at 
any  point  of  the  neutral  axis  distant  x  from  the  supported  end 


Hence,  from  (VIII), 


Integrating,  since  for  x  =  /,  -~-  =  o,  we  have 

dx 


dy 


wl* 


Integrating  again,  since  for  x  —  o,  y  =  o, 


_^          wx4  [_  Rfx      wl* 
:    6      "   24          2  6 


Since  for  x  —  /,  ^  =  o,  we  have 


^_^_V'      ?£=0,     or    *]=3 
6         24          2          6 


CHAP.  V.]    BEAMS  SUPPORTED  AT  ONE  END  AND  FIXED  HORIZONTALLY  AT  THE  OTHER.         545 

This  is  the  same  value  we  have  already  found  (page  524)  by  the  principle  of  least  work. 
Inserting  this  value  of  R^ ,  we  have 

jiy  _  $ivlx*      wx*  __  a//8 
dx         16          6~       48"' 


48 


wxt 

"24          48 


If  we  put  the  first  of  these  equations  equal  to  zero,  we  find  for  the  point  at  which  the 
deflection  is  a  maximum 


x  = 


10 


or     x-  0.42  1  5/. 


Inserting  this  value  of  x  in  the  second  equation,  we  have  for  the  maximum  deflection 

39+55  ^33    w/4 


/"\f  dv 

If  we  put  -p2~  0>  we   ^ave  f°r  tne  point  at  which  the  moment  is  zero,  or    — ^   changes 
sign,  that  is  for  the  point  P  of  inflection, 


—  =  o,     or    x  =  —  * 

2  4 


Beam  Supported  at  One  End  and  Fixed  Horizontally  at  the  Other— Uniform  Cross 
section — Concentrated  Load. — Let  the  load  P  be  distant  IR,  j 

from  the  supported  end  a  distance  kl,  where  k  is  any  given  W -f- 

fraction.     Then  the  distance  from  the  fixed  end  is  (i  —  k)l.    M/ 
Take  the  origin  at  the    supported  end.     Then   for   any 
value  of  x  less  than  kl  we  have 

for  x<kl        M  =  —  R^Xy 
and  for  x  greater  than  kl  we  have 

for  x>kl         M=-Rlx+P(x-  kl). 
Hence,  from  (VIII), 

for  x<kl         £f-j~=Rxt  iorx>kl        EIj^ 

CIX  l*X> 

Integrating  we  have 

EI-j-  =  —^—  -|-  C^      and     Efj-  =  — - 

dy 
For  x  =  kl  these  two  values  of  ~  are  equal,  and  hence 


STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  V. 

Inserting  this  value  of  £*2  and  integrating  again,  we  have 

6  6622  4* 

In  the  first  of  these  equations,  when  x  =  o,  y  =  o  and  hence  C±  =  O. 
For  x  =  kl  these  two  equations  are  equal,  hence  £"4  ==  —  —  . 

For  x  =  l,y  in  the  second  equation  is  zero,  hence 


6 


Also  for  4r  =  /,   —  =  o,  and  hence 
ax 


Hence 

d—  ^4=fl,     and    *1  =  f(I-Wz+*). 

This  is  the  same  value  for  R^  that  we  have  already  found  (page  526)  by  the  principle  of 
least  work. 


-f-  &} 

We  have  then  (7,  =  —  —  ,  and  all  the  constants  are  determined. 

4 

If  we  insert  the  values  of  C\  and  C2  in  the  equations  for  —  and  put  -j-  =  o,  we   have 
for  the  value  of  x  which  makes  the  deflection  a  maximum 

when  x  <  kl  x  —  I     ' 


when  x  >  kl 

When  x  =  kl  in  these  equations  the  maximum  deflection  will  be  at  the  load  and  will  be  the 
greatest  possible.    Putting,  therefore,  x  =  kl,  we  obtain  from  both  these  equations  the  condition 

k=   4/2  —  i  =  0.414213. 
That  is,  the  greatest  maximum   deflection  is  at  the  load  when  the   load  is  at   a  distance  of 

^2  —  i  =0.414213  of  the  span  from  the  supported  end.  For  any  other  position  of  the 
load  the  maximum  deflection  is  between  the  load  and  the  supported  end  when  k>  4/2  —  1, 
and  between  the  load  and  the  fixed  end  when  k  <  4/2  —  I . 

Inserting,  then,  these  values  of  x  in  the  values  of  j>,  we  have  for  the  maximum  deflection 
in  general 


P(\-KfkP      I      k 

* 


P(\  - 

when  k  <  4/2  —  i       A  =  - 


3  £7(3  - 


CHAP.  V.] 


BEAM  FIXED  HORIZONTALLY  AT  BOTH  ENDS. 


547 


Both  of  these  are  equal  and  have  their  greatest  value  when  k  =  i/J  —  I.     Inserting  this 
value  of  k,  we  have  for  the  greatest  maximum  deflection  at  the  load 


3^7 


or  only  about  -  as  much  as  for  beam  supported  at  the  ends. 
100 

If  the  load  is  at  the  centre  of  the  span,  k—    —  and 


,       -  _ 

and  since  k  >   4/2  —  I,  we  have  the  maximum  deflection  in  this  case 


when  k  •=•  — 


and  this  maximum  deflection  is  at  a  point  between  the  load  and  the  supported  end  given  by 


For  the  point  of  inflection,  if  we  put  —~  =  o,  we  have  the  distance  of  the  point  of 

inflection  from  the  supported  end 

2/ 


If  the  load  is  at  the  centre  of  the  span  this  becomes  —  /. 


Beam  Fixed  Horizontally  at  Both  Ends— Uniform   Cross-section — Uniform  Load. — 


wl 
For  a  load  wj)er  unit  of  length  the  reaction  is  —  at  each  end, 

and  the  bending  moment  at   any   point  of  the  neutral  axis 
distant  x  from  the  left  end  is 


wlx        wx* 
—  +  —. 


Hence,  from  (VIII), 


wlx       wx* 


Integrating,  since  f  =  o  for  x  =  o,  we  have 


548  STATICS   OF  ELASTIC  SOLIDS. 

Integrating  again,  since  for  x  =  o,  y  =  o,  we  have 


wlx*      wx* 

'~~+-' 


[CHAP.  V 


Since  for  x  =  -  we  have  also  -j-  =  o,  we  have 


,    , 

--  L  4-  —  £-  -      ^  =  o,     or 
2      "  16        48 


Inserting  this  value  of  Mv  ,  we  have 


12 


wPx      wlx* 


_ 
dx  ~          12   "       4         ~~6~' 


a/;*"4 

-. 
24 


24 


12 


If  we  put  the  first  of  these  equations  equal  to  zero,  we  find  for  the  point  at  which  the 
deflection  is  a  maximum  x  '—  -.     The  maximum  deflection  is  then  at  the  centre  and  given  by 


A  -  - 


or  only  one  fifth  as  much  as  for  beam  supported  at  the  ends. 
If  we  put  -7-3  =  o,  we  have  for  the  points  of  inflection 

x  =  --—=.     and      x  =  -  + 


21/3 


21/3 


=  o.2ii3/     and     x  =  o.  78877. 
Beam  Fixed  Horizontally  at  Both  Ends— Uniform  Cross-section— Concentrated  Load. 
— Let  the  load  P  be  distant  from  the  left  end  a  distance  kl,  where  k  is  any  given  fraction. 

Then  the  distance  from  the  right  end  is  (i  —  £)/. 

Take  the  origin  at  the  left  end.     Then  for  any  value  of 
x  less  than  kl  we  have 

for  x<kl        M  =  Mv  —  R^xt 
and  for  x  greater  than  kl  we  have 

for  x> kl         M  =  MI  -  R^x  +  P(x  —  kl). 
Hence,  from  (VIII), 


for  x<kl  E          =  -  M,  +  RI*\ 


for  x>kl 


lX  -  P(x  - 


<  V']  BEAM  FIXED  HORIZONTALLY  AT  BOTH  ENDS.  549 

Integrating,  we  have,  since  ~-  =  o  for  x  =  o 
ax 


For  x  =  M  these  two  values  of  -£_  are  equal,  and  hence 


Inserting  this  value  of  C2  and  integrating  again,  we  have,  since  y  —  o  for  x  =  o, 


For  x  =  £/  these  two  equations  are  equal,  hence 


For  x  =  /,  y  in  the  second  equation  is  zero,  hence 

Also  for  x  =  /,  ~  =  o.      Hence 

From  these  two  equations  we  obtain 

Ml  =  Plk(i  -  £)2,      and     R^  =  P( I  -  £)2(i  -f  2k\ 

This  is  the  same  value  for  Rl  that  we  have  already  found  (page  526)  by  the  principle  of 
least  work. 

dy 

If  we  put  the  values  of  ~-  —  o,  we  have  for  the  value  of  x  which  makes  the  deflection 

a  maximum 

when  x  <  kl  x  = 


when  x  >  kl 


3  -2k' 


When  x  =  kl  in  these  equations  the  maximum  deflection  will  be  at  the  load  and  will  be  tne 
greatest  possible.    Putting,  therefore,  x^kl,  we  obtain  from  both  these  equations  the  condition 


*.i. 


55°  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  V. 

That  is,  the  greatest  maximum  deflection  is  at  the  load  when  the  load  is  at  the  centre  of 
the  span.  For  any  other  position  of  the  load  the  maximum  deflection  is  between  the  load 
and  the  farthest  end  from  the  load. 

Inserting,  then,  these  values  of  x  in  the  values  of  y,  we  have  for  the  maximum  deflection 
in  general 

i  2/W(i  -  Kf 

>-  J=     - 

I 


2  3£/(3  _  2^- 

Both  of  these  are  equal  and  have  their  greatest  value  when  k  =  — .     Inserting  this  value 
of  k,  we  have  for  the  greatest  maximum  deflection  at  the  load 

I  7^/8 

when  k  =  -  A  =  - 


or  only  one  fourth  as  much  as  for  beam  supported  at  the  ends. 
For  the  load  at  the  centre 

PI  P 

M,  =  +^,     and    R^=  -. 

For  the  points  of  inflection,  if  we  put  —  ~  =  o,  we  have  for  the  distance  of  the  points 

tix  ' 

of  inflection  from  the  left  end 

kl  (2  -  K)l 

and     x  — 


l+2k  '     3-2/1'  ' 

If  the  load  is  at  the  centre  of  the  span,  k  =  --  and  these  values  of  x  become  -  /  and  -  /. 

2  44 

Examples.— (i)  A  rectangular  beam  of  wrought  iron  5  feet  long ,  3  inches  wide  and  3  inches  deep  is  deflected 
—  of  an  inch  by  a  load  of  3000  pounds  applied  at  the  centre.     Find  E. 
ANS.  We  have  (page  543) 

PI*  _        /Y«          />/» 

,    hence     E  = 


Inserting  numerical  values 

^  =  3000  x  60  x  6o_x_6oj<jos  20000000  pounds  per  square  inchf 
4  x  3  x  27 

provided  the  elastic  limit  has  not  been  exceeded. 

In  order  to  find  whether  this  is  the  case  we  have  (page  497)  for  the  unit  stress  in  the  outer  fibre 

Sf  =  ^=™L  =   3oo,>  x  60  x  3  x  .2      ]Qooo 

/  4/  4x2x3x27 

Since  the  elastic  limit  (page  476)  is  25000.  the  elastic  limit  is  not  exceeded. 

(2)  An  iron  rectangular  beam  whose  length  is  12  ft.,  breadth  l\  inches,  coefficient  of  elasticity  24  ooo  ooo,  has 

a  load  of  10000  pounds  at  the  centre.    Find  the  depth  in  order  that  the  deflection  may  be  -      of  the  length. 


PHAP  v  n 

EXAMPLES-FLEXURE. 
ANS.  We  have 

d*  =  -^L  =    I000°    X    144    X    144   X    144    X    2    X    480 

tfEA  4  x  3  x  24000000  x  144 —         =691.2  inches,  hence  d  =  8.8  inches, 

provided  the  limit  of  elasticity  is  not  exceeded. 

In  order  to  find  whether  this  is  the  case,  we  have  as  before 

C,  _  Plv  _  lOOOOX  144  X  4.4  X  12  X  2 

~  4T  -  ~T^7^8^8T8^8-  =  l8595  P°unds  P 
Since  the  elastic  limit  is  25000  pounds  per  square  inch,  the  elastic  limit  is  not  exceeded. 

(3)  Find  the  depth  of  a  rectangular  beam  so  loaded  at  the  centre  that  the  elongation  of  the  lowest  fibre 
shall  equal  ~^o  of  its  original  length. 

ANS.  We  have  A  =  _L,  and  (page  477)  A  =  ^.     We  also  have  (page  497)  Sf  =  ^.     Hence 


Since  v  —  —,f=  —bd*  and  M  —  — ,  we  have 


'2IOO/Y 


"T       Eb     ' 
provided  that  Sf  is  less  than  the  elastic  limit. 

(4)  Find  the  radius  of  curvature  at  the  middle  point  of  a  wooden  beam  when  the  load  at  middle  is  3000 
pounds,  the  length  to  feet,  breadth  4  inches,  depth  8  inches  and  E  is  i  ooo  ooo  pounds  per  square  inch. 

ANS.  We  have  (page  532) 

El       4£Y       4x1000000x4x8x8x8 

P  =  -irr  =  ^r  =  -  — -=  1896  inches, 

M         PI  1 2  x  3000  x  1 20 

provided  the  elastic  limit  is  not  exceeded.     To  find  whether  this  is  the  case  we  have 
c        Plv       3000x120x4x12 

5'  =  77  =! '  4x4x8x8x8    =  2I°9  pounds  per  sc*uare  inch- 

Since  the  elastic  limit  is  3000  pounds  per  square  inch  (page  476),  the  elastic  limit  is  not  exceeded. 

(5)  A  wrought-iron  13  inch  I  beam,  whose  moment  of  inertia  is  6os  in  inches,  has  a  length  of  30  feet  and 
E  =  24  ooo  ooo  pounds  per  square  inch. 

If  supported  at  the  ends  with  a  uniform  load  of 7 5  pounds  per  inch  of  length  over  the  first  to  feet,  find 
the  deflection  at  the  end  of  the  load. 

ANS.  Deflection  =0.23444  inch. 
Find  the  deflection  at  the  centre. 

ANS.  Deflection  =  0.24421  inch. 
Find  the  deflection  10  feet  from  the  unloaded  end. 

ANS.  Deflection  =  0.19537  inch. 
Where  is  the  point  of  greatest  deflection  and  what  is  the  greatest  deflection  ? 

ANS.  At  13.1676  feet.     Greatest  deflection  =  0.24847  inch. 
//  the  weight  of  the  beam  itself  is  5.57 3  pounds  per  inch  of  length,  find  the  deflection  at  the  centre. 

ANS.  Deflection  =  0.07349  inch. 
If  the  same  io-foot  load  is  moved  to  the  centre,  find  the  deflection  at  the  centre. 

ANS.    Deflection  =  0.50063  inch. 
If  the  uniform  load  of  73  pounds  per  inch  covers  the  whole  span,  find  the  deflection  at  the  centre. 

ANS.  Deflection  =0.98905  inch. 

If  the  same  beam  is  half  loaded  with  75  pounds  per  inch,  find  the  deflection  at  the  centre,  the  maximum 
deflection  and  the  point  at  which  the  deflection  is  a  maximum. 

ANS.   Deflection  =  0.494525  inch.      Max.  deflection  =  0.49855  inch  within  the  loaded  portion  at   14.48 
inches  from  centre. 

If  the  same  beam  has  three  weights  of  4500  pounds  each  placed  at  intervals  of  bo  inches  beginning  at  one 
end,  find  the  deflection  at  centre. 
ANS.   Deflection  =  0.6154  inch. 


552  S7V/7YCS  OF  ELASTIC  SOLIDS.  [CHAP.  V. 

If  tht  beam  is  fixed  horizontally  at  both  ends  and  loaded  uniformly  with  75  pounds  per  inch,  find  the 
deflection  at  10  feet  from  either  end  and  at  the  centre. 

ANS.  Deflection  =  0.1563  inch  ;  at  centre  =  o.  19781  inch. 

If  only  one  end  is  fixed,  the  other  supported,  find  the  dfflection  at  10  feet;  at  centre;  at  20  feet.  Find  ///<• 
maximum  deflection.  IV here  is  it? 

ANS.  At  10  feet  =  0.39074  inch  ;  at  centre  —  0.39563  inch  ;  at  20  feet  =  0.27352  inch.  Maximum  deflec- 
tion =  0.41018  inch  at  151.7524  inches  from  supported  end. 

If  the  beam  is  fixed  horizontally  at  both  ends,  with  a  load  of  27000  pounds  at  the  centre,  find  the  deflection 
at  the  quarter  points  and  at  the  centre.  Where  are  the  points  of  inflection  ? 

ANS.  At  quarter  points  =  0.19781  inch;  at  centre  =  0.39562  inch.  Point  of  inflection  at  90  inches  froqa 
each  end. 

If  only  the  right  end  is  fixed  and  the  other  supported,  and  the  load  of  27000  pounds  is  at  the  centre,  finil 
the  deflections  d  at  the  quarter  points  and  at  the  centre,  and  the  maximum  deflection. 

ANS.  At  the   quarter   points  0.5316  and  0.3091    inch;    at  centre   0.69234  inch;    maximum  deflection 

=  0.70732  inch  at  /y  —  from  supported  end.. 


CHAPTER   VI. 

SHEARING     STRESS. 
Shearing  Stress  in  Beams.— Let  A&  and  A2B2  be  two  consecutive  cross-sections,  so 


that  the  distance  C^C^  ,  between  them  is  ds,  and  let  Ml  be  the 
bending  moment  at  Cl  and  M2  the  bending  moment  at  Cr 

Then  by  equation  (II),  page  497,  the  unit  stress  S/  in 
any  fibre  at  a  distance  v'  is 


M?' 


and  the  unit  stress  S,  in  the  same  fibre  at  the  other  cross-      C, 
section  is 


The  difference 


v-. 


is  the  horizontal  unit  shear  at  this  fibre.     But  J/2  —  Ml  is  */J/,  and  if  Fis  the  shear,  dM  =  Vds. 
Hence  the  horizontal  unit  shear  at  this  fibre  is 

Vv'ds 


and  if  the  area  of  cross-section  of  the  fibre  is  a,  then  the  horizontal  shear  for  this  fibre  is 

Vds  .  av' 

For  the  total  horizontal  shear  at  the  neutral  axis  we  have  then 

Vds  .  2av' 

H  =     ~7 — • 

the  summation  extending  to  all  the  fibres  above  C. 

Let  £0be  the  breadth  of  cross-section  at  the  neutral  axis.    Then  b^ds  is  the  area  sheared, 
and  the  horizontal  unit  shear  at  the  neutral  axis  is 


553 


554  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  VI. 

Thus  for  a  rectangular  beam  of  breadth  b  and  depth  d  we  have  7  =  —  ,  2ai/  =  —  x  -  =  -g-  and  hence 


For  a  cylindrical  beam  of  radius  r,  I  —  —  and  2av  =  —  x  4-  =  ^-,  £„  =  ir,  and  hence 


Take  the  elementary  fibre  at  the  neutral  axis  of  length  ds  and  depth  dv'  and  area  of 

cross-section  a  —  b0dv'  '.      Let  Sv  be  the  vertical  unit  shear 
at  the  neutral  axis.      Then  the  vertical  shearing  force  is 
_  v     Sva  =  Svb0dv'  and  the  horizontal  shearing  force  is  Shb0ds. 
* 


We  have  then  for  equilibrium 

Skd0ds  X  dv'  =  Svb0dv'  X  ds. 
Hence 

V2av' 


That  is,  the  horizontal  and  vertical  unit  shear  at  the  neutral  axis  are  equal. 

Work  of  the  Shearing  Force  in  Beams.  —  We  have  from  equation  (I),  page  477,  for 
the  strain  due  to  shear 

_  Sjts  _ 

=  -= 


where  E,  is  the  coefficient  of  elasticity  for  shear. 

If  then  V  is  the  shear,  we  have  the  work  of  the  shear  between  two  consecutive  sections 

V\ 

2     = 
The  total  work  of  shear  is  then 


For  straight  beams  we  have  s  =  /,  and  dx  in  place  of  ds. 

Influence  of  the  Shearing  Force  upon  Deflection. —  If  we  divide  (X)  by  -,  we  have  for 
the  deflection  due  to  the  shear,  for  straight  beam, 

deflection  = 


and  this  d«  flection  should  be  added  to  the  deflection  due  to  bending,  as  already  found  in  the 
preceding  pages. 


CHAP.  VI]  DETERMINATION  OF  COEFFICIENT  OF  ELASTICITY  FOR  SHEAR.  555 

Thus  for  a  beam   of  constant  cross-section,  fixed  at  one  end,  with  a  load  P  at  the  free 
end,  we  have  already  found  the  deflection  due  to  bending  (page  533) 


where  E  is  the  coefficient  of  elasticity  for  tension  or  compression. 
For  the  deflection  due  to  shear  we  have,  since  V=P 


A  = 


where  Es  is  the  coefficient  of  elasticity  for  shear. 
The  total  deflection  is  then 


For  rectangular  cross-section  bn  =  b,  I  =•  — bdz,  2av  =  — — ,  and  hence 

12  8 


A  — 


AJP/ 

while  for  bending  only  we  should  have  A  =  £-75  • 
If  we  assume  the  ratio—  =  3,  we  have 


. 

Ebd*_        8/2 

/  A.P'3 

We  have  then  for-  =  10,  ^  —  i.oi  i  •^r-«.  We  see  at  once  tixati.  for  depth  small  compared 


to  the  length,  the  effect  of  the  shear  can  be  disregarded.  In  all  practical  cases  where  /  is  greater 
than  io</,  the  deflection  due  to  bending  only  is  then  sufficiently  accurate,  and  the  results 
already  obtained  can  be  taken  as  practically  correct. 

Determination  of  Coefficient   of   Elasticity  for  Shear.—  From   equation  (i)  we  have 


at  once 

PI 


} 


For  rectangular  cross-section  7=  ^fc/8  and  2av  =  —  ,  and  (3)  becomes 


f.\ 


From  (0  or  (4),  then,  we  can  determine  experimentally  the  coefficient  of  elasticity  for 
shear  E  ,  when  the  coefficient  of  elasticity  E  for  tension  or  compression  is  known,  by  mea 
uring  the  deflection  A  for  a  given  load  P  at  the  end  of  a  beam  of  known  length  /  and  known 


cross-section. 


556  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  VI. 

Influence  of  the  Shearing  Force  upon  Reaction.— For  all  cases  of  beams  fixed  at  one 
end  and  free  at  the  other,  or  simply  supported  at  both  ends,  the  reactions  are  entirely  deter- 
mined by  statical  conditions  and  are  therefore  independent  of  the  shearing  force.  But  for 
beams  fixed  at  both  ends  or  fixed  at  one  end  and  supported  at  the  other  the  shearing  force 
has  an  influence  on  the  reaction. 

Let  us  take,  for  instance,  a  beam  of  uniform  cross-section  fixed  horizontally  at  the  right 
end,  supported  at  the  left  end  and  uniformly  loaded. 

R/  I  The   work  due    to   bending  is    from    equation  (IV), 

— *—  — *|     page  522, 

I 

and  the  work  due  to  shear  is,   from  equation  (X),  page  554, 

flVldx2ai/ 


The  total  work  is  then 

C'M*dx         rV*dx2av' 
work  =     /  — -=j  +    / ,  g  ,     . 

Jo    2  El       Jo       2&0EJ 

In  the  present  case,  if  w  is  the  load  per  unit  of  length,  we  have 
and  the  value  of  R^  must  make  the  work  a  minimum.      Hence 


and 


rk)  A"  w*r\xdx 

T         =  J,  I   '   '  '  ^  J  £/ 


If  we  omit  the  last  term  and  perform  the  integration,  we  have  Rl  =  -wl,  just  as  already 

8 

found  on  page  524  for  bending  only. 

If,  however,  we  perform  the  integrations  as  given,  we  have 


' 


R  - 
1        8 


For  rectangular  cross-section  60  =  b,  2av'  =  -5-  and 

o 

,-,   _3^/      l  + 

"  '~ 


CHAP.  VI.]  MAXIMUM  INTERNAL  STRESSES  IN  A  BEAM.  557 

We  see  at   once  that  .for  -large,  or  length  great  compared  to  the  depth,  this  becomes 

RI  =  —~- .      Hence  in  all   practical   cases  where  /  is  large  compared  to  d,  the  effect  of  the 

shear  upon  the  reaction  can  be  neglected. 

Maximum  Internal  Stresses  in  a  Beam. — From  page  484  we  have  already  proved  that 
if  s  is  the  direct  unit  shear  and  t  the  direct  unit  tension,  the  maximum  combined  unit  shear  is 


and  the  angle  which  it  makes  with  /  is  given  by 


Also,  the  maximum  combined  unit  tension  is 


and  the  angle  which  it  makes  with  /,  /?  =  90  —  or,  where  a  is  given  by 


2S 

tan  2a=  —  — . 


Now  the  direct  unit  shear  in  a  beam  is,  from  page  553, 


S-~JT' 


and  the  direct  unit  tension  is,  from  page  497, 

Mv' 


We  have  then  for  a  beam  in  general 


~'  and 


a. 


where  a  is  given  by 


Let  us  take  a  beam  of  uniform  cross-section,  fixed  horizontally  at  one  end,  with  a  load  P 
at  the  free  end.       Then  #  =  **,/  =  ^,   V  =  -  P,  and,  taking  7  =  ,, 


55* 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.   VI. 


Hence  we  have  in  this  case,  for  the  maximum  unit  shear  at  any  point  given  by  x  zndy, 
the  origin  being  at  the  free  end  of  the  neutral  axis,  • 


and  its  angle  a  with  the  neutral  axis  is  given  by 


tan  2a  = 


XV 


For  the  maximum  unit  tension  at  any  point  we  have 


+ 


bd*         V     |_         M»        _  -r  L  ^TJ  ' 
and  its  angle  /?  with  the  neutral  axis  is  given  by  ft  —  90  —  a  where  a  is  given  by 


tan  2ot  —  — 


(0 


(2) 


(3) 


(4) 


At  the  neutral  axis  y  =  o,  and,  from  (i),  the  maximum  unit  shear  at  the  neutral  axis  is 
S,  =  -^j,  and  its  direction,  from  (2),  is  a  —  o.  At  the  upper  surface  y  =  -  and  S,  —  -r-^ , 
and  its  direction,  from  (2),  is  a  —  45°. 

The  maximum  unit  tension  at  the  neutral  axis  is,  from  (3),  when  7  =  0,  St  =  — 7-7,  and 

its    direction,   from  (4),   is  ^  =  45°.       At  the  upper  surface  y  =  -  and  St  =  j^,  and  its 

direction,  from  (4),  is  ft  =  o. 

We  see,  then,  that  equation  (II),  page  497, 

Mv 


gives  the  maximum  unit  stress  St  in  the  outer  fibre,  but  for  other  fibres  this  maximum  unit 

stress  St  is  greater  than  that  given  by  equation  (II) 
unless  the  shear  is  zero  or  is  disregarded.  We  have 
just  seen  that  the  shear  can  be  disregarded  in  practical 
cases  (pages  555  and  557). 

In  the  figure  the  full  lines  above  the  neutral  axis 
represent  the  direction  of  maximum  tension  St ,  and 
those  below  the  neutral  axis,  the  direction  of  maximum 

compression  Sc. 

The  dotted  lines  represent  the  direction  of  maximum  shear  S,. 


CHAPTER   VII. 


STRENGTH   OF   LONG  COLUMNS. 

The  Ideal  Column. — The  ideal  column  is  supposed  to  be  perfectly  homogeneous  so  that 
the  coefficient  of  elasticity  E  is  constant  for  every  portion,  to  have  a  uniform  cross-section 
A,  a  perfectly  straight  axis,  and  to  have  a  load  P  applied  exactly  in  that  axis. 

Such  an  ideal  column,  ideally  loaded,  has  no  tendency  to  bend  in  any  direction.  It  is 
simply  compressed  by  the  loading. 

Theory  of  the  Ideal  Column. — Suppose,  then,  an  ideal  column  whose  original  length  is  I 
to  be  compressed  by  the  load  Pin  its  axis.  The  new  length  /x  is,  from  equation  (I),  page 477, 


PI 


°r  - 


(i) 


T 


Now  suppose  this  compressed  column  of  length  /x  to 
be  bent  very  slightly  by  a  horizontal  force,  and  suppose 
the  column  does  not  spring  back  completely  when  the 
horizontal  force  is  removed,  but  takes  a  certain  position 
of  equilibrium  as  shown  in  the  figure. 

Let  x  and  y  be  the  co-ordinates  of  any  point  of  the 
elastic  curve  of  the  axis,  taking  the  free  end  of  the  axis 
as  origin.  Then  the  bending  moment  at  any  point  is 
M=Py. 

Let  dx  be  the  distance  cC  between  two  consecutive  cross-sections  ab  and  AB  before  the 

load  P  is  applied.      Then  when  the  load  P  is  applied 
b  p 

the  unit  stress  is—;-,  and   the   shortening  of  the  axis 
A 

>  A  =  cc-i  is,  by  equation  (I),  page  477, 

Pdx 

*-~AE' 

If  now  the  column  deflects  towards  the  left,  the  unit  stress  on  the  inner  compressed  fibre 
at  a  distance  v  from  the  axis  is,  from  equation  (II),  page  497, 

P    .    Mv 


and  hence  the  compression  ad  =  A/  of  that  fibre  is 

_Pdx   ,   Mvdx 

-zr  H    m  - 


559 


560  STATICS.  OF  ELASTIC  SOLIDS.  [CHAP.  VII. 

The  distance  a^d  is  then 

Mvdx 

Old  =  \  -&-• Tip. 

If  now  p  =  OC  is  the  radius  of  curvature  of  the  axis,  we  have  at  once,  from  the  figure, 

p  i  dx  —  A  : :  v  :  a^dt 
or 

"•^--JE]  •••"• 

P  / 

But,  from  (i),  i  —  -j-g  =  j.     Hence 

I  _        d*y 
Now,  from  Calculus,  ;;  —    ~  -£-%•      Hence 


(3) 


Comparing  with  equation  (VIII),  page  531,  we  see  that  when  we  take  /x  =  /  we  have 
bending  only  without  compression. 

If  we  multiply  both  sides  of  (2)  by  2dy  and  integrate,  we  obtain,  since  M=  Py, 


When  y~A  =  the  maximum  deflection,  ~-  —  o.      Hence  C  —      .      ,  and  we  have 


dx 

dy 


Integrating  again,  we  have 


When  y  =  o,  x  =  o.     Hence  C  =  o,  and  we  have 


CHAP.  V'll.] 


THEORY  OF  THE  IDEAL   COLUMN. 


56i 


(a)  COLUMN  FIXED  AT  ONE  END,  FREE  AT  THE  OTHER. — For  a  column  fixed  at  one 
end  and  free  at  the  other  we  have  from  (3),  when  x  =  ilt  y  =  J,  and  hence 


IP 


or,  since  Ia=  ,/4/c2,  where  K  is  the  radius  gyration  of  the  cross-section, 

P 


(b)  COLUMN  WITH  Two  PIN  ENDS  (Fig.  i). — In  this      FIG.  i.          FIG.  a .         FIG.  3. 
case  we  have   only  to  make,   in  (3),  y  =  A  when  x  =  —  • 
We  thus  obtain 


(c)   COLUMN    FIXED  AT  ONE    END,    PIN  AT    THE 
OTHER  (Fig.  2).  —  In  this  case  we  have,  in  (3),  y  =  A  when 


x  =  -  /r      We  thus  have 


COLUMN  FIXED  AT  BOTH  ENDS  (Fig.  3). — In  this  case  we  have,  in  (3),  y 


when  x  —  -  /lt  and  hence 


GENERAL  EQUATION. — All  these  equations  are  of  the  form 

P 


(4) 


where  we  have  for  n  the  values^,  \n,  |TT,  4r  for  one  fixed  and  one  free  end,  two  pin  ends, 

2          £          &          £ 

one  fixed  and  one  pin  end,  and  two  fixed  ends. 

p 
EQUATION  OF  THE  ELASTIC  CURVE. — Substituting  the  value  of  -^  from  (4)  in  (3),  we 

have  for  the  equation  of  the  elastic  curve 


and  hence 


A         '         "* 

=  A  sin  -j-  , 


dy      nA 

-r  =  ~r  cos 

dx        / 


(5) 
(6) 


5&*  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  Vlf. 

WORK  OF  P  DURING  BENDING.  —  During  the  direct  compression  of  the  column  from 

P  P*l 

the  length  /  to  /,  ,  the  work  done  is      (/  —  /J,  or,  from  (i),      A  c. 

2.  J  .  /  /: 

If  now,  when  the  column  is  pushed  slightly  to  one  side,  it  continues  to  bend  and 
assumes  a  deflection  A,  we  have,  from  equation  (6),  for  the  moment  M  at  any  point  of  the 
neutral  axis 

M=  Py  =  PA  sin^. 
'\ 

From  equation  (IV'),  page  522,  we  have  then  for  the  work  of  P  during  this  bending 


C'lM*dx         Cl* 
I      —  ^—  =    / 
J0      2hl       J0 


DJ     .       .  nx 

work  of  P  during  bending  =    I      —  ^—  =    /      —  —  -  sin2  -=-  •  dx  = 

2EI  / 


WORK  OF  P  NECESSARY  TO  PRODUCE  THE  DEFLECTION  A.  —  The  horizontal  compo- 

dy  Pdy    dy 

nent  of  P  at  any  point  of  the  neutral  axis  is  P~r-      Its  work  is  -y*-  •  —  ,   and  hence,  from 

(6),  we  have 

work  of  P  necessary  to  produce  the  deflection  ==   /      —  f-  =  —  cos*  —  .  <£r  =  -  —  . 

J9      2dx        J0        2l?  /,  4/t 

Deportment  of  the  Ideal  Column.  —  If  the  work  of  P  during  bending  is  greater  than  the- 
work  of  P  necessary  to  produce  the  deflection  A,  that  is,  if 


the  compressed  vertical  column  when  pushed  slightly  to  one  side  will  continue  to  bend  until 
the  deflection  A  is  reached.  The  compressed  vertical  column  is  then  in  unstable  equilib- 
rium, and  if  pushed  slightly  to  one  side  by  a  horizontal  force//,  will  not  return  to  its  original 
vertical  position  when  H  is  removed. 

If  the  work  of  P  during  bending  is  just  equal  to  the  work  of  P  necessary  to  produce  the 
deflection  A,  that  is,  if 


44    ' 

the  compressed  vertical  column  when  pushed  slightly  to  one  side  will  remain   in  its  new 
position.      The  compressed  vertical  column  is  then  in  indifferent  equilibrium. 

If  the  work  of  P  during  bending  is  less  than  the  work  of  P  necessary  to  produce  the 
deflection  A,  that  is,  if 


44' 

the  compressed  vertical  column  is  in  stable  equilibrium.      If  pushed  slightly  to  one  side  by  a 
horizontal  force  .//,  it  will  return  to  its  original  position  when  //is  removed. 

If  we  put  for  /  its  value  Ax2,  where  K  is  the  radius  of  gyration,  we  have   in  these 
three  cases 


If  then  the  ratio  -±  of  the  length  /t  to  the  radius  of  gyration  K  is  greater  than  A  /  —  ^—  , 


CHAP.  VII.]  DEPORTMENT  OF  THE  IDEAL   COLUMN,  563 

the  column  is  in  unstable  equilibrium,  and  if  pushed  slightly  to  one  side  will  not  return.     If 


-i  is  equal  to 


.    ln*EA 
V— ' 


the  column  is  in  indifferent  equilibrium,  and  if  pushed  slightly  to 


one  side  will  remain  where  placed.      If  -i  is  less  than  \j  — — ,  the  column  is  in  stable  equi- 

K  V  -L 


kt 


librium,  and  if  pushed  slightly  to  one  side  will  return  to  its  original  position. 

EXPERIMENTAL  VERIFICATION.— These  theoretic  conclusions  as  to  the  deportment  of 
the  ideal  column  can  be  verified  by  the  following  experiment.*  Let  a  thin  bar  of  wrought 
iron  AB  be  placed  with  one  end  on  a  spring  balance, 
and  apply  a  load  P  in  the  axis  by  means  of  a  hinged 
lever  AC  upon  which  weights  can  be  placed.  After  a 

P 
little  adjusting  it    will   be  found   that   for  —.   less   than 

the   column  will  not  deflect,  and  if  we  deflect  it 

by  applying  a  lateral  force   at  the  middle,  the  column 
will  straighten  when  this  force  is  removed. 

P  n2j£f^ 

If,  however,  ^  is  equal  to  — jj-    the    column    will 

not  straighten,  and  if  -  is  greater  than  this  by  a  very  small  amount,  for  the  least  disturbance 

it  will  be  bent  by  the  load  to  a  very  great  extent,  and  even  bent  almost  double  or  broken. 

P      1$ E* K 

CRIPPLING  LOAD — EULER'S  FORMULA. — We  see,  then,  that  -^  =  —^~  gives  the  crip- 
pling load  for  the  ideal  column.  For  a  load  less  than  this  for  slight  disturbance  the  column 
recovers. 

Now  from  equation  (i),  page  559,  we  have 


But  —  can  never  exceed  the  elastic  Hmit  Se  ,  and  hence  -^  is  always  a  small  fraction 
which  can  be  disregarded  with  respect  to  unity.     Thus  (pages  476  and  478)  for  wrought 

,  and  for  timber         . 


iron 


it  is^-,  for  steel  .^,  for  cast  iron 

We  have  then  practically  for  the  crippling  unit  stress 


(E) 


Formula  (E)  is  known  as  Ruler's  formula  for  long  struts.  As  we  see,  it  neglects  the 
change  of  length  due  to  direct  compression.  It  is  usually  deduced  directly  by  writing  in 
place  of  equation  (2),  page  560,  the  equation 


and  then  integrating  twice. 


*  Given  by  T.  Claxton  Fidler  in  "  A  Practical  Treatise  on  Bridge  Construction"  (London,  Charles 
Co.),  page  158. 


564 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  VII. 


Diagram  for  Ideal  Column. — If  we  lay  off  -  as  abscissa,  and  the  corresponding  value  of 

p 

-j  as  given  by  (E)  as  ordinate,  we  have  the  curve  DE  of  Ruler's  formula.     The  ordinate  to 

this  curve  to  scale  gives  the    crippling    unit    stress 

P  I 

-T-  for  any  given  — ,  and,  as  we  have  seen    for   a    less 

load,  the    column  will    recover    if  slightly  disturbed. 

p 

But  when  -3  becomes  equal    to  the    elastic   limit 
A 

St,  the  corresponding  ratio  is 


For  less  values  of  -  than    this  the    column    can 
/c 

be    loaded   up  to  the   elastic  limit  St.     The  ordi- 
nates,  then,  to  ADE  give  the  crippling  unit  stress  for  the  ideal  column  for  any  value  of  — . 

Actual  Column. — The  preceding  discussion  and  the  conclusions  and  deportment  of  the 
ideal  column  do  not  hold  good  for  actual  columns,  because  in  such  columns  the  ideal  con- 
ditions are  not  realized.  Thus  no  actual  column  is  perfectly  homogeneous,  has  a  perfectly 
straight  axis  or  has  the  load  exactly  centred. 

*Lack  of  ideality  in  any  of  these  conditions  will  cause  a  column  of  any  length  to  deflect 
when  loaded,  and  this  is  in  accord  with  common  experience. 

We  can,  then,  only  load  the  actual  column   up  to  the  elastic  limit  St  when  it  is  very 

p 

short — theoretically  only  when  the  length  is  zero.      For  any  finite  length  —.  must  always  be 

A 

less  than  given  by  the  preceding  diagram. 

The  actual  curve,  then,  for  any  actual  column  will  be  some  curve  such  as  represented 
by  the  broken  curve  in  the  preceding  diagram,  which  is  tangent  at  A  to  the  line  AD,  and  at 
an  infinite  distance  to  the  curve  DE. 

We  see  at  once  that  any  such  curve  which  should  give  the  actual  values  of  the  crippling 

P 
unit  stress  -.  for  any  one  actual  column  must  depend  upon  the  actual  eccentricity  of  the  load 

and  upon  all  other  actual  deviations  from  ideal  conditions. 

As  all  such  deviations  can  never  be  identical  for  any  two  actual  columns,  the  actual 
curve  must  be  a  different  one  for  each  column. 

It' is  therefore  obvious  that  any  one  curve  which  gives  the  average  experimental  values 
P 
of  -j  for  any  number  of  actual  columns  must  rest  at  bottom  upon  the  average  deviations 

from  ideal  conditions.  A  single  curve  must,  then,  be  based  upon  average  experimental 
results,  and  to  try  to  deduce  any  single  theoretic  curve  which  shall  give  actual  results  for  all 
columns  is  to  attempt  the  impossible.  Any  practical  formula  must  be  directly  based  upon 
average  experimental  results. 

Practical  Values  for«. — The  theoretic  values  of  n  given  on  page  561  disregard  friction. 
Also,  the  ends  in  practice  cannot  be  perfectly  "  fixed."  We  have  to  do  practically  with  two  pin 


CHAP.  VII.] 


PRACTICAL   FORMULAS  FOR  LONG   COLUMNS. 


565 


ends  with  friction,  or  one  pin  end  with  friction  and  one  flat  end,  or  two  flat  ends.      Hence 
the  practical  values  of  n  should  be  different  from  the  theoretic  values  given  on  page  561. 

Experiments  show  that   Ruler's  formula  (E)  gives  the  average  crippling  unit  stress  -^- 


with  very  good  accuracy  for  very  long  columns,  i.e.  when  -is  very  great,  provided  we  take 
for  n  the  following  values: 


/. 

K 


Two  pin  ends. 

~  =  4'°53' 


One  pin  end  and  one  flat  end. 
—^=  =  4.524, 


Two  flat  ends. 


V    2    =  4' 


These  values  of  n  should  be  taken  when  Euler's  formula  is  used. 

Practical  Formulas  for  Long  Columns.  —Experiment  also  shows  that  actual  values  of 
p 
-£-  for  short  columns,  instead  of  following  the  ideal  diagram  ADE,  page  564,  are  so  scattered, 

that  almost  any  curve  through  A  and  tangent  to  Euler's  curve  DE  gives  very  satisfactory 
average  results.      On  this  fact  all  our  practical  column  formulas  are  based. 

STRAIGHT-LINE  FORMULA. — Of  these  prac- 
tical formulas  one  of  the  best  known  is  the 
"straight-line  formula,"  first  given  by  Thomas 
H.  Johnson,  C.E.  (Trans.  Am.  Soc.  C.  £.,  July, 
1886).  It  consists  in  drawing  a  straight  line 
through  A  tangent  to  Euler's  curve  EE.  The 
point  of  tangency  D  is  at  a  distance  from  O  (see  A 

figure)  given  by  — .      Both  curve  EE  and  straight 

line  AD  have   then  a  common   ordinate  at  the Se 
point  D. 

The  equation  of  a  straight  line  through  A  is 

y  —  St-\-bx (i) 

The  equation  of  Euler's  curve  EE  is 

n*£ 


(2) 


If  we  differentiate  (i)  and  (2),  and  equate  the  values  of  -j-,  making*  =  — ,  we  have  for 
the  condition  of  a  common  tangent  at  D 


(3) 


If  we  equate  (i)  and  (2),  making*  =  -,  we  have  fcr  the  condition  of  a  common  ordinate 


bL 


(4) 


566  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  VII. 

From  (3)  and  (4)  we  find  for  the  limiting  value  of  — 

—  =n\/^-,    and  hence     b— *       * . 


P  I 

Inserting  this  value  of  b  in  (i),  and  putting  y  =  —  and  x  =  — ,  we  have  for  the  straight- 

./i  A" 

line  formula, 

when  '-<  ,,/'lK  P_rf.          **S. 


where  /£  is  the  /ra.tf  radius  of  gyration  of  the  cross-section. 
This  formula  holds  for  any  value  of  -  so  long  as 


Beyond  this  limit  we  use  Ruler's  formula,  and  have, 

P 


The  straight-line  formula  is  simple  and  easily  applied,  and  contains  no  experimental 
constants  except  Se,  E,  and  w. 

P  I 

It  gives  values  for      .    for  small  values  of  —  considerably  less  than  the  average  of  experi- 
A  K 

ments,  owing  to  the  fact  that  the  tangent  at  A  is  not  horizontal. 

PARABOLA  FORMULA.  —  This  formula  is  given  by  Prof.  J.  B.  Johnson  (Theory  and 
Practice  of  Modern  Framed  Structures  —  Wiley  &  Sons).  The  curve  AB  (figure,  page  565) 
is  assumed  as  a  parabola  tangent  to  Euler's  curve  at  B.  We  have  then 


(i) 


where  b  must  be  determined  by  the  condition  of  tangency. 

This  equation  gives  y  =  Se  for  x  =  o,  and  the  tangent  at  A  is  horizontal. 
From  Euler's  formula  we  have 


Differentiating  (i)  and  (2),  and  proceeding  as  before,  we  have  the  parabola  formula, 


where,  as  always,  K  is  the  least  radius  of  gyration  of  the  cross-section. 
This  formula  holds  for  any  value  of  -  so  long  as 


CHAP.  VII.  ]  LONG   COLUMNS.  567 

Beyond  this  limit  we  use  Euler's  formula,  and  have, 


The  parabola  formula  is  as  simple  and  easily  applied  as  the  straight-line  formula.  It 
also  contains  no  experimental  constants  except  Se ,  E  and  «.  It  gives  on  the  whole  better 

P 
average  values  for  -7,  owing  to  the  fact  that  the  tangent  at  A  is  horizontal. 

p 
Remarks  on  these  Formulas.— Formula  (P)  gives  greater  values  for-^  than  formula  (S), 

and  is  to  be  preferred,  therefore,  for  very  perfect  columns  with  load  very  accurately  centered. 

Both  the  straight-line  and  the  parabola  formulas  are  lines  tangent  to  Euler's  curve  EE 
at  points  D  and  B  (figure,  page  565).  This  means,  in  the  light  of  our  remarks,  page  564, 
that  both  assume  ideal  conditions  for  all  columns  at  and  beyond  a  certain  length  L,  which  is  a 
different  lengtlt,  for  each  formula. 

Such  an  assumption  is  of  course  incorrect.  There  is  no  one  length,  to  say  nothing  of 
two  different  lengths,  at  which  ideal  conditions  can  be  considered  as  existing.  Experiments, 

p 
however,  sho\v  that  average  values  of  -.  approach  at  and  beyond  these  lengths  very  closely  to 

Euler's  curve,  being  always,  however,  slightly  below,  and  hence  the  assumption  is  practically 
justified. 

The  actual  average  curve,  however,  as  we  have  seen  (page  564),  should  run  through  ^4  as 
shown  by  the  broken  curve  in  the  figure  page  564,  should  have  a  horizontal  tangent  at  A, 
and  should  then  run,  as  shown,  somewhat  below  Euler's  curve,  and  be  tangent  to  it  at  an 
infinite  distance. 

Rankine's  Formula. — Such  a  curve  is  Rankine's  formula.  Let  A  be  the  deflection. 
Then  the  maximum  moment  is  PA. 

From  equation  (II),  page  497,  we  have  for  the  unit  stress  Sf  due  to  bending  in  the  most 
compressed  fibre  at  a  distance  v  from  the  axis 

_  PAv  -  PAv 

P 
We  have  in  addition  a  direct  compressive  unit  stress^. 

If  then  Se  is  the  elastic  limit  unit  stress,  we  have  for  the  crippling  unit  stress 

P      PAv                         P          St  /,x 

—  -4-  -T  -r  =  S* ,      or      -j  •=  —      -r- V 1 1 


Equation   (i)    is  rational   in    form,  and   if  we    knew  A   it    would    give    accurately    the 
crippling  unit  stress. 

If  we  suppose  for  small  deflections  the  curve  of  deflection  to  be  practically  a  circl, 

radius  of  curvature  p,  we  should  have 

A:  I  ::  /:  p—  A, 


568  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  VII. 

or,  since  A  is  small  compared  to  p, 


In  general  whatever  the  curve  of  deflection,  we  can  assume  A  to  be  some  function  of 
and  to  vary  inversely  as  v,  since  p  increases  with  v.     We  can  then  write 

cP  Av      cP 

v'  /c8         x» 


Inserting  this  value  of  —  ^  in  (i),  we  have 


<*> 


where,  as  always,  K  Is  the  least  radius  of  gyration  of  the  cross-section,  and  c  is  a  constant  to 
be  determined  by  experiment,  depending  upon  the  material  and  the  end  conditions.  Since 
the  column  bends  easiest  in  the  direction  of  its  least  dimension,  ive  take  for  K  the  least  radius 
of  gyration. 

Equation  (R)  is  Rankine's  formula  for  long  columns.      It  holds  for  all  values  of  —  . 

P  I 

We  see  that  Rankine's  formula  gives  —  =  St  for-  =  o.      The  tangent  at  A  (figure,  page 

A  K 

p  i 

556)  is  horizontal,  and  we  have  -7  =  0  for  -  =  oo  .     It  therefore  complies  with  the  conditions 

A  K 

for  the  average  actual  curve  given  on  page  564. 

It  is  not  so  simple  or  easily  applied  as  the  straight-line  or  the  parabola  formula,  and 
the  experimental  constant  c  must  be  determined  before  it  can  bo  used  in  any  case. 

Gordon's  Formula.  —  Since  K  is  a  function  of  the  least  dimension  d  of  the  cross-section, 
we  may  also  write  for  the  crippling  unit  stress 


where  c  is  again  a  constant,  to  be  determined  by  experiment.      Equation  (G)  is  known  as 
Gordon's  formula  for  long  struts.      It  also  holds  for  all  values  of  -,  and  the  same  remarks 

apply  as  for  Rankine's  formula. 

Merriman's  Formula.  —  The  equation  of  the  curve  AB  (figure,  page  564)  has  been 
assumed  by  Prof.  Merriman  (Engineering  News,  July  19,  1894)  as  identical  in  form  with 
Rankine's  formula.  We  have,  then, 

S, 


Instead,  however,  of  regarding  b  as  an  experimental  constant,  Prof.  Merriman  deter- 
mines b  precisely  as  in  the  case  of  the  straight-line  and  parabola  formulas,  by  the  condition 
of  tangency. 


CHAP.  VII.]  ALLOWABLE   UNIT  STRESS—  FACTOR   OF  SAFETY.  569 

We  thus  obtain 


where,  as  always,  K  is  the  least  radius  of  gyration  of  the  cross-section. 

Equation  (M)  is  Merriman's  formula  for  long  columns.  Like  Rankine's  fornvila,  it 
complies  with  the  conditions  of  the  average  actual  curve  given  on  page  564.  It  is  preferable 
to  Rankine's  in  that  it  contains  no  experimental  constant.  It  is  therefore  probably  nearer 
the  true  curve  for  an  average  actual  column  than  any  of  the  formulas  thus  far  given. 

'     Allowable  Unit  Stress  —  Factor  of  Safety.  —  The  preceding  formulas  will  enable  us  to 
find  tne  crippling  unit  stress. 

In  practice  only  a  portion  of  this  is  taken  as  the  allowable  working  unit  stress.  This 
portion  is  called  the  factor  of  safety  (page  481).  For  quiescent  loads  (buildings,  etc.y  this 
factor  is  taken  at/=  4  for  wrought  iron  and  steel,  and/=  6  for  cast  iron  and  wood, 

For  variable  loads  a  variable  factor  of  safety  is  used  equal  to 

f  =  4  +  -  ,         for  wrought  iron  and  steel, 


f  =•  7  -I  --  for  cast  iron, 

2Oa 

f  =  6  -I-  -          for  wood, 
^  2Qd 

where  /is  the  length  in  inches  and  d  the  least  dimension  in  inches  of  the  rectangle  which 
encloses  the  given  cross-section. 

We  have  then  in  general  for  the  working  unit  stress 


where  —  is  found  by  any  one  of  the  preceding  formulas,  and  the  value  of  /  is  taken  as  just  given. 

^1 

Examples.—  (i  )  Let  the  ratio  of  the  length  of  a  steel  column  to  the  least  radius  of  gyration  of  its  cross- 
section  A  be  -  =  100,  and  to  the  least  dimension  of  the  enclosing  rectangle  be  -  =  25.     Let  Se  =  40000  and 

K 

E  —  jo  oooooo  pounds  per  square  inch.     Find  the  crippling  and  working  unit  stress  by  the  straight-line 
formula.  - 

ANS.  We  have,  using  the  practical  values  of  n  on   page  565,    ^r  =  IO°  less   than  "  V  ^  fn  a11  Cases' 
Hence  by  the  straight-line  formula  the  crippling  unit  stress  is 


7i* 

Hence  for 

Two  pin  ends ^  =  0.655,  =  26000  pounds  per  square  inch  ; 

P  «  M 

One  pin  end  and  one  flat  end -.  =  0.695,  =  27600 

75 

Two  flat  ends ^  =  0.725,  =  28800 

The  factor  of  safety  is  /  =  4  +  ^  =  5-25-     Hence  the  working  stress  in  these  three  cases  is 
Sw  =  4952,     5257,     5486     pounds  per  square  inch. 


STATICS  OF  EL/tSTIC  SOLIDS.  [CHAP.  VII. 


(2)  Find  the  crippling  and  working  unit  stress  by  the  parabola  formula. 
ANS.  We  have  by  this  formula 


570 


- 

j».- 

Hence  for 

p 
Two  pin  ends  ......................     —.  =  0.805",  =  32000  pounds  per  square  inch 

A 

p 
One  pin  end  and  one  flat  end  .......      -  =  0.845,  =  33600 

A 

Two  flat  ends   .....................     ^  =  o.865,  =  34400       "          "         "          " 

A 

The  factor  of  safety  is  as  before/"=  5.25.     Hence  the  working  stress  in  these  three  cases  is 
5w  =  6ioo,     6400,     6550     pounds  per  square  inch. 

It  will  be  seen  that  the  values  given  by  the  parabola  formula  are  greater  than  those  given  by  the 
straight-line  formula.  For  very  perfect  columns  and  load  very  accurately  centred  the  oarabola  formula 
results  are  preferable;  for  poorer  columns,  the  straight-line. 

(3)  Find  tht  crippling  and  working  unit  stress  by  the  Merriman  formula. 

ANS.  We  have  by  this  formula 

P  =      S.  _ 

A       i  +  < 
3«' 
Hence  for 

p 
Two  pin  eods  ...................      —  =  0.555,  =  22000  pounds  per  square  inch; 

fl 

p 
One  pin  end  and  one  flat  end  .......     —  =  o.6o5,  =  24000       "          "        "          * 

V  •'* 

p 
Two  flat  ends  .................      ...      -=  0.655,  =  26000       "          "         "          " 

A 

The  factor  of  safety  is  as  before/  =  5.25.     Hence  the  working  stresses  in  these  three  cases  are 
5,  =*•  4190,     4570,     4950     pounds  per  square  inch. 


CHAPTER   VIII. 

THE   PIVOT   OR  SWING   BRIDGE. 

Pivot  or  Swing  Bridge. — The  pivot  or  swing  bridge  is  a  girder  continuous  or  partially 
continuous  over  three  or  four  supports.  If  over  three  supports,  it  is  a  pivot  span.  If  over 
four  supports,  the  length  of  the  short  centre  span  is  the  width  of  the  turn-table,  and  loads  in 
this  span  come  directly  on  the  turn-table  and  hence  cause  no  stresses  in  the  truss  members. 

If  the  bracing  is  carried  through  the  centre 
span  as  shown  in  the  figure,  it  is  evident  that  a 
load  in  one  end  span,  as  AB,  tends  to  lift  the 
span  from  the  support  C.  It  will  be  found  in 
general  difficult  to  hold  the  span  down  at  C. 

For  this  reason  the  bracing  in  the  centre 
span  is  omitted.  The  continuity  in  such  case 
is  partial,  but  the  span  can  be  held  down  at  C. 

Centre     Span    without     Bracing.  —  By 
means  of  the   principle   of  least   work   we  can 
find  the  reactions  at  the  supports  for  a  load  P  placed   anywhere,  and  then  the  stresses  in  the 
truss  members  for  this  load  are  easily  found. 

ist.  LOAD  P  IN  SPAN  AB  =  lv. — Let  the  span  be  AB  =-.  /,,  BC=  /2,  CD  =  /3.  Let  the 
load  P  be  at  the  distance  k^  from  the  left  end,  where  k^  is  any  given  fraction.  Let  J/2  be 

the  moment  on  the  left  of  the  second 


S, 


I, 


t—  kTlr 


S2 


It 


ss 


b 


VD 


, 
3 


support.  Let  the  pressure  on  the 
right  of  A  be  Slt  on  the  left  of  B  be 
S2',  on  the  right  of  C,  S3,  and  on  the 
left  of  D,  Ss'.  < 

There  are  no  braces  in  the  span 

BC  and  hence  no  pressure  on   right  of  B  or  left  of  C,  and  the  moment  on  left  of  C  will  also 
be  My  the  same  as  on  left  of  B. 

Taking  moments  about  B,  we  have 


Taking  moments  about  D   we  have 


From  these  equations  we  have 


>,  =  ^«=_.S,'.    .    (i) 


571 


STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  VIII. 


Equations  (i)  give  the  pressures  at  the  supports  in  terms  of  Mv 
If  then  we  can  determine  J/2,  we  can  find  these  pressures. 
For  any  point  distant  x  from  A  we  have  the  moment, 


between  A  and  P,     M  =  -  S,x  =          -  P(i  -  £>  ; 


between  P  and  B,     M=  -  S,x  +  P(x  -  £  /  )  =  -         -  P(  i  -  k,)x  +  P(x  -  £/,). 

li 

For  any  point  between  B  and  C  we  have 


For  any  point  distant  x  from  D  we  have, 

between  C  and  D,  M=  S^x  = 

3, 

Let  v  be  the  lever-arm    for  any  truss  member,  and  M  the  moment  at  the  centre  of 

moments  for  that  member.      Then  the  stress  in  that   member  is  —  .      Let  a  be  the  cross- 

v 

section  of  the  member,  and  s  its  length.     Then,  from  equation  (III),  page  515,  the  work  of 
straining  that  member  is 


the  total  work  of  straining  all  the  members  is  then 

work  — 


and  this  work  must  be  a  minimum  by  the  principle  of  least  work,  page  517. 
We  have  then,  in  the  present  case,  inserting  the  value  of  M  just  found, 


work  =  -  /Xl  -  *  +  2  -  /M  -  «•  +  ^  - 


~  ^  2 


Suppose  the  span  AB  to  be  fixed  horizontally  at  B  and  the  support  at  A  removed,  and 
let  «  be  the  stress  in  any  member  due  to  a  unit  load  at  A,  and  /  the  stress  in  any  member 
due  to  a  unit  load  at  P.  Then  we  have 

uv  =  I  X  x,     or     -    =  K2 


CHAP-   Vin-]  SMNG   BRIDGE. 

•J ' 

If,  then,  we  insert  these  values,  and  put  ^^  =  o,   we    have  for 
which  makes  the  work  a  minimum 

.  — -A/TI  -  *,)«> 


573 
Q{ 


or,  solving  for  Mv 


„.„ 

.....  •'  '  '" 


Equation  (2)  gives  the  value  of  Mv  or  the  moment  at  the  support  B,  for  a  load  P  at 
any  distance  k^  from  the  left  end  A  in  the  first  span.  If  M2  is  known,  then  equations  (i) 
give  the  pressures  at  the  supports.  When  these  are  known  the  stresses  can  be  easily  calcu- 
lated. Note  that  in  equation  (2)  u  is  the  stress  in  any  member  due  to  a  unit  load  at  A,  and 
/  the  stress  in  any  member  due  to  a  unit  load  at  P,  considering  the  span  AB  as  fixed  hori- 
zontally at  B  and  without  anv  support  at  A  . 

2d.  LOAD  P  IN  SPAN  CD  =  /3.—  Let  the  load  P  be  in  the  third  span  /„  at  the  distance 
£3/3  fr°m  support  A  where  £3  is  any  given  fraction. 

Then  we  have 


c 

~ 


X-X 

4     A  JA 


5 


1+      AA  4  /\  Z\[ 

M.  B      M- c  ^-H^H 

and  proceeding  now  precisely  as  before,  we  find  in  the  same  way 


^,  =  *y, /.,_/. ^      11777-    LTR (4) 


Equation  (4)  gives  the  value  of  7J/2,  or  the  moment  at  the  support  B,  for  a  load  /*  at 
any  distance  -£3/3  from  the  right  end  D  in  the  third  span.  If  M2  is  known,  then  equations  (3) 
give  the  pressures  at  the  supports.  When  these  are  known  the  stresses  can  be  easily  calcu- 
lated. Note  that  in  equation  (4)  u  is  the  stress  in  any  member  due  to  a  unit  load  at  D,  and 
/  the  stress  in  any  member  due  to  a  unit  load  at  P,  considering  the  span  CD  as  fixed  hori- 
zontally at  C  and  without  any  support  at  D. 


574 


STstTICS  OF  ELASTIC  SOLIDS, 


[CHAP.  VIII. 


Special  Cases.  —  The  preceding  equations  are  general  and  include  all  cases.  Thus  if 
the  two  end  spans  are  equal,  we  have  only  to  make  /,  =  /3  =  /  and  kl=  k^  =  k.  If  we  have 
only  two  spans,  we  make  /2  =  o.  Hence  we  have 

THREE  SPANS,  END  SPANS  EQUAL.  —  Making 
|S'  t*  _  j  '  /,  =  /3  =  /  and  £,  =  £3  =  k,  we  have  the  following 

A  --  '  --  B       *•       C  *  0    ecluations: 

1st.   Load  in  span  AB  : 


-• 


2d.  Load  in  span  CD  : 


And  in  both  cases 


Mt  = 


2d.  Load  in  span  BC  : 


tfs  .  l*'*u*s 


(5) 


(6) 


M  —  PI 

o  a        —  */  a 

(7) 

l*£+t^'JL 

\ 

Two   SPANS  ONLY.  —  Making  /2  — 

°»      WC    AS/                                                                    *H     t^* 

ls; 

have  the  following  equations  : 

1                                      //                                        II?* 

1 

1st.  Load  in  span  AB  : 

A,                                                   B 

C 

5i  =  -y2  +  ^!-^i). 

^=T2+/^'        5,=  -^'=^,    . 

•        (8) 

(9) 


),     .     .     (10) 


M2  = 


/2a_^«^ 


*./. 


|s«         Two  EQUAL  SPANS.— Making  /2  =  o  and 
I     /,  =  /s  =  /,  we  have  the  following  equations  : 
C  1st.   Load  in  span  AB  : 


2d. 


in  span  BC  : 


CHAP.  VIII.] 

And  in  both  cases 


SWING  BRIDGE. 


575 


(14) 


Values  Of  a  Indeterminate.— It  will  be  noted  that  our  equations  for  M2  require  that  the 
area  of  cross-section  a  for  each  member  shall  be  known,  while  it  is  our  object  to  determine 
these  areas  by  first  finding  the  stress  in  each  member  and  then  dividing  this  stress  by  the 
allowable  unit  stress. 

It  is  necessary,  then,  to  make  a  first  approximation  by  supposing  a  the  same  for  each 
member.  It  will  then  cancel  out  of  our  equations  for  My  We  can  then  find  M2  approxi- 
mately, and  determine  the  stresses  and  corresponding  areas  a.  With  these  values  of  a  we 
can  again  determine  J/0. 

A  short  example  will  thoroughly  illustrate  the  application  of  our  equations. 

Example. — Find  the  stresses  for  a  swing  bridge  of  two  equal  spans  /=  So  ft.,  each  span  divided  into 
four  panels  of  20  ft.    Let  the  spans  be  symmetrical,  the  centre  height  Be  =  10  ft.  and  the  height  at  ends  bi  =  7  //. 


I 


---  ......  ............. 


I 


*-- -k-l- 


V 
M,  B 


For  these  dimensions  the  lever-arm  v  for  any  upper-chord  member,  as  de,  will  not  differ  appreciably  from 
the  height  ^3,  and  the  length  s  for  any  upper-chord  member,  as  cd,  will  not  differ  appreciably  from  the  panel 
length. 

We  have  then  the  following  lever-arms 

A\      1-2     2-3      3-Z?       be       cd      de          Ab  b\          e\          t2  d2  d$  e^ 

lever-arms          7         8         9         10        7         8        9         39.64         140         52         160        65.66         180        80.5 

We  can  now  calculate  the  stresses  u  in  every  member  due  to  a  unit  load  at  A,  and  the  stresses  p  in 
every  member  due  to  a  unit  load  at  P,  considering  the  span  AB  as  fixed  horizontally  at  B  and  without  sup- 
port at  A .  We  can  then  form  the  following  table  : 


(0 

(2) 

(3) 

(4) 
/i 

(5) 
/a 

(6) 

A 

(7) 
•»J 

A) 

/.«* 

(9) 
tea 

(.0) 

/»«« 

163  25 

20 

250.00 

6  666 

883  88 

CQ2  CO 

2-3  
3-Jf  

20 

—  8.000 

_|_  o  8e7 

—  6.000 

—  4.000 

—  2.  COO 

1280.00 

163  35  " 

960.00 

64O.OO 

320.OO 

500  oo 

250.00 

_i_  6  5^5 

888  88 

592  59 

206  to 

5.  14 

7 

1  2  6o2 

114  64 

*33  77 

21.54 

5  25 

f-2  

-4-2  di6 

72  87 

85  42 

Q7  60 

z-d  

21.93 

o  888 

4  66 

^3  • 
3-f  
eB  

9 
22.36 

10 

+  1.490 
—  0.600 

+  1-739 
—  0.700 

+  1-987 
—  0.800 

+  2.236 
—  O.gOO 

49.64 
3.60 

57-94 
4.20 

66.20 
4.80 

74-49 
5.40 

• 

4833.90 

2936.42 

1406.53 

399  39 

576  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  VIII. 

In  the  first  column  we  place  the  designation  of  each  member  ;  in  column  (2)  the  length  s  of  each  member  ; 
in  column  (3)  the  stress  u  in  each  member  for  a  load  of  unity  at  the  end  A,  considering  the  span  AB  as  a 
semi-girder  fixed  horizontally  at  fi\  in  columns  (4),  (5),  (6)  the  stress^  in  each  member  for  a  load  of  unity  at 
apex  i,  2  and  3,  considering  the  span  AB  as  a  semi-girder  fixed  horizontally  at  B.  In  column  (7)  we  have  the 
values  of  u*s  for  each  member;  in  columns  (8),  (9),  (10),  the  values  oi  PUS  for  each  member.  At  the  bottom 
of  columns  (7),  (8),  (9),  (10)  we  give  the  summation  of  the  values  in  each  column. 

We  have  then,  from  equation  (14), 


where  k  has  the  values  —  ,  —  ,—  at  apex  i,  2,  3,  and  the  summation  2pus  is  given  by  columns  (8),  (9),  (10). 

We  have  then  for  load  P  at  apex  i,  from  equations  (12), 

M,  =  +  0.0712  PI,  Si  =  +  0.67  S8P. 

For  P  at  apex  2, 

Mi  =  +  0.1045  PI,  &  =  +  0.395  5/>. 

For  P  at  apex  3, 

Mt  =  +  0.0836  PI,  Si  =  +  o.i664/>. 

For  P  at  apex  4,  St  will  be  the  same  as  S»  for  P  at  3,  or 

Si  =  —  0.0836  P. 
For  P  at  apex  5,  5i  will  be  the  same  as  S3'  for  P  at  2,  or 

Si  =  —  O.IO45/*. 
For  P  at  apex  6,  Si  will  be  the  same  as  S*  for  P  at  i  ,  or 

5,  =  -  0.07  12P. 

Negative  values  denote  that  5,  acts  downwards. 

We  can  now  find  the  stresses  in  each  member  for  each  load,  and  then  by  tabulation   can   find   the 
loading  which  gives  the  maximum  stress  and  the  maximum  stress  itself  in  each  member. 

Solid  Beam  —  Uniform  Cross-section.  —  We  can  easily  find,  from  the  equations  already 
deduced,  the  value  of  M2  for  a  solid  beam  of  uniform  cross-section.  Thus,  let  x  be  the 
distance  of  the  point  of  moments  for  any  member  from  the  left  end  A,  and  v  its  lever-arm. 

Insert  these  values  in  equation  (2)  and  we  have 


=  PL 


Now  if  the  girder  is  a  solid  beam  of  uniform  cross-section,  we   have  s  =  dx,  and  at? 
constant.      Hence  we  have 


-  /o  /*' 

JQ 


_  __  __ 

rw^+A8  rv.*+f>  r*** 

Jo  Jo  's  »/o 


Aft  =  PI, 


If  we  perform  the  integrations,  we  obtain 

for  load  in  span  AB          ff^—L         (I5) 


CHAP.  VIII.]  SWING  BRIDGE.  577 

The  pressures  of  the  supports  are  then  given  by  equations  (l). 
In  the  same  way  we  have,  from  equation  (4), 

for  load  in  span  CD          M2  =  Pl*k*( *  ~_jyl (16) 

2(/i  ~T  3fz  ~r  ^3) 

The  pressures  of  the  supports  are  then  given  by  equations  (3). 

For  three  spans,  end  spans  equal,  the  pressures  on  the  supports  are  given  by  equations 
(5)  for  load  in  span  AB,  and  by  equations  (6)  for  load  in  span  CD,  and  in  both  cases 


For  two  spans  only  we  have 

for  load  in  span  AB         M2  =  — ^~~^~T~r\*     •••••••••     0&) 

2  (A  ~T~  ^2) 

The  pressures  of  the  supports  are  given  by  equations  (8). 

-^ 


for  load  in  span  BC         M2= 


The  pressures  of  the  supports  are  given  by  equations  (10). 

For  two  equal  spans  the  pressures  on  the  supports  are  given  by  equations  (12)  for  load  in 
span  AB,  and  by  equations  (13)  for  load  in  span  BC,  and  in  both  cases 


(20) 


Example.—  In  the  example  given  on  page  575,  consider  the  girder  as  a  solid  beam  of  uniform  cross-section. 
In  this  case  we  have,  from  equation  (20), 

M*  =  zoPk(l  -/&'), 

where  k  has  the  values  —  ,   —  ,    —  • 
444 

We  have  then  for  load  P  at  —  /  from  left  end,  from  equations  (12), 


4-687  $P, 


For  load  P  at  -/  from  left  end, 


For  load  P  at    /  from  left  end, 

M  a  =  +  6.5625P,        5,  =  +  o.i68o/>. 

For  P  at—  from  right  end,  from  equations  (13), 

M»  =  +  6.5625^,        5,  =  -  o.o820/>. 

For  P  at  -/  from  right  end, 
4 

Mt  =  +  7-  $P,  &  =  —  0.0938^. 

For  P  at  —  /  from  right  «•<  1  , 

M»  =  +  4.6875/),        Si  =  -  o.o$86P. 


CHAPTER   IX. 


THE    METAL    ARCH. 

Three  Kinds  of  Metal  Arch. — We  may  distinguish  three  kinds  of  metal  arch,  viz.,  arch 
hinged  at  crown  and  ends;  arch  hinged  at  ends  only;  arch  without  hinges. 

If  the  arch  is  a  framed  structure,  the  stresses  in  the  members  can  be  found  in  any  case, 
if  for  a  given  load  we  can  find  the  horizontal  thrust  and  vertical  reactions  at  the  ends  and 
the  moments,  if  any,  which  exist  at  the  ends. 

Framed  Arch  Hinged  at  Crown  and  Ends.— This  form  of  construction  is  an  arch  only 

in  form,  but  in  principle  is  simply  two 
braced  rafters  the  thrust  of  which  is 
taken  by  the  abutments  instead  of  by  a 
tie-rod.  It  is  therefore  a  very  simple 
matter  to  find  the  end  reactions  for  a 
given  load. 

Let  the  span  AB  be  /,  and  P  the 
load  at  the  distance  £/from  the  left  end, 
where  k  is  any  given  fraction.  Let  the  rise,  measured  always  from  the  chord  AB  to  the 
hinge  C  at  the  crown,  be  denoted  by  r. 

Then  taking  moments  about  the  right-hand  hinge  at  Bt  we  have  for  the  reaction  V^  at 
the  left  end  for  any  position  of  P 


-  k)  =  o,     or     Vl=P(i  —  £). 


(0 


Taking  moments   about  the  hinge  Cat  the  crown,  we  have  for  the  horizontal  thrust  H 
at  the  left  end,  when  kl  is  less  than  -,  that  is  when  P  is  on  the  left  of  the  centre, 


•)=o, 


or          = 


V£ 
2r 


PI 
2r 


-  2k)  = 


Pkl 

2r  ' 


When  kl  is  greater  than  -,  that  is  when  P  is  on  the  right  of  the  centre, 

VI      7V(i  -*) 
or     H=it  = Tr 


(3) 


These  values  of  V^  and  H  are  independent  of  the  shape  of  the  arch. 

Change  of  temperature  causes  no  stresses  in  the  arch  hinged  at  crown  and  ends.  Each 
half  is  free  to  turn  about  the  hinges  and  accommodate  itself  to  any  change  of  shape  due  to 
change  of  temperature. 

Equations  (i),  (2)  and  (3).  hold  for  a  solid  as  well  as  for  a  braced  arch.  We  can  then 
determine  the  stress  in  each  member  for  each  load,  and  then  by  tabulation  can  find  the 
loads  which  give  the  maximum  stress  and  the  maximum  stress  itself  in  each  member. 

578 


CHAP.  IX.] 


FRAMED  ARCH  HINGED  AT   ENDS   ONLY. 


579 


Framed  Arch  Hinged  at  Ends  Only.  —  In  this  case  we   have   just    as   before,  taking 
moments  about  B,  for  any  position  of  P 


It    remains    to    find    the    horizontal 
thrust  H. 

Let  v  be  the  lever-arm  for  any  member,  as  determined  by  the  method  of  sections,  page 
401,  and  M  the  moment  at  the  centre  of  moments  for  that  member.      Then  the  stress  in 

that  member  is—  .      Let  a  be  the  cross-section  of  the  member,  and  s  its  length.     Then  from 
equation  (III),  page  515,  the  work  of  straining  that  member  is 


The  total  work  of  straining  all  the  members  is  then 

work  =  * 


and  by  the  principle  of  least  work,  page  517,  this  work  must  be  a  minimum. 

Let  x  and  y  be  the  co-ordinates  of  the  point  of  moments  for  any  member,  as  deter- 
mined by  the  method  of  sections,  page  401. 

Then  for  any  member  on  the  left  of  P  we  have  the  moment 

M=  Hy  —  V^x  =  Hy  —  P(i  —  k}x,  . . 

and  for  any  member  on  the  right  of  P  we  have 

M  =  Hy  —  Vjc  +  P(x  —  £/)  =  Hy  —  P(i  —  k)x  -f  P(x  —  kl). 
We  have  then  for  the  work  of  straining  all  the  members 

work  =  ^\Hy-  P(i  -  ^XP^^  +  2E     {_Hy  -  P(\  -  K)x  +  P(x  -kl^J^p 

If  we  differentiate  with  reference  to  H  and  put  the  differential  coefficient  equal  to  zero, 
we  have  for  least  work 

</(work)  _      —^,klrff-»        P  (    —  i 
dH  o 

Hence,  since  E  is  constant, 


5&o 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  IX. 


Let  h  be  the  stress  in  any  member  due  to  a  negative  unit  horizontal  force  at  the  left 
end  A.  Let/  be  the  stress  in  any  member  due  to  a  unit  load  at  P,  considering  the  arch  a 
simply  supported  at  the  ends.  Then  we  have  for  any  member  on  the  left  of  P 

±  hv  =  i  X y,         ± pv  =  i  X  (i  -  k)x. 
Multiplying  these  two  equations,  we  have  for  any  member  on  the  left  of  P 


hence 


For  any  member  on  the  right  of  P  we  have 

±  hv  =  i  X y,          ± pv  =  I  X  (i  —  K)x  —  I  X  (x  —  kl)\ 


We  see,  then,  that  equation  (2)  can  be  written 

phs 
H=P- 


(3) 


where  p  is  the  stress  in  any  member  with  its  proper  sign,  (-}-)  for  tension  and  ( — )  for  com- 
pression for  a  unit  load  at  P,  considering  the  arch  as  simply  supported  at  the  ends,  and  //  is 
the  stress  in  any  member  for  a  unit  negative  thrust  at  the  left  end  A,  also  taken  with  its 
proper  sign,  (-}-)  for  tension  and  (— )  for  compression. 

From  equations  (i)  and  (3)  we  can  find  l\  and  H  for  a  load  P  placed  anywhere,  and 
can  then  determine  the  stresses  for  this  loading. 

We  can  then  easily  determine  the  stress  in  each  member  for  each  load,  and  then  by 
tabulation  can  find  the  loads  which  give  the  maximum  stress,  and  the  maximum  stress  itself, 
in  each  member. 

It  will  be  noted  that  equation  (3)  requires  that  the  area  of  cross-section  a  for  each 
member  shall  be  known  in  advance,  while  it  is  our  object  to  determine  these  areas  by  first 
finding  the  stress  and  then  dividing  by  the  allowable  unit  stress.  It  will  in  general,  then,  be 
necessary  to  first  find  values  for  H,  assuming  a  to  be  constant.  It  then  cancels  out.  Using 
these  values  of  H,  we  can  find  the  areas  a,  and  then  with  these  areas  can  find  new  values  for 
H  and  new  areas. 

Example.— A  short  example  will  illustrate  the  use  of  equations  (i)  and  (3). 

Let  us  take  a  circular  arch,  the  radius  of  the  outer  chord  being  44  feet,  and  of  the  inner  cord  36  feet.  The 
apices  A,  a,  b,  d,  e,  B  are  on  the  outer  circle  and/",^,  h,  /on  the  inner  circle,  the  hinges  being  at  A  and  /?, 
and  bracing  as  shown, 

The  members  a/,  bg,  dh,  et\ 
are  radial,  and  the  chords  ab, 
i>d,  de  subtend  an  angle  of  30° 
at  the  centre,  while  Aa,  eB 
subtend  an  angle  of  15°. 

We  can  now  find  the  stress 
h  in  each  member  for  a  nega- 
tive thrust  of  unity  at  the  end 
A,  and  also  the  stress  p  in  each 
member  for  a  unit  load  at  a,  b 
d,  e,  considering  the  arch  as 
simply  supported  at  the  ends. 
We  can  then  draw  up  the  following  table.  In  the  first  column  we  have  the  members;  in  column  two  the 


CHAP.  IX.] 


TEMPERATURE  STRESS-FRAMED  ARCH  HINGED  AT  ENDS. 


58r 


lengths  s  of  the  members  ;  in  the  third  column  the  stresses  h  ;  in  the  fourth  column  the  quantities  h*s,  and  in 
the  following  columns  the  quantities /,/fo,  p^hs,  p*hs,  p^hs. 

The  minus  sign  for  a  stress  denotes  compression,  and  the  plus  sign  tension. 


• 

h 

A*s 

jM* 

>•** 

P*hs 

p<hs 

Aa  

11.4866 
22.7762 
22.7762 
22.7762 
11.4866 
13-1133 
18.6350 
18.6350 
18.6350 
13-1133 
8 
22.1005 
8 
22.1005 
8 
22.1005 
8 

—  0.4357 
-  0.4472 
—  1.6530 
-  0.4472 
—  0-4357 
+  1-3116 
+  2.6530 
+  2.6530 
+  2.6530 
+  1.3116 
+  0.1726 
-  1.4300 
+  1-3733 
o 
-r*  1-3733 
—  1.4300 
-f-  0.1726 

2.1805 
4-5550 
62.2339 
4-5550 
2.1805 
22.5587 
131.1615 
131.1615 
131.1615 
22.5587 
,   0.2383 
45-1933 
15.0876 
O 
15.0876 

45-1933 

0.2383 

+  4-0073 
-j-  7.6816 
-j-  21.1919 
+  1-5255 
+   0.7302 
-f-  15.7202 
+  39-0452 
+  15-6671 
+  15.6671 
+  1-5789 
0-4457 
7.1709 
+  2.2972 
o 
+   1.8017 
+   7-1835 
+  0.0802 

+   2.8652 
-f  10.8253 
+  85.9151 
5.1662 
2.7966 
+  H.24I5 

+  rn.'oooo 
+  59-9J47 
h  59.9I47 
+   6.0628 
0.5665 
+  50.9041 
+   8.6188 
o 
+   6.8917 
h  29.9284 
+  0.2719 

4-  2.7966 
+  5.1662 
+  85.9151 
+  10.8253 
+  2.8652 
+  6.0628 
h  59-9I47 
h  59-9I47 

+  III.  0000 

-  11.2415 
+  0.2719 
+  29.9284 
+  6.8917 

o 

+   8.  6188 
-  50.9041 
-f   0.5665 

+   0.7302 
+    1-5255 
+  21.1919 
-f    7.68l6 
+   4-0073 
+    L5789 
+   15.6671 

+  15  6671 
h  39-0452 
+  15-7202 
-f   0.0802 
+   7-1835 
+   1.8017 

0 

+   2.2972 
-   7-1709 
-   0.4457 

ab  

bd     

de 

eB  

Af.  .. 

fff 

/I..:::.::::::::: 

hi  

iB 

af.  .  . 

%  ::: 
bg  

/d... 

dh 

di  

635.3252 

+  126.561 

+  452.8835 

+  452.8835 

+  126.561 

From  equation  (3)  we  have  then  for  the  horizontal  thrust,  assuming  the  cross-sections  a  constant, 


H=P- 


. 

635.3252 

where  ^phs  is  126.561,  452.8835,  452.8835  and  126.561  for  the  load  P  at  a,  b,  dfaad  /. 
We  have  then  for  the  loads  Pi,  Pi,  P»,  P^  at  a,  b,  d,  e : 


P, 
0.712SP, 


0.7I2&P, 


0.2P. 


P* 

0.3506/", 


P< 

0.091 8P. 


H  =  o.2P, 
•  For  the  vertical  reaction  we  have,  from  equation  (i), 

P,  P* 

Fi=  0.90827',  0.6494^, 

With  these  values  of  H  and  V,  we  can  easily  find  the  stresses  in  every  member  for  each  load  P,  either 
by  computation  or  diagram.  Then  by  tabulation  we  can  find  the  loading  which  gives  the  maximum  stress 
and  the  maximum  stress  itself  in  each  member. 

Temperature  Stress— Framed  Arch  Hinged  at  Ends.— While  in  the  arch  with  three 
hinges  there  are  no  temperature  stresses,  in 
the  arch  hinged  at  the  ends  only  the  stresses 
due  to  change  of  temperature  may  be  con- 
siderable. 

The  effect  of  a  change  of  temperature 
is  to  cause  a  horizontal  thrust  at  the  ends. 
For  a  rise  of  temperature  we  have  a  positive  thrust  Ift.  For  a  fall  of  temperature  we  have 
a  negative  thrust  Ht.  If  Ht  is  known,  the  resulting  stresses  are  easily  found  cither  by  com- 
putation or  diagram. 

Let  e  be  the  coefficient  of  expansion,  and  t  the  number  of  degrees  rise  or  fall  of  temper- 
ature.     Then  the  change  of  length  of  the  span  is  elt.      The  work  is  then 


582  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  IX. 

The  moment  at  any  point  is  M  •=•  Hty.      If  v  is  the  lever-arm  for  any  member,   and  M 

the  moment  at  the  centre  of  moments  for  that  member,  the  stress  in  that  member  is  --  ' 

v 

Let  a  be  the  cross-section  of  the  member,  and  s  its  length.     Then,  from  equation  (III),  page 
515,  the  work  of  straining  that  member  is 

Ms 


The  total  work  for  all  the  members  is  then 

Htelt  =  ^ 
2         ^ 

Hence  we  have 


Suppose  the  arch  to  be  rigidly  fixed  at  the  right  end  and  free  at  the  left  end,  and  let  h 
be  the  stress  in  any  member  due  to  a  unit  horizontal  force  at  the  left  end.     Then  we  have 

V 

hv  =  I  X  y,  or  h  =  —  ,  and  hence  we  can  write 


<* 


where  the  summation  ^>    _  is  made  as  in  the  preceding  example,  /  is  the  number  of  degrees 

o  a 

rise   or   fall    of   temperature    above   or   below   the  mean   temperature   of   erection,    e  is  the 
coefficient  of  expansion  or  the  change  of  length  per  unit  of  length  for  one  degree,  and  E  is 
the  coefficient  of  elasticity  (page  478).      The  plus  (+)  sign  is  taken  for  rise  of  temperature, 
and  the  minus  (  — )  sign  for  fall  of  temperature. 
We  have  for  one  degree  Fahrenheit : 

For  cast-iron.  ...  e  =  0.00000617, 
wrought-iron.  .  e  =  0.00000686, 
steel e  =  0.00000599. 

Values  of  E  are  given  on  page  478. 

Equation  (3)  requires  that  the  areas  of  the  members  should  be  known  in  advance, 
whereas  these  are  what  we  wish  to  find.  We  must  in  general,  then,  first  assume  a  constant 
value  fora  in  (3)  equal,  say,  to  the  section  at  the  crown.  Denote  this  assumed  constant 
section  by  «0.  We  have  then  for  our  first  calculation 


CHAP.  IX.]  SOLID  ARCH  HINGED  AT  ENDS  ONLY.  583 

Example,  —  In  the  preceding  example,  page  580,  let  the  cross-section  at  crown  be  aa  =  2  square  inches 
Let  the  arch  be  of  steel,  and  let  us  take  £=  30000000  pounds  per  square  inch.  Let  the  change  of  temper- 
ature be  /  =  40°. 

Then    from    the    table   page    581   w.e    have    2  o  Ws  =  635.297.;  and  since  /=  76.21024  ft.,  we    have    for 

€  —0.00000599 

Ht  =  ±  1728  pounds. 
The  stresses  due  to  this  thrust  can  now  be  found. 

Solid  Arch  Hinged  at  Ends  Only.  —  For  any  load  P  we  have,  just  as  for  the  framed  arch 

page  579,  the  left  vertical  reaction 


We  have  found  on  page  579,  equation  (2),  for  the  horizontal  thrust  of  a  framed  arch 


If  the  arch  is  a  solid  beam,  we  can  put  ds  for  s,  and  ai?  =  I  =  moment  of  inertia  of  the 
cross-  section.     Hence  if  x  and  7  are  the  co-ordinates  of  any  point  of  the  neutral  axis,  we  have 


This  equation  is  general.      If  the  moment  of  inertia  /  of  the  cross-section  is  constant, 

/cancels  out. 

Instead  of  performing  the  integrations  indicated  in  (2)  we  can  in  any  case  divide  the 
neutral  axis  into  a  number  of  equal  arcs  of  length  s.      We  have  then,  since  s  cancels  out, 


.       T  *.  T 

0/         2-—     —  •     .  ....     (3) 


If  7  is  constant,  it  cancels  out. 

From  (i)  and  (3),  then,  we  can  find  Vl  and  H  for  any  given  load,  and  can  then  find  the 
moment  M  at  any  point  of  the  neutral  axis  for  a  load  anywhere.  Then  by  tabulation  we  can 
find  the  loading  which  gives  the  maximum  moment  at  any  point  of  the  neutral  axis,  and  this 
maximum  moment  Mm.A*.  itself. 

We  have  then,  from  equation  (II),  page  497, 


(4). 


where  ^Tmax  is  in  pound-feet,  Sf  is  the  allowable  unit  stress  in  pounds  per  square  inch  in  the 
most  remotS  fibre  at  a  distance  v  infect,  and  /  is  given  for  dimensions  in  inches. 


584 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  IX. 


'P. 


P, 


Example.— Let  us  take  a  circular   arch  of  radius  of  neutral  axis  40  ft.  and  central  angle  120°,   so  that 
69.282  ft.  and  rise  r  =  20  ft.     We  first  suppose  /  constant  so  that  it  cancels  out. 

Divide  the  neutral  axis  into  a  sufficiently  large  number  of  equal  segments,  say  six,  so  that  the  length  of  a 

segment  is  s  =  13.963  ft.,  and  let  the 
end  segments  Aa  and  Bf  be  each  one 
half  of  s. 

Let  the  ordinates  to  the  middle 
points  a',  V,  <t ',  etc.,  of  each  segment 
for  origin  at  A  be  x,y,  and  take  the 
loads  Pi,  F\,  etc.,  acting  half  way 

/  ,  x"|     .^^^  ^^<  ^XSV     X.  between  A  and  a',  a'  and  &,  V  and  c1, 

/  \  /  \     X.  X         etc 

In  order  to  apply  equation  (3), 
page  583,  we  can  then  draw  up  the 
following  table. 


Jf 

y 

xy 

s 

4a  

I  87* 

8  661 

ab 

be 

^68  6dJ. 

cd.  

34  641 

2O 

692  820 

400 

de  

48  ^22 

17  588 

849  887 

1OQ  1^8 

ef... 

60  SS2 

jo  642 

642  266 

113  252 

fR... 

8  661 

2750.590 

1253.841 

In  the  first  column  we  have  the  segments  Aa,  ab,  be,  etc.;  in  the  second  and  third  columns  the  values 
of  x  and^x  for  the  middle  points  of  these  segments;  in  the  last  two  columns  the  values  of  xy  and  7*. 


We  have  then  2txy  =  2750.590  and 


=  1253.841.     Note  that  in  finding  these  summations,  since  the 


end  segments  Aa  andyz?  are  only  half  length,  we  take  in  the  summations  one  half  the  values  for  Aa  andfB. 
We  have  now  for  the  values  of  kl  and  i  —  k  for  each  load : 

r>  r>  p>  T) 


54-337 
0.216 


A 

64.817 

0.065 

0.0657*.. 
[n  order  to  find  the  summations  2 kly(x  —  kl}  for  each  load  we  can  draw  up  the  following  table. 


Pi  Pi  P*  P* 

kl  =  4.465  14-945  27.8  41.481 

i  —  k  =  0.935  0.784  0.599  0.401 

Hence  from  equation  (i).  page  583,  the  values  of  Ft  for  each  load  are 

V\  =  o.93$Pi,        O.784/V         O.599/V        0.40  i/V        o.2\6Pt, 


PI- 

fk 

ft 

A 

fk 

Ft 

X    -  kl. 

X-r  -*/)• 

X  -   kl. 

X-r  -  */)• 

x-  kl. 

y(*-kt). 

x-kl. 

y(-«—  *0. 

x-kl. 

**-*/). 

x-kl.      \Ax-U). 

ab 
be 
ed 
de 
<f 

4.464 

16.495 
30.176 
43-857 
55-887 

47.506 
290.114 
603.520 
771-357 
594-749 

6.015 

19.696 

33-377 
45-407 

105.792 
393-920 
587.03< 
4S3.22J 

6.841 
20.522 
32.552 

136.820 
360.941 
346-418 

6.841 
18.871 

120.309 

200.825 

6.015 

64.911 

fB 

63.042 

179-533 

52.462 

154-396 

39.607 

116.563 

25.926 

76.300 

13.070 

38.465 

2-59 

7.622 

2397-012 

1653.366 

902.460 

359-284 

84.14- 

3.811 

We  have  then,  from  equation  (3),  page  583,  for  /  constant 


a 


2750.59(1  -k}  —  2 
1253.841 


CHAP.  IX.]  SOLID  SEMICIRCLE,  HINGED  AT  ENDS   ONLY.  585 

where  the  summation  2lkly(x  -  kl)  for  each  load  is  given  by  the  preceding  table.  Note  that  in  finding  these 
summations,  since  the  end  segments  Aa  and/£  are  only  half  length,  we  take  in  the  summations  one  half  the 
values  for  fB. 

We  have  then  for  each  load 

//  =  o.i39A,         0.401  A,        o.6A,        0.6  A,        0.401  A,        0.139  A. 

Since  we  now  know  //and  V\  for  each  load,  we  can  find  the  moment  M  at  the  centre  of  each  segment 
for  each  load.  By  tabulation,  then,  we  can  find  the  maximum  moment  J/max.  at  each  of  these  points.  Then 
from  equation  (4),  page  583,  we  can  find  /at  each  of  these  points. 

Solid  Semicircle,  Hinged  at  Ends  Only  —  Constant  /.  —  The  preceding  method  applies 
to  any  solid  arch  of  any  shape  and  any  loading.  In  the  case  of  a  semicircle  of  constant  / 
the  integrations  of  equation  (2),  page  583,  are  easily  made. 

ds       R  Rdx 

Thus  let  R  be  the  radius  of  the  neutral  axis.      Then  we  have  -7-  =  —  ,      or     ds  =  --  -. 

dx      y  y 

Inserting  this  value  of  ds  in  (2),  we  obtain 

H  I  ydx  =  P(i  —  k)    I    xdx  -  p  I    (x  —  kl)dx. 
J§  Jo  J  ki 

Cl  TfR2 

Now    /  ydx  is  the  area  A  =  —  :  —  of  the  semicircle.     Performing  the  other  integrations, 

Jo  2 

we  have,  since  /  =  2R, 

H=*Pk^-k\  .     .    ,     .     .     (4) 

7t 

Equation  (4)  gives  H  directly  for  a  load  P  anywhere. 

Temperature  Thrust—  Solid  Arch  Hinged  at  Ends.  —  We  have,  just  as  on  page  582,  for 
a  framed  arch 

Htelt  =        'fftfs 
2         ^jtE&F 

For  a  solid  arch  we  can  put  ds  for  s,  and  /  for  anr  ;   hence 


. 


where  /,  y,   ds  are  in  feet,   E  in  pounds  per  square  inch,  and  /  is  given  for  dimensions  in 


inches. 


. 
Instead  of  performing  the  integration,  we  can,  as  before,  divide  the  neutral  axis  into  a 

number  of  equal  arcs  of  length  s.     We  have  then 


These  equations  are  general. 


586 


STATICS  OF  EL  A 'STIC  SOLIDS. 


[CHAP.  IX. 


If  the  arch  is  a  semicircle,  we  have  ds  =  — — ,  and  (5)  becomes 


•44* 


(7) 


Let  70  be  the  moment  of  inertia  at  the  crown.       Then  for  our  first  calculation  we  have- 
in  general,  from  (6), 


144 


and  for  a  semicircle  from  (7),  since  JQydx  =  A  =  n- — , 


(9) 


In  all   equations,  distances  are  in  feet,  E  in   pounds  per  square  inch,  and  /  is  given  for 
dimensions  in  inches. 

Example.  — In  the  example  page  584,  let  the  moment  cf  inertia  at  the  crown  be  /„  =  2000  in.4.  Let  the 
arch  be  of  steel,  and  let  us  take  E=  30000000  pounds  per  square  inch.  Let  the  change  of  temperature 
/  =  40°. 


Then,  from  the  table  page  584,  we  have 
for  e  =  o.  00000599 


*s  =  17307.06  ft. s,  and  since  7  =  69.28  ft.,  we  have,  from  (8,1, 


Ht  =  ±  400  pounds. 
For  a  semicircle,  since  J?  =  40  ft.,  we  have,  from  (9), 

Ht  =  ±  72  pounds. 

Framed  Arch  Fixed  at  Ends. — Let  the  arch  be  fixed  at  the  ends,  and  a  load  P  act  at  the 

distance  /•/  from  the  left  end 
A  of  the  upper  or  lower  chord. 
In  this  case  we  have  at  A  not 
only  a  vertical  reaction  V ^ 
and  a  horizontal  thrust  H, 

\       but  also  a  moment  J/.. 

1 1. 

^    *  That   is,  the  resultant  A' 

of  l\  and  //,  instead   of  pass- 
ing through  A  as  in  the  case 
of  end  hinges,  page  579,  must 
now  pass  through  some  point 
Alat   a  distance  y\  vertically 
below  A,  so  that  —  Hyl  =  J/,. 
Let  T'  be  the  lever-arm  for  any  member  as  determined  by  the  method  of  sections,  page 
401,  and  a  the  cross-section  of  any  member,  and  s  its  length.      Let  .r,  y  be  the  co-ordinates 
for  origin  at  A  of  the  point  of  moments  for  any  member  as  determined  by  the  method  of 
sections,  page  401. 

Let  O  be  a  point  whose  co-ordinates  for  origin  at  A  are  given  by 

^^L  ^'* 

=t^ (•) 


av1 


CHAP.  IX.]  FRAMED  ARCH  FIXED  AT  ENDS.  5^7 

so  that  we  have 

^  (         ~\    S  "S'f         ~     S 

For  arch  symmetrical  with  respect  to  the  centre  we  have 

1  ^  ~    'S 

=  ~2J  o  ^~a^~  °' 

Draw  through  this  point  O  a  parallel  OA2  to  the  resultant  R  of  Vv  and  H. 

If  now  we  consider  O  as  a  fixed  point  rigidly  connected  to  the  arch  at  A2  by  members 
OA2  and  A2A,  we  can  remove  the  abutment  at  A  and  the  equilibrium  of  the  arch  will'not  be 
affected.  We  shall  then  have  at  O  the  reactions  Vl  and  H  and  a  moment  M0. 

For  any  member  on  the  right  of  P  we  have  the  moment 


(2) 

z  / 

For  any  member  on  the  left  of  P  we  have 

-~)  +  M^'       V    •     '•     '    /    ...-     (3) 
We  have  then,  just  as  on  page  579,  for  the  work  of  straining  all  the  members 


Since  H,   1\  and  M0  must  have  such  values  as  to  make  the  work  a  minimum,  we  place  the 
first  differential  coefficients  of  the  work  with  reference  to  //,  l\  and  M0  equal  to  zero.      Hence 


aT(work) 
dH 


But  since  for  symmetrical  arch 


ST MT1CS  OF  EL  A STIC  SOLIDS. 


[CHAP.  IX. 


these  equations  reduce  to 


=  -  p 


<(y-y}(x-kl)s 


(4) 


These  equations  give  M0,  Vl  and  H  for  a  load  P  anywhere  at  a  distance  £/  from  the 
left  end  A,  as  shown  in  the  figure,  page  586. 

Let  the  arch  be  supposed  to  be  fixed  at  the  right  end  and  free  at  the  left  end,  the  left 
support  being  removed,  and  in  this  condition  let  u  be  the  stress  in  any  member  for  a  unit 
load  at  the  left  end  A,  h  the  stress  in  any  member  for  a  negative  unit  horizontal  force  at  A, 
p  the  stress  in  any  member  due  to  a  unit  load  at  P,  m  the  stress  in  any  member  due  to  a 
negative  unit  moment  at  the  point  of  moments  for  that  member. 

Then  we  have  for  any  member 


±«=-,    ±*=£, 


±«=  —  -, 


X—  kl 


I 

2V 


I  y 

Hence    w2  =  —5,    mh  =  -~, 
3  % 


V1 

p(h  - 


V* 

(y  -  J](x  - 


We  have  then,  from  equations  (i)  and  (4),  for  a  load  P  any  where  on  a  symmetrical  arch 
at  a  distance  kl  from  the  left  end 


mhs 


y  = 


(5) 


CHAP.  IX.]  FRAMED  ARCH  FIXED  AT  ENDS.  589 

Equations  (5)  give  MQ  and  Vv  and  H  at  the  left  end  A  for  a  load  P  anywhere  on  a  sym 
metrical  framed  arch  at  a  distance  kl  from  the  left  end  A. 

If  we  take  moments  about  the  left  end  A  (figure,  page  586),  we  have 

for  load  P  on  right  of  the  centre       Ml=~^ ffy  +  M0, (6) 

where  M1  is  the  moment  at  the  left  end  A  for  a  load  anywhere  on  the  right-hand  half,  and 
H,   Vv ,  MQ  are  given  by  (5). 

If  we  take  moments  about  the  right  end  B,  we  have 

for  load  P  on  right  of  the  centre       M2  =  Ml  -  F/  +  Pl(i  —  k\        .......     (7) 

where  Ml  is  given  by  (6).      This  value  of  M2  is  the  same  as  the  moment  Mf  at  the  left  end 
A  for  a  similarly  placed  load  on  the  left  of  the  centre. 

If  Vl  as  given  by  (5)  is  the  reaction  at  the  left  end  for  a  load  P  on  the  right-hand  half, 
we  have  for  the  reaction  VJ  for  a  similarly  placed  load  on  the  left-hand  half 


(8) 


The  value  of  H  is  the  same  in  both  cases  whether  the  similarly  placed  load  is  on  the 
right  or  left  half. 

From  equations  (5)  and  (6)  we  can  find  Ml,  V^  and  //for  a  load  anywhere  on  the  right- 
hand  half  of  a  symmetrical  arch.  From  equations  (5),  (7)  and  (8)  we  can  then  find  V^  ,  H 
and  MI  at  the  left  end  A  for  a  similarly  placed  load  anywhere  on  the  left-hand  half. 

We  can  then  determine  the  stresses  in  the  members  for  a  load  P  on  either  half.  For  a 
first  approximation  we  can  take  the  cross-section  a  constant  for  all  members,  so  that  it  can- 
cels out  in  equations  (5). 

We  can  thus  find  the  stress  in  each  member  for  each  load,  and  then  by  tabulation  the 
maximum  stress  in  each  member  for  all  the  loads. 

Example.  —  Let  us  take  a  circular  arch,  the  radius  of  the  outer  chord  being  44  ft.  and  of  the  inner  chord 
36  ft.  The  apices  Ai,  a,  b,  d,  t,  Bi  are  on  the  outer  circle,  and  A*,f,gt  h,  /,  B*  on  the  inner  circle.  The 
bracing  as  shown. 

b      _  * 


The  members  of,  bg,  dh,  ei  are  radial,  and  the  chords  ab,  bd,  de  subtend  an  angle  of  30°  at  the  centre, 
while  Aia,  eBi  subtend  an  angle  of  15°. 

Considering  the  arch  fixed  at  the  right  end  and  free  at  the  left  end,  we  can  now  find  the  stress  u  in  every 
member  due  to  a  unit  load  at  A, ;  the  stress  m  in  every  member  due  to  a  negative  unit  moment ;  the  stress 
h  in  every  member  due  to  a  negative  unit  horizontal  force  at  Ai ;  and  the  stresses/, ,  p<  due  to  a  unit  load 


49°  STATICS  OF  ELASTIC  SOLIDS.  [CHAi>.  IX. 

at  </and  e  on  the  right-hand  half  of  the  arch.     We  can  then  fill  up  the  following  table.     The  plus  (+)  sign 
for  a  stress  denotes  tension,  and  the  minus  (— )  sign  compression. 


S 

m 

M 

M 

/I 

A 

mkt 

M* 

jt.a 

II  d86i 

ab 

22  7761 

4-  i.  6^60 

—  o.  1  204 

1.3164 

O.38O4 

bd 

A  87l8 

o  ^804 

de 

+1  820^ 

*B\ 
**f 
ft 

11.4863 

9-3979 
18  6140 

+  8.0149 
o 

-  0.4357 

O 

4-  2  6^O 

—  0.1260 
-f-  0.1260 

+  1-7736 

-  0.7132 

O.63O6 

O 

6  3974 

0.1929 
0.1579 

0.31  12 

fl 

18  6140 

6  4048 

6  3974 

O.31  12 

hi 

I  8  6l4Q 

6  3974 

O.3II2 

iBt 

Atf 

9-3979 
13  1128 

-  9.6085 

+  0.5043 

+  o.  i  260 

-  3-3684 

-  0.8815 

0.5967 
o 

0.1579 

o 

af 

g 

o  0690 

O.O2OO 

% 

+  2  4768 

o 

o 

z 

g 

O.O359 

A 

-f-  2  86OO 

o 

dh 

g 

_|_  o  0669 

o  0359 

di 

2  4768 

o 

ft 
iBi 

8 
13.1128 

-  3-1746 
+  1.0064 

+  0.1726 
+  1.3116 

+  0.0499 

o 

—  0.7026 
+  1.0064 

+  1.0064 

0.0690 
o 

O.O2OO 
0 

30.1625 

2.8882 

In  the  first  column  we  have  the  members;  in  column  two  the  lengths  s  of  the  members;  in  the  third 
column  the  stresses  u ;  in  the  fourth  and  fifth  columns  the  stresses  h  and  m;  in  the  next  two  columns  the 
stresses /i  and /« ;  finally,  in  the  last  two  columns,  the  products  tnhs  and  m*s.  We  see  from  the  table  that 

21  tnhs  —  30.1625      and       ^  tn*s  =  2.8882. 
We  have,  then,  from  the  first  of  equations  (5),  page  588, 

*=;=+  38,05ft.,       7=^=  +  .c.4433f, 

For  the  quantities  (u  +  -m),  (h  —Jm),  (u  +  -m)*s,  (h  —ym)*s,  pms,  p(u  +  -ni)s,p(h  —  ym)s,  we  can  now 
fill  up  the  following  table  : 


i 

/ 

, 

, 

lm 

2 

/4 

a 

>«<   42 

/.(   y  ) 

A,a 

-  3.2065 

+  0.8801 

118.0680 

8.8968 

ab 

-  3-2938 

+  0.9041 

247-0949 

18.6170 

bd 

—  1-2053 

-  0.3417 

2-6593 

de 

+  3  .  2947 

+  0.9041 

247.  2350 

18.6170 

—  5.3654 

+  I36.6IIO 

i   --  J.874 

+  3-2947 

+  0.8801 

124.6811 

8.896S 

-  2.5669 

+  1.0322 

+  67.1200 

—  26.9902 

+  17-9295 

—  7.2098 

A  if 

+  4.8012 

-  1.3158 

216.6358 

16.2710 

fg 

+  -4733 

+  1.3017 

40.4494 

31.5760 

-  -4741 

+  1-3017 

40.4930 

31.5760 

hi 

—  .4741 

+  1-3017 

40.4930 

31.5760 

iBt 

—  -8073 

—  0.8115 

217.6870 

6.1888 

-  3-9885 

-  1-0437 

+  152.1792 

+  39.8248 

+  25.6887 

+  6.7227 

Aif 

—  .0064 

+  1.3116 

13.2183 

22.5580 

•f 

+  -2735 

-  0-3495 

12.9744 

.0-9772 

fr 

+  -4768 

—  1.4300 

135.5730 

45-1920 

tf 

-  -2758 

+  0.6747 

0.6085 

3-6418 

g* 

+  .8600 

o 

180.7691 

0 

dh 

-  .7660 

+  0.6747 

4.6940 

3-6418 

di 

-  .4768 

-  1.4300 

135-5730 

45.1920 

o 



+  135-5755 



+  78-2755 



n 

-  .  2694 

-  0.3495 

12.8910 

0-9772 

—  0.3085 



+    7.1350 



+  o.  1964 



iB\ 

+  .0064 

+  1.3116 

13.2183 

22.5580 

o 

0 

+  13.2811 

-fI3.28lI 

+  I7-30S7 

1-17.30*7 

1835.5346 

319.6127 

I  12.-J  1M,; 

—  0.0115 

+  5II.90I8 

+  26.1157 

+  176.8662 

-»  r6.8sx6 

We  have  then,  from  equations  (5),  page  588  and  the  tables, 


2".  8882 


CHAP.  IX.] 


TEMPERATURE  STRESS— FRAMED  ARCH  FIXED  AT  ENDS. 


59' 


and  hence  for  the  apex  loads  A  ,  A  at  apices  d,  e 

apex  d 

M0  =  —  4.  2342  A 
We  have  also 


apex  * 
-  0.004  A 


I835.5346 


and  hence  for  the  apex  loads  A  and  A  at  apices  d,  e 

apex  d 
Fx  =  +0.2788  A 


We  have  also 


apex* 
+  0.01427*4 


2U  (*  -J 


apex  b 
+  0.55347*,, 

apex  d 
+  0.5  534  A 

apex  e 
+  o.o  5  26  A 

apex  b 

+  0.721  27*i 

apex  d 
+  0.2788  A 

apex  e 
+  o.o  1  42  A 

apex  b 
26.7172 

apex*/ 

apex  e 

+  6.0801  7*9 

+  o.6ioi7*t 

—  O.O  I  22  A 

319-0127 
and  hence  for  the  apex  loads  A  and  A  at  apices  'd,  e 

apex  d  apex  e 

H  =  +  0.5  5  34  A  +  0.05  26  A 

We  have  the  same  values  of  H  lor  similarly  placed  loads  at  apices  a,  b.     Hence 

apex  a 

H  =  +  0.0526A 
Also,  from  (8)  we  have 

apex  a 

F,  =  +  0.98  58  A 
From  (6)  and  (7)  we  now  have 

apex  a 

(/-*/)=       6.9925 
Mi  =  +  5.89827*. 

We  can  now  find  the  stresses  in  each  member  for  each  apex   load,  and  then  by  tabulation  find  the 
maximum  stress  in  each  member. 

Temperature  Stress— Framed  Arch  Fixed  at  Ends. — The  effect  of  a  change  of  tempera- 
ture is  to  cause  a  horizontal  thrust 
Ht  at  the  point  O,  or  a  horizontal 
thrust  HI  and   moment  Mt  at   the 
left  end  A,  where 

For  the  moment  at  the  point         '* 
of  moments  for  any  member  we  have 

We  have  then  for  the  work  of  straining  all  the  members 

work  = 


If  e  is  the  coefficient  of  expansion  and  /  the  number  of  degrees  rise  or  fall  of  tempera- 
ture, the  change  of  length  of  the  span  is  elt.      The  work  is  then 

Htelt 


work  = 


Hence 


Htelt 


592 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  IX. 


or 


Eelt 


or,  just  as  on  page  588, 


Eelt 


Where  considering  the  arch  fixed  at  the  right  end  and  free  at  the  left,  h  is  the  stress  in 
any  member  for  a  negative  unit  horizontal  force  at  the  left  end  A,  and  m  the  stress  in  any 
member  due  to  a  negative  unit  moment  at  the  point  of  moments  for  that  member. 

1  s 

The  summation^  (h  — ymf-  is  made  as  in  the  preceding  example.     The  plus(-}-)  sign 
o  « 

is  taken  for  rise  of  temperature,  and  the  minus  (— )  sign  for  fall  of  temperature. 

The  values  of  e  are  given  on  page  582,  and  of  E  on  page  478.  The  value  of  Mt  at  the 
left  end  A  is  given  by  (i)  when  Ht  is  known. 

Equation  (2)  requires  that  the  areas  a  of  the  members  should  be  known  in  advance. 
We  must  then  in  general  assume  a  constant  value  for  a  in  (2)  equal,  say,  to  the  section  at  the 
crown.  Denote  this  assumed  constant  cross-section  by  a0.  We  have  then  for  our  first  cal- 
culation 

Eeajt 


Ht=  ± 


(3) 


(h  -  ynifs 


Example. — In  the  preceding  example,  page  589,  let  the  cross-section  at  crown  be  a  =  2  square  inches. 
Let  the  arch  be  of  steel,  and  let  us  take  E  =  30000000  pounds  per  square  inch.  Let  the  change  of  temper- 
ature be  /  =  40°. 

Then  from  the  table  page  590  we  have   ^>  (h  —  ~ym?s  =  319.6127,  and  since  /=  76.21  ft.,  we  have  for 

i 
e  =  o.ooooo  599,  from  (3), 

Ht  =  ±  3428  pounds, 

and  from  (2),  since  jT  ;=  10.4433  feet, 

Mt  =  T  35800  pound-feet. 

Solid  Arch — Fixed  at  Ends. — If  the  arch  is  a  solid  beam,  we  can  put  in  equations  ^4), 
page  588,  ds  for  s  and  /  for  av*,  where  7  is  the  moment  of  inertia  of  the  cross-section.  Hence 
if  x  and  y  are  the  co-ordinates  of  the  neutral  axis,  we  have  for  a  load  P  anywhere  on  a  sym- 
metrical arch 

A*-*)* 


~ 


—       »/o 

y 


ds 


yds 


? 


(0 


CHAP.  IX.] 


SOLID  ARCH— FIXED  AT  ENDS. 


593 


These  equations  are  general.      If  the  moment  of  inertia  /  is  constant,  it  cancels  out. 

Instead  of  performing  the  integrations  indicated  inequations  (i),  we  can  in  any  case  divide 
the  neutral  axis  into  a  number  n  of  equal  segments  of  length  s.  We  have  then,  since  s  is 
constant, 


or,  if  /  is  constant, 


_  P 


where  n  is  the  number  of  segments; 


=  -P 


If  /is  constant,  it  cancels  out  in  these  two  equations. 


or,  if  /  is  constant, 


y  = 


= 


•2 

•»/ 


where  n  is  the  number  of  segments. 

From  (i)  or  (2),  then,  we  can  find  M0  and  Vl  and  H  at  the  left  end  A  of  the  neutral 
axis  for  a  load  P  anywhere  on  the  right-hand  half  of  a  symmetrical  arch.  We  have  then  for 
the  moment  M^  at  the  left  end  for  a  load  on  the  right-hand  half,  just  as  on  page  589, 


(3) 


and  for  the  moment  M2  at  the  right  end,  or  for  a  similarly  placed  load  on  the  left-hand  half, 
the  moment  M,'  at  the  left  end, 


(4) 


594  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  IX. 

The  reaction  l\'  at  the  left  end  for  a  similarly  placed  load  on  the  left-hand  half  is 

(5) 


V  —  P  —  V 
y  \  —  *         v  r 


The  value  of  H  is  the  same  in  both  cases,  whether  the  similarly  placed  load  is  on  the 
right-  or  left-hand  half. 

From  equations  (i)  or  (2)  and  (3)  we  can  find,  then,  J/,,  Vl  and  //at  the  left  end  for  a 
load  anywhere  on  the  right-hand  half  of  a  symmetrical  solid  arch.  From  (i)  or  (2)  and  (4) 
and  (5)  we  can  find  Mt't  V{  and  //at  the  left  end  for  a  load  anywhere  on  the  left-hand  half. 

We  can  now  find  the  moment  M  at  any  point  of  the  neutral  axis  for  a  load  anywhere. 
Then  by  tabulation  we  can  find  the  loading  which  gives  the  maximum  moment  at  any  point 
of  the  neutral  axis,  and  this  maximum  moment  Mm^  itself.  We  have  then,  from  equation 
(II),  page  497, 


r  


(5) 


where  ^/max.  is  in  pound-feet,  S/is  the  allowable  unit  stress  in  pounds  per  sqiiare  inch  in  the 
most  remote  fibre  at  a  distance  v  in  feet,  and  /is  given  for  dimensions  in  inches. 

Example.  — Let  us  take  a  circular  arch  of  radius  of  neutral  axis  40  ft.  and  central  angle  120°,  so  that  the 
span  /  =  69.282  ft.  and  rise  r  =  20  ft.     We  first  suppose  /constant,  so  that  it  cancels  out. 

Divide  the  neutral 
axis  into  a  sufficiently 
large  number  of  equal 
segments,  say  six.  so  that 
the  length  of  a  segment  is 
s=  13.963  ft.,  and  let  the 
end  segments  Aa  and  fit 
be  each  one  half  of  s. 

Let  the  ordinates  to 
the  middle  points  a',  b',  tf, 
etc.,  of  each  segment  for 
origin  at  A  be  x,  y,  and 
take  thejloads  A,  P\,  etc., 
acting  half  way  between  A 
and  a',  a'  and  b',  etc. 

We  can  now  draw  up  the  following  table. 


X 

y 

/ 

jr  —  — 

2 

('-;)' 

y  -y 

OK  -7)" 

Aa 

1.875 

2.943 

-  32-766 

1073.611 

—  10.291 

105.005 

01 

be 
fd 
de 
<f 

8.929 

20.960 
34-641 
48.322 
60.3525 

10.642 
17.588 
20 
17.588 
10.642 

-25.711 
-  13.681 
O 
-t-I3.68l 

-1-25-7" 

661.081 
187.170 

0 

187.170 

66:.o8i 

-2.592 
-f  4-354 
+  6.766 

+  4-354 
-2.592 

6.718 
18.957 
45-779 
18.957 
6.718 

fB 

67.407 

2.943 

+  32-766 

1073.611 

—  10.291 

105.905 

79-403 

2770-113 

203.034 

We  have  then  "2,*^  =  79.403.  Note  that  in  taking  this  summation  and  the  other  summations  of  the 
table  since  the  end  segments  Aa  and/7?  are  only  half  length,  we  take  in  the  summations  one  half  the -values 
for  Aa  an<ifB. 


CHAP.  IX.] 

We  have  then 


TEMPERATURE   THRUST— SOLID  ARCH  FIXED  AT  ENDS. 


,3-34  ft.. 


595 


and  can  fill  out  the  last  two  columns. 

For  the  values  of  kl,  (i  —  k)  and  /(r  —  k)  for  each  load  we  have  now 


Pi 

A 

P, 

P. 

P* 

P. 

kl=  4.465 

14-945 

27.8 

41.481 

54-337 

64.817 

I  -  k  =  0.935 

0.784 

0.599 

0.401 

0.216 

0.065 

/(l  —  k)  =  64.817 

54-337 

41.481 

27.8 

14.945 

4.465 

We  can  now  draw  up  the  following  table  for  loads  P4 ,  Pt ,  Pf. 


pt 

PI 

/'. 

x  —  kl 

H«-> 

(y-y)(x-kl) 

x  -  kl 

H<~"> 

(y-J)(.x-kl) 

x  -  kl 

H<-» 

(y  --&*-*!) 

de 
<f 

fB 

+    6.841 
+  18.872 

+  93-592 
+485.237 

+    29.786 
—    48.916 

+    6.016 

+154-683 

~    15.593 

+  25-Q26 

+849.491 

+266.802 

+  13.070 

+428.252 

-I34.503 

+   2.590 

+  84.864 

-  26.654 

+38-676 

+1003.574 

—  152.531 

+  12.551 

+368.809 

—    82.845 

+    1-295 

+  42.432 

-  13-327 

—  6.446^4, 


—  2.092/V 


-O.SS8P.; 
+  2.8467%; 
+  0.98477%. 


Note  that  in  taking  the  summations,  since  fB  is  of  half  length,  -we  take  one  half  the  values  for/5  in 
summing  up. 

We  have  then  from  these  tables  and  equations  (2),  page  593,  and  (3),  (4),  (5),  page  594,  for  loads  7%, 7% ,7% : 

-  o.2i67%; 
+  0.01537%; 

7/ =  +  0.75137%,  +  0.40817%,  +  0.06587%; 

M ,  =  —  3.8387%',  —  2.8827%  , 

Mi'=  —  1.1387%,  +  2.8427%, 

Vi   =  0.6377  7% , 

Hence  we  have  at  the  left  end  A  for  each  load  : 
M i  =  +  2.8467%  ,  +  2.8427%  , 

Vi  —  +  0.98477%  ,         +  0.86697%  ,         + 
H  =  +  0.06587% ,         +  0.408 1  Pt,         +0.75137%,         +0.75137%,         +0.40817%,         +  0.06587%. 

We  can  now  find  the  moment  M  at  the  centre  of  each  segment  for  each  load.  Then  by  tabulation  we 
can  find  the  maximum  moment  J/max.  at  each  of  these  points.  Finally,  from  equation  (5),  page  594,  we 
can  find  7  at  each  of  these  points. 

Temperature  Thrust — Solid  Arch  Fixed  at  Ends. — We  have,  as  on  page  591,  for  a 

framed  arch 

Htelt  =  ^lH?(y  -JJs 
2      ~  *o       zEav* 


-  2.882P., 


For  a  solid  arch  we  can  put  ds  for  s,  and  the  moment  of  inertia  /  for  the  cross-section  in 
place  of  atf.     Hence 

Ht=±- 


Belt 


where  /,  y,  ds  are  in  feet,  E  in  pounds  per  square  inch  and  /  is  given  for  dimensions  in  inches. 


596  STATICS  OF  ELASTIC  SOLIDS.  [Cn,\r.  IX. 

Instead  of  performing  the  integration  we  can,  as  in  the  preceding  example,  divide  the 
neutral  axis  into  a  number  of  equal  parts  of  length  s.     We  have  then 


For  the  moment  at  the  left  end  we  have,  as  on  page  591, 

Mt=-HtJ.     .     .'    „'     .-    .     .     .     ......     (3) 

Let  70  be  the  moment  of  inertia  at  the  crown,  then  for  our  first  calculation  we  have 


t  ..      ........ 

144  ;>     (y   -  yJS 

0 

Example,  —  In  the  example  page  594  let  the  moment  of  inertia  at  the  crown  be  2000  in.4.  Let  the  arch 
be  of  steel,  and  let  us  take  E  =  30000000  pounds  per  square  inch.  Let  the  change  of  temperature  be 
/  =  40°. 

Then  from  the    preceding   example   we  have  ^   (y  —  yf  s  =  4714.18,  and  since  /=  69.28  ft.,  we  have, 

from  (4),  for  e=  0.00000599 

Ht  =  ±  1467  pounds. 
Since  y  =  10.38  ft.,  we  have,  from  (3), 

Mt  =  T  1  5227  pound-ft., 

the  top  signs  being  taken  for  expansion  and  the  bottom  signs  for  contraction. 


CHAPTER   X. 

THE     STO  NE    ARCH. 

Definitions. — The  stone  arch  consists  of  a  number  of  arch-stones  or  voussoirs  which 
press  upon  each  other.  The  central  one  of  these  is  the  keystone.  The  extrados  is  the 
exterior  outline  of  the  arch  proper.  The  intrudes  is  the  interior  line,  and  the  corresponding 
surface  of  the  arch  is  the  soffit.  The  sides  of  the  arch  are  the  haunches,  and  the  spaces 
above  are  the  spandrels.  The  ends  of  the  arch  or  the  area  between  in  trades  and  extrados 
are  \\\Q  faces.  The  inclined  surfaces  or  joints  upon  which  the  arch  rests  at  the  ends  are  the 
skewbacks.  The  permanent  load  supported  by  the  arch  in  addition  to  its  own  weight  is  the 
surcharge.  The  masonry  or  other  material  which  supports  two  successive  arches  is  the  pier; 
at  the  extreme  ends  this  is  the  abutment. 

Thus  in  the  figure  cdd ' c'  is  a  voussoir  or  arch-stone,  and  k  is  the  keystone.  The  line 
abed,  etc.,  is  the  extrados,  and  a'b'c'd' ',  etc.,  the  intrados,  and  the  interior  surface  corre- 
sponding the  soffit.  The  line  rrr  marks  the  upper  limit  of  the  surcharge.  Between  this 


and  the  haunches  on  either  side  is  the  spandrel  space,  and  the  material  with  which  this 
space  is  filled  is  the  spandrel  filling  or  surcharge.  The  area  between  the  extrados  abed, 
etc.,  and  the  intrados  a'b'c'd' ,  etc.,  is  the  face,  and  aa'  is  the  skewback.  At  A  and  B  \ve 
have  the  abutments  or  piers.  Arch  and  abutments  or  piers  are  stone.  The  surcharge  may 
be  stoue  or  filled  in  with  rubble  or  lighter  material. 

The  upper  limit  of  surcharge  may  be  level  or  on  any  desired  grade.  The  extrados 
and  intrados  may  be  circular,  elliptic  or  any  desired  curve,  and  may  or  may  not  be  parallel. 
Often  the  depth  at  key  is  less  than  at  ends. 

In  all  investigations  and  calculations  we  suppose  the  width  to  be  one  foot  and  the  effect 
of  mortar  between  the  joints  is  disregarded.  The  neutral  axis  A CB  or  centre  line  is  the 
curve  passing  through  the  centre  of  the  voussoirs.  The  rise  r  of  this  axis  is  the  rise  of  the 
arch  and  the  span  /  of  this  axis  is  the  span  of  the  arch. 

597 


598 


ST/1TICS   OF  EL/1  STIC  SOLIDS. 


[CHAP.  X. 


Reduced  Surcharge. — The  arch  proper  is  constructed  of  cut  stone.  The  material  above 
it  may  also  be  of  stone  or  of  some  lighter  material.  The  density  or  weight  per  cubic  foot 
of  the  surcharge  is  then  in  general  less  than  the  density  of  the  arch  proper. 

Thus  in  the  figure  let,  abed,  etc.,  be  the  extrados,  and  rrr  the  roadway.  Between  the 
roadway  and  the  extrados  the  surcharge  may  have  a  less  density  than  for  the  arch  proper. 


Say,  for  instance,  that  the  density  of  the  surcharge  is  —  of  that    of  the    arch. 


r. 


Then  if  we 


•jflf^"*^ 


2  "* 

draw  verticals  aa',  bb'  ,  cc'  ,  etc.,  and  lay  off  aa"  equal  to  —  of  aa1  ',  bb"  equal  to  —  of  bb'  and  so 

on,  we  obtain  the  line  a"b"c",  etc.,  which  marks  the  limit  of  the  reduced  surcharge.  We 
can  then  treat  and  discuss  all  the  area  below  this  line  as  if  it  were  homogeneous  and  of  the 
same  density  as  the  arch  itself. 

Pressure  Curve.  —  Let  Fig.  (a)  represent  one  half  an  arcli  with  its  surcharge  rr,  and  a'f 

the  line  of  reduced  surcharge,  so  that 

all  the  area  below  can  be  considered 

t 

as  homogeneous  and  of  the  same  den- 
sity ai  the  arch. 

Let  us  take  the  width  as  one 
foot,  and  divide  the  area  into  a  suit- 
able number  of  slices  by  vertical  lines. 
In  this  case  we  have  five.  The 
weight  and  centre  of  mass  of  each 
slice  can  be  found.  Let  I,  2,  3,  4, 
5  represent  the  weights  acting  at  the 
centres  of  mass.  Let  H  be  the  hori- 
zontal thrust  at  the  crown  due  to  the 
pressure  of  the  other  half  of  the  arch. 
Let  the  magnitude  and  point  of  action 
a  of  //  be  known.  In  Fig.  (b)  lay 
off  the  weights  I,  2,  3,  4,  5  to  scale, 
let  Oil  be  the  known  thrust  to  scale, 

Then  O\   is   the  resultant   of  H  and  weight  i  ;    (92   is  the 
is  the   resultant  of  (?2  and  weight  3.  and  so  on. 


and  draw  Oi,  02,  #3,  04,  O$. 

resultant  of  O\  and  weight  2  ; 


In  Fig.   (rf)  produce  //acting  at  a  till   it  meets  weight  I.      From  I  draw  1-2  parallel  to 
Ol  till  it  meets  weight  2  ;  from  2  draw  2-3  parallel  to  02  till  it  meets  weight  3,  and  so  on 


CHAP.  X.]  CONDITIONS   OF  STABILITY  OF   THE  ARCH.  599 

We  thus  have  a  polygon  a  I2$4.$f,  each  segment  of  which  is  in  the  direction  of  the 
resultant  of  the  forces  acting  at  its  end.  Thus  the  resultant  of  H  and  weight  I  is  in  the 
direction  1-2,  the  resultant  of  Oi  and  weight  2  is  in  the  direction  2-3  and  so  on. 

As  we  increase  the  number  of  divisions  this  polygon  approaches  a  curve  tangent  at  the 
points  of  division  a,  b,  c,  d,  e,  f.  This  curve  is  the  curve  of  pressures  in  the  arch. 

Conditions  of  Stability  of  the  Arch. — The  conditions  for  stability  of  the  arch  are  the 
same  as  for  walls,  page  425.  Thus: 

ist.  The  joints  of  the  voussoirs  should  be  so  arranged  that  the  tangent  to  the  curve  of 
pressure  at  each  joint  shall  make  an  angle  less  than  the  angle  of  friction  with  the  normal  to 
the  joint,  otherwise  there  is  danger  of  sliding. 

2d.  The  curve  of  pressures  must  lie  entirely  within  the  arch,'  otherwise  there  is  danger 
of  rotation. 

3d.  The  curve  of  pressure  must  not  approach  too  near  the  edge  of  a  joint,  otherwise 
there  is  danger  of  crushing. 

Let  d  be  the  depth  of  any  joint,  N  the  normal  pressure  on  the  joint,  and  e  the  least  edge 
distance  of  N.  Then  for  a  width  of  one  foot  we  have,  as  already  shown  page  426,  for  the 
maximum  unit  pressure/, 

when  e  is  greater  than  -d,          p  =  —7- (2  —  -j  j  ; 

I  2N 

when  e  =  -a,          p  =  -j- 

I  2N 

when  e  is  less  than  -d,         p  — . 

0  o 

In  any  case  the  value  of  /  must  not  exceed  the  allowable  compressive  unit  stress  C, 
which  for  stone  may  be  taken  at  the  average  value  of  25  tons  per  square  foot  or  50000 
pounds  per  square  foot.  (See  table  page  424.) 

VALUE  OF  N. If  we  make  the  joints  nearly  at  right  angles  to  the  curve  of  pressure, 

the  first  condition  of  stability  is  complied  with,  and  we  have  with  sufficient  accuracy,  if  we 
denote  by  Pn  the  sum  of  all  the  loads  between  the  crown  and  any  joint  on  the  right, 

VALUE  OF  e. It  Mis  the  moment  at  any  point  of  the  neutral  axis,  then  ^  is  the  distance 

of  N  from  that  point.  If  we  subtract  this  from  ^  we  have  for  the  edge  distance  c'  from  the 
intrados,  with  sufficient  accuracy, 


If  ,'  at  any  joint  is  negative  or  is  positive  and  greater  than  4,  the  equilibrium  cun  e  runs 
outside  of  the  arch. 

If  M  is  negative,  /  is  greater  than  -  and  the  least  edge  distance  is 


II  Mi*  positive,  /  is  less  than  j  and  the  least  cd-e  distance  is 

e—  <?', 


6oo 


STATICS   OF  ELASTIC  SOLIDS. 


[CHAP.  X. 


We  see,  then,  that  if  in  any  case  we  can  find  the  values  of  77  and  the  moment  M  at  any 
point  of  the  axis,  we  can  find  the  normal  pressure  TV  at  any  joint,  the  distance  of  the  normal 
pressure  from  the  intrados  and  the  maximum  unit  pressure/  on  the  joint. 

If  the  normal  pressure  at  every  joint  is  within  the  arch  ring,  and/  does  not  exceed  the 
allowable  unit  stress,  and  the  joints  are  taken  perpendicular  or  nearly  so  to  the  pressure  line, 
the  arch  is  stable  at  every  point. 

Determination  of  77 and  M. — From  equations  (2),  page  593,  we  have  already  for  a  solid 
arch  fixed  at  the  ends 

/*, 


y  = 


, 

'of 


or,  if  7  is  constant, 


X. 

n 


where  n  is  the  number  of  segments; 


=  -  P — 


or,  if  7  is  constant, 


(4) 


If  7  is  constant,   it  cancels  out  in  the  last  two  equations.      In  these  equations  7  is  the 
,  —*£---  moment   of  inertia  of   the 

I  p 

\ £_^^  cross-section,  77  the  hori- 

zontal thrust,  F,  the  reac- 
tion at  the  left  end  A  of 
the  neutral  axis  for  a  load 
T'at  a  distance  kl  from  A. 
The  neutral  axis  is  divided 
*i  into  a  number  n  of  equal 
segments,  Aa,  nb,  etc.,  of  length  s,  and  x,  y  are  the  co-ordinates  of  the  middle  point  of  a 
segment  for  origin  at  A.  The  load  Tracts  half  way  between  the  ends  of  a  segment. 


r 


CHAP-  X  -1  THE  STONE  ARCH.  60  1 

We   have  then  for  the  moment  Ml  at  the  left  end  A  for  a  load  P  on  the  right-hand 
half,  just  as  on  page  593, 


M0t      .-.     .     .  v     .    .:.     .     .     (5) 

and  for  the  moment  M{  at  the  left  end  for  a  similarly  placed  load  on  the  left-hand  half 

M{=  Mv—VJ+Pl(\-  k}  ...........     (6) 

The  reaction   F/  at  the  left  end  for  a  similarly  placed  load  on  the  left-hand  half  is 


The  value  of  H  is  the  same  in  both  cases,  whether  the  similarly  placed  load  is  on  the 
right-  or  the  left-hand  half. 

From  equations  (4)  we  can  find  MQ  and  V^  and  //at  the  left  end  for  a  load  on  the  right- 
hand  half,  and  then  from  (5),  (6)  and  (7)  can  find  //  and  Ml  and  Vl  at  the  left  end  for  each 
load.  By  summation,  then,  we  can  find  2Ml  ,  ^'Ft  and  2H  at  the  left  end  for  all  the  loads. 

For  the  moment  at  any  point  of  the  neutral  axis  we  have  then 


}kl  ......     (8) 

This  is  the  value  of  M  to  be  used  in  equation  (3),  page  599,  and  ^H  is  the  value  of  H 
to  be  used  in  equation  (2),  page  599.  We  thus  finally  have  N  and  e,  and  then  from  equa- 
tions (i),  page  599,  can  see  if  the  maximum  unit  pressure/  is  less  than  the  allowable  unit 
stress  at  any  joint. 

It  will  generally  be  necessary  to  first  assume  some  constant  depth,  so  that  /  is  constant 
and  cancels  out  of  equations  (4).  We  can  then  determine  for  this  assumed  case  H,  e  and 
/  at  the  crown  and  N,  e  and/  at  the  skewback.  If/  at  either  point  is  too  great,  we  can 
increase  the  depth  assumed;  if  very  small,  we  can  decrease  it.  If  H  or  N  takes  effect  out- 
side of  the  middle  third,  then,  as  we  have  seen  page  426,  the  entire  joint  is  not  effective 
and  we  can  assume  a  depth  of  3^  at  crown  and  skewback.  We  thus  obtain  proper  depths  at 
orown  and  skewback  and  can  then  make  a  second  and  final  calculation. 

In  many  cases  much  computation  can  be  avoided  by  drawing  the  arch  and  reduced  sur- 
charge to  scale  and  taking  off  many  of  the  required  quantities  directly  from  the  draw- 
ing. Having  found  H  and  e  at  the  crown,  we  can  then  construct  the  curve  of  pressures  as 
in  the  figure  page  598. 

The  method  of  calculation  thus  outlined  adapts  itself  to  any  shape  of  arch  with  any 
surcharge.  Its  detailed  application  wiH  be  perfectly  illustrated  by  the  following  examples. 

Examples.—  1  1)  Required  to  design  a  circular  arch,  radius  of  neutral  axis  40  ft.  and  central  anqle  120°, 
so  that  I  —  69.282  ft.  The  allowable  unit  stress  to  be  23  tons  per  square  foot,  and  the  surcharge  level  with  the 
crown  of  extrados  and  cf  the  same  material  as  the  arch,  density  of  both  165  Ibs.  per  cubic  foot. 

Since  we  first  assume  a  constant  depth.  /cancels  out  in  equations  (4).  page  600.  and  we  proceed  precisely 
as  for  the  solid  arch  of  the  same  dimensions  given  in  the  example  page  594.  It  will  not  be  necessary,  there- 
fore, to  repeat  here  this  portion  of  the  calculation.  Referring  to  the  results  there  obtained,  we  have  for  the 
values  of  M\  ,  V\  ,  H  for  loads  P\,  Pi,  etc., 


Mi  =  +  2.8467',  +2.842T3,  -i.itfP,  -3-838A  -2.8827%  »o5587>. 

V,  =  +  0.98477',  +o8669/>3  +  0.6377  K  +  o.3623/>4  +  o.i33i7>.  +0.01537'. 

H  =  +  0.065873,  +o.4o8i7>,  +0.7513^  +  o.75r37\  +  0.4081  P.  +  0.06  587>. 


602 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  X. 


Let  us  first  assume  the  constant  depth  at  2  feet.     If  we  take  one  foot  width,  we  have  for  the  loads 

J\  =  P.  =  25966, 
P*  =7'§=  16253, 
Pt  =  Pt=  7169  pounds. 

Hence 

2Afi  =  +  23087  pound-feet, 
2  V\  =  +  49388  pounds, 
2H  =  +  27455  pounds. 

For  the  values  of  2  f(x  ~  M)  we  have  for  the  points  of  division  A,  a',  b' t  cf,  d'.  e1 ',  B 


4.4647*1  16.4957*1  +  6.0157*,  30.1767*,  +  19.6967*,  + 


115912 


526071 


1152712 


+  33-377A  +  27. 3637*, 
1877432 


55.881  A  +  51.422(7',  +  P») 
2655412 


2oP(*-W  =  69.282(7',  +  P,  +  71,) 
3421700 

We  have  then  from  equation  (8),  page  601,  for  the  moment  at  any  point  of  the  axis 

M  =  +  23087  -  49388-r  +  27455^  +  ^9P(X  ~  ft). 
For  the  points  ^ ',  d' ,  /,  B  we  have  then 


x  =  34.641 
=  20 


d' 
48.322 

17.588 


60.3525 
10.642 


B 
69.282  ft. 

o          ft. 


M  =  +  14050  —  3130  —  10014  +  23087  pound-feet. 

From  equation  (2),  page  599,  we  have  then 


d' 

28375 


B 

56506  pounds. 


N  =  27455 
For  the  edge  distance  of  N  from  the  tntrados  we  have  now,  from  equation  (3),  page  599, 


e1  —  0.488  1. 1 1  1.27 

and  hence  the  least  edtfe  distance  of  N  at  each  point  is 


t  =  0.488 


d' 
0.99 


0.73 


B 
0.592  ft,, 


B 

0.592  ft. 


CHAP.  X.]  THE  STONE  ARCH— EXAMPLES.  603 

We  have  then,  from  equations  (i),  page  599,  for  the  maximum  unit  pressure p 

<?  d'  e'  B 

P  =  375°7  H7-55  32840  63633  pounds  per  square  foot. 

Taking  the  allowable  unit  stress  at  25  tons  or  50000  pounds  per  square  ft.,  we  see  that  this  is  exceeded 
only  at  B.  Also,  we  see  that  the  least  edge  distance  at  crown  and  springing  is  less  than  one  third  the  depth. 
We  ought,  then,  to  have  the  depth  at  springing  greater  than  2  ft.  At  the  crown  we  can  have  a  less  depth 
than  2  ft.  if  we  wish.  As  we  have  seen,  page  426,  if  N  is  outside  of  the  middle  third,  the  entire  joint  is  not 

brought  into  action.     If  then   we  take  the  depth  at  springing  B  equal  to  —  =  —  —  =  2.26  ft.,  the 

whole  joint  there  will  act  and  the  allowable  unit  stress  not  be  exceeded.     We  may  take  this  constant  depth, 

*2,N        2    X    °74.^^ 

or  take  the  depth  at  crown  equal  to        = _£±i2  =  n  ft.     The  loads  Pi.  7*,   etc.,  will  now  he  somewhat 

changed.  To  allow  for  this  let  us  take,  say,  a  uniform  depth  of  2.5  ft.  Or,  if  preferred,  we  may  take  2.5  ft. 
at  springing  and,  say,  1.5  ft.  at  crown.  In  the  latter  case  /  will  vary.  The  calculation  can  now  be  repeated 
to  be  sure  that  the  allowable  unit  stress  is  not  exceeded  at  any  joint,  and  that  the  curve  of  pressures  does  not 
pass  outside  of  the  middle  third. 

(2)  In  the  preceding  example  suppose,  in  addition  to  the  surcharge,  a  moving  load  cf  8000  pounds  per  lineal 
foot  moves  over  the  arch,  the  width  of  arch  being  20  feet. 

We  still  have,  just  as  before,  for  the  values  of  Mi,  Vi,  //at  the  left  end  for  loads  7*,,  P,,  etc., 

Mi  —  +  2.8467*1  +  2.8427*3  —  1.1387*3  —  3.8387*4  —  2.882/)8  —  0.5587*6 

Vi  =  +  0.98477*1  +  0.86697*.  +  0.63777*1  +  0.36237*4  4-  0.13317*.  +  o.oi53/'0 

H  —  +  o.o6s8P1  +0.40817*,  +  o.75i3JP3  +0.75137*4  +  0.408 1/5.  +0.06587*. 

Since  the  moving  load  is  8000  pounds  per  lineal  ft.  and  width  20  ft.,  we  have  -^-  =  400  pounds  per 
lineal  ft.  for  width  of  one  foot.  Hence  the  live  loads  are 

Pl  =  />.  =  8.93  x  400  =  3572  pounds, 
7*3  =  7*6  =  12.03  x  4°°  =  4812  pounds, 
P,  —  7*4  =  13.681  x  400  =  5472  pounds. 

Let  us  find  the  moment  at  c1  and  B  due  to  each  of  these  loads. 

For  the  moment  at  c'  we  have  for  any  load  on  left  of  cf,  if  r  is  the  rise  of  the  neutral  axis, 


and  for  any  load  on  right  of  c1 


M  =  Mi  -  —  +  Hr. 


For  the  moment  at  B  we  have  for  any  load 

M  -  Mi  -  VJ  +  P(l  -  kl). 

We  have  then 

at  the  crown  c' 

M  =  +  2.8467*1  -  0.98477*.  x  34.641  +  0.06587*.  x  20  +  7*.  x  30.176  =  +  810    pound-feet, 

M  =  +  2.8427*,  -  0.86697*,  x  34.641  +  0.40817*,  x  20  +  7*,  x   19.696  =  +  3220 

M  =  -  1.1387*.  -  0:63777*,  x  34-641  +  0.7513^3  x  20  +  7*.  x     6.841  =  --  7450 

M  =  -  3.8387*4  -  0.36237*4  x  34.641  +  0.75137*4  x  20  =  -  745° 

M  —  —  2.882  A  -0.13317*5  x  34.641  +  0.40817*5  x  20 

M-  -  0.5587*8  -  0.01537*.  x  34.641  +  0.06587*.  x  20  =  +    8l° 


6o4 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  X. 


at  the  springing  7? 

M  =  +  2.8467%  —  0.98477*,  x  69.282  +  Pi  x  64.817  =  —     1996  pound-feet, 
M  =  +  2.8427%  -  0.86697%  x  69.282  +  7%  x  54.337  =  -  13866 
M  =  —  1.1387%  -  0.63777%  x  69.282  +  7%  x  41.482  =  —  20997          " 
M  =  -  3.8387%  -  0.36237%  x  69.282  +  7%  x  27.8      =  -    6232 
J/=  —  2.8827%  -  0.13317%  x  69.282  +  Pi  x  14.945  =  +  13674 
M  =  —  0.5587%  —  0.01537%  x  69.282  +  7%  x    4.465  =  +  10169 

Since  for  the  surcharge  alone  we  have  at  c",  M  =  +  14050,  we  see  that  there  can  never  be  a  negative 
moment  at  c',  and  the  maximum  moment  at  c1  is  when  the  live  loads  Pi,  7%,  7%,  7%  only  act  together  with  the 
surcharge,  and  is 

•A/mm*,  at  <?  =  +  14050  +  8060  =  +  221 10  pound-feet. 

Also,  since  for  the  surcharge  alone  we  have  at  B,  M  •=•  +  23087,  we  see  that  the  maximum  moment  at  B 
is  when  the  live  loads  7%  and  7%  act  and  is  given  by 

^'/max.  at  B  =  23087  +  23843  =  +  46930  pound-feet. 
When  Pt,  Pi,  7%,  7%  act  together  with  the  surcharge  we  have 

H  —  +  27455  +  3927  =  +  31382  pounds. 

Hence  the  edge  distance  of  H  from  the  intrados  at  the  crown  is,  from  equation  (3),  page  599, 
c'  =  i  —  '- — jr^  =  0,7.  This  is  the  least  edge  distance,  and  hence,  from  equations  (i),  page  599,  we  have  at 

the  crown 

/  =  29813  pounds  per  square  foot. 
When  7%  and  7%  act  together  with  the  surcharge  we  have 

H  =  +  27455  +  2I98  =  +  29653  pounds. 
Hence,  from  equation  (2),  page  599, 

.V  =  4/29653"  +  57772"  =  64937  pounds. 

The  edge  distance  of  TV  from  the  intrados  at  B  is,  from  equation  (3),  page  599, 

469^0 

64937 

This  is  the  least  edge  distance  f,  and  hence,  from  equations  (i),  page  599,  we  have  at  the  springing  B 

P  —  59742  pounds  per  square  foot. 

Taking  the  allowable  unit  stress  at  50000  pounds  per  square  foot,  we  see  that  tliis  is  exceeded  at  B. 
The  dimensions  chosen  in  the  preceding  example  will  be  ample  for  both  surcharge  and  live  load. 
(3)  Investigate  the  conditions  of  stability  for  a  circular  stone  arch,  rise  of  t  lit  intrados  35  ft.,  span  of  the 
intrados  140  ft.,  uniform  depth  2.5  ft.,  surcharge  level -with  the  crown  of  extrados  and  of  the  same  material 
as  the  arch,  density  of  both  i  bo  Ibs.  per  cubic  foot.  •   « 

This  arch  ivas  actually  erected,  and  fell  on  the  removal  of  the  centre,  the  crown  rising.  Show  that  this 
might  have  been  anticipated,  and  design  the  arch  so  as  to  be  stable. 

Since  the  depth  is  constant,  7  cancels  out  in  equations  (4),  page  600,  and  we  proceed  precisely  as  for  the 


solid  arch  given  in  the  example,  p.ige  594.     Lot  us  divide  the  neutral  axis  into  twelve  equal  segments  ab,  be, 
fd,  etc.,  of  length  s,  and  let  the  end  segments  Aa  and  Bm  be  one  half  of  s.     Let  the  ord mates  to  the  middle 


CHAP.  X.] 


THE  STONE  ARCH-EXAMPLES. 


605 


points  a',  V  d,  etc.,  of  each  segment  for  origin  at  A  be  x,y,  and  take  the  loads  A,  P*.  etc.,  acting  half  way 
between  A  and  a',  a'  and  V t  etc. 

We  have  then  the  following  table. 


X 

y 

jr  

2 

(—0- 

y  —  y 

<,-;>• 

Aa 

2.n 

2.70 

—  68.89 

4746.0321 

—  20.69 

428.0761 

at 

9.04 

10.29 

-  61.96 

3839.0416 

—  13.10 

r7I.6TOO 

cd 
de 

19.56 

44.00 
57-34 

19.07 
26.13 
31.29 
34-44 

~  51-44 

—  27.OO 
-  13.66 

2646.0736 
1575.2961 
729.0000 
186.5956 

-  4-32 
h  2.74 
+  7-90 
+  11-05 

18.6624 
7.5076 
62.4100 
I22.IO25 

JS 

71.00 

35-50 

0 

0 

+  12.  II 

146  6521 

hi 
ik 
kl 
Im 

84.66 
98.00 
110.69 
122.44 
132-96 

34-44 
31.29 
26.13 
19.07 
10.29 

+  13.66 

+  27.00 
+  39-69 
+  51.44 
+  61.96 

186.5956 
729.0000 

1575.2961 
2646.0736 
3839.0416 

+  H.05 
+   7.90 
+   2.74 
-   4-32 
—  13.10 

122.1025 
62.4100 
7.5076 
18.6624 
I7I.6IOO 

mB 

139.89 

2.70 

+  68.89 

4746.0321 

—  2O.69 

428.0761 

280.64 

22698.0459 

I339-3032 

We  have  then  2^y  =  280.64.  Note  that  in  taking  this  summation  and  the  other  summations  of  the 
table,  since  the  end  segments  Aa  and  mB  are  only  half  length,  we  take  in  the  summations  one  half  the 
•values  for  Aa  and  mB. 

We  have  then 

280.64 

y  =  —^  =  23-39  ft., 

and  can  now  fill  out  the  last  two  columns. 

For  the  values  of  kl,  (i  —  K)  and  /(i  -  K)  for  each  load  we  have 

Pi  P*  P*  Pt          P*          P,          Pi 

kl=      4.52        14.30        25.435      37.655    50.67      64.17      77.83 
/(i  —  £)  =  137.48      127.70      116.565    104.345    91.33      77.83      64.17 
i  —  k  =      0.968        0.899        °-82 1        0.734      0.643      0.548      0.452 
We  can  now  draw  up  the  following  table  for  the  loads  P-,  to  Pi, 


P, 
91-33 
50.67 

°-357 


P»  Pi.           P.i        Ptt 

104.345  116.565  127.70  137.48 

37.655  25.435  14.30      4.52 

0.266  0.179        °-I0i     0.032 


/>, 

f, 

f» 

x-kl 

H<~«> 

(y  -yte  -  *0 

X   -  kl 

H)*-«> 

(>-»(*-*/ 

X-kl 

(*-;)  fcr-*0 

O  -  >x*  -  *f) 

gh 
hi 
ik 
kl 
Im 

mB 

6.83 
20.17 
32.86 
44.61 
55-13 

93.2978 
544.5900 
1304.2134 
2294.7384 

3415.8548 

+    75.4712 
+  159.3430 
+    90.0364 
-  192.7152 
—  722.2030 

6.67 
19.36 
31.11 
41.63 

180.0900 
768.3984 
1600.2984 
2580.3948 

+      52-6930 
+      53-0464 
—    I34-3952 
-    545-3530 

6-35 
18.10 
28.62 

252.0315 
93L0640 
W-2952 

+     I7.399C 
-     73.1920 
-  374-9220 

62.06 

4275.3136 

—1300.7776 

48.54 

3343.9206 

—  1017.3984 

35-54 

2449.039^, 

-   745-1280 

+  190.63 

+9790.3512 

—  1240.6564 

+123.04 

+6801.1419 

—  1080.7080 

+  70.84 

+4180.9105!—  808.2700 

/>„ 

**n 

A« 

x  -  kl 

H<-*> 

(y-yte  -W 

x-ti 

H<-o 

(,-yXx-M 

X  -    kl 

H)<~«, 

(r-JX-r-  */) 

kl 
Im 

mB 

5-87 
16.39 

302.0702 
1015-5244 

-       25.3584 
—    2I4.7090 

5.26 

325.9096 

-          68.9006 

23-32 

1606.5148 

-    488.7872 

12.  2O 

839.7690 

-      255.5024 

2.40 

1 

166.0248 

-  50.5136 

+  33-32 

+2120.8520 

—    484.4610 

+    11-36 

+    745-7941 

-       196.6608 

+    I.2o|    +    83.0124 

1 

—  25.2568 

6o6 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  X. 


Note  that  in  taking  the  summations,  since  tnB  Is  of  half  length,  we  take  one  half  the  values  for  mB  in 
summing  up. 

We  have  then,  from  these  tables  and  equations  (4),  page  600, 

Aft  =  —  15.8867%,         —  10.2537*. ,         —  5.9037*.,         —  2.8277*1., 
Vi  =  +    0.4317*7,         +    0.2997%,         +0.1847*., 
H  —  +    0.9267*1,         +    o.8oi/%,         +0.6037*., 

and  from  equations  (5),  (6)  and  (7),  page  601, 

Mi  =  —  6.9447%.          —  7-759^.          —  6.9437%. 
Mi'=  -  10.927*.,          -7.3067%.          -2.357%. 
V\  =  +  0.5697*.,          +  0.7017%,         +  0.8167*4, 

Since  the  depth  is  2.5  ft.  and  density  160  Ibs.  per  cubic  foot,  we  have  for  one  foot  width  the  loads 

7%  =  7%,  =  48200,        7*,  =7%i  =  40000,        7%  —  7*,o  =  29400,        7*4  =  7*.  =  19100, 
7%  =  7*.  =i  looo,        7*.  =  7*i    =  6600. 
Hence 

2Aft  =  +  272750  pound-feet,          ^F,  =  154300  pounds,          ~2.H  =  87750  pounds. 


9870685 


—  2.8277%.  , 

—  0.9477%,, 

—  o.ioo7%,; 

+  0.0937%.  , 

+  0.0337*.  ,  . 

+  0.0047*,,; 

+  0.3627%., 

+  0.1477%,, 

+  0.0197*1,  ; 

—  4.6917%., 

—  2.0427%  ,. 

—  o.26o/',,; 

+  2.8427%, 

+  5-53A  . 

+  3432/>,  1 

+  0.9077%, 

+  0.9677',, 

+  0.9967V 

For  the  values  of  2*  P  (x  —  £/)  we  have  for  the  points  of  division  A,  a',  &,  etc., 

A  a'  V  (f  d'  /  /' 

2?p(x  —  kr)  =  o   217864   935328   2144550  .3758000   5654936   7717664 

h'        i         k'         I'        B 

=  12090728   14392982   16809712   19338720   21910600 

We  have  then,  /rom  equation  (8),  page  601,  for  the  moment  at  any  point  of  the  axis 

M  =  +  272750  -  154300*  +  877507  +  2^  P(x  —  kl). 
For  the  points/',^',  K ',  etc..  we  have  then 

1 10.69 

y  =        35-50       34.44  31-29  26.13 

M  =  +  150500     +102500          —12500 

From  equation  (2),  page  599,  we  have  then 


g1 
84.66 


h1 
98 


122.44    132.96     142 
19.07     10.29      ° 

—  1 21000  -136650  -1300  +272750. 


g 
88010 


h' 
89550 


f 
95»5 


k' 
109860 


/' 
137685 


B 

177510 


N-  87750 
For  the  edge  distance  of  N  from  the  intrados  we  have  now,  from  equation  (3),  page  599, 


=  -  0.47 


+  0.08 


k' 
+  1.39 


+  2.52 


t' 
+  2.50 


/'  B 

1.26         —  0.29 


We  see  that  the  curve  of  equilibrium  passes  outside  of  the  arch  and  below  the  axis  at  the  crown  /'  and 
springing  77,  and  outside  of  the  arch  and  above  the  axis  at  /'.  The  arch  is  then  unstable  and  will  fall,  the 
joints  at  the  crown  and  springing  opening  at  the  extrados,  and  at  /"  opening  at  the  intrados.  In  other  words, 
the  haunches  sinking  and  the  crown  rising. 

This  is  precisely  what  happened  when  the  arch  was  erected.  In  order  to  make  the  arch  stable  we  should 
malte  one  of  two  changes  in  the  design.  We  can  either  increase  the  depth  or  make  the  surcharge  lighter 
over  the  haunches,  by  building  up  the  surcharge  with  hollow  spaces  at  the  haunches  or  lightening  the 
surcharge  there  l>y  filling  in  with  gravel  instead  of  stone. 

Tlie  arcli  was  actually  rebuilt  with  hollow  spaces  in  the  surcharge  over  the  haunches. 


CHAP.  X.] 


THE  STRAIGHT  ARCH. 


607 


The  Straight  Arch — The  straight  arch  is  a  stone  beam  fixed  at  the  ends,  subjected  to 


compression  and  bending.  The  beam  is 
composed  of  voussoirs  and  therefore  will 
not  resist  tension  at  any  joint. 

Let  the  loading  be  uniform  and  equal 
to  w  pounds  per  foot  of  length,  and  the 
length  of  the  span  be  /. 

We  have  at  the  left  end  the  reaction 

wl 
Pi  =  —  and  the  thrust  H  acting  at  the 

distance  y^  below  A. 

The  moment  at  any  point  of  the  axis 
is  then 

wx 

1          2  ' 

The  work  is 


work 


r' 
=:  / 

•-/O 


C*-— r-^-"::---/ 


\ 


-2EA+ 


V 
\ 


rir  wx 

J  L-^-T 

»/  0 


If  we  differentiate  with  respect  to  H  and  put    ^          '  =  o,  we  have  for  the  value  of  H 
which  makes  the  work  a  minimum 

..dx 


If  we  substitute  /  =  Ax2,  where  /c2  is  the  square  of  the  radius  of  gyration  of  the  cross- 
section  A,  and  integrate,  we  have 


1  TT  ~*  ./I  /      \ 

—*  =  o,     or      -"  =  —  —,«   ,   „. av (0 


Let  the  angle  of  the  ends  with  the  vertical  be  «,  and  let  the  ends  be  at  right  angles  to 
the  curve  of  equilibrium.     Then  the  normal  pressure  on  the  ends  is 

_     H 
cos  of 

and  the  end  areas  are  A,  =  ^-^.      Hence  —  =  -^-,  and  if  we  take  the  breadth  unity 


We  have  then  from  equations  (i),  page  599,  for  the  maximum  unit  pressure  p  at  the  end 


where  e  is  the  edge  distance  of  H,  or,  disregard™     signs,  e  =  -  -  y 


6o8  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  X. 

Hence 


The  work  will  be  a  minimum  when/  is  a  minimum. 

If  we  differentiate  and  put  —y—  =  o,  we  have  for  the  value  of  _yt  which  makes/  a,  mini- 

mum, since  /c3  =  —  , 


Substituting  in  equation  (i),  we  have,  since  yc2  =  —  , 

*=-£-.   ...  ;<V  .'  .....  w 

8  V$d 
We  have  then  for  the  angle  a 


tan  «  =  j±  =  2  i/3  .  -^  =  3.464  -j 
The  maximum  unit  pressure  is,  since  /c2  =  —  , 


"> 


Let  the  maximum  allowable  unit  stress  be  Sf,  then  we  have 
w/2(^3~-0 

__._^  or  rf= 


Equation   (c)   gives  the  depth  </  for  the  maximum  allowable  stress,  equation  (b]  gives 
the  angle  of  the  ends  with  the  vertical,  equation  (a)  gives  the  horizontal  thrust  //. 

Example.  —  Design  a  straight  arch  of  20  feet  span  to  sustain  a  load  of  4000  pounds  per  foot  of  length. 
The  allowable  unit  stress  is  50  ooo  pounds  per  square  foot. 
From  (c)  we  have 

d=  6.5  1/5°°^  =  1.8  ft. 
r    50000 

From  (&}  we  have  for  the  angle  a  of  the  ends  with  the  vertical 

tan  a  =  3.464—  =  0.31176,     or     a=  17°  19'. 
20 

The  ends  intersect,  then,  at  a  point  O  at  a  vertical  distance  CO  below  the  centre  C  given  by 

co=     l     =  —  !£-,  =  32.07  ft. 

2  tan  a        0.31176 


CHAPTER    XI. 


COMPOSITED  STRUCTURES.     SUSPENSION   SYSTEM    WITH   STIFFENING   TRUSS. 

EACH  of  the  structures  of  Chapter  IX  may  be  inverted,  and  constitutes  in  such  case  an 
inverted  arch  or  rigid  suspension  system.  The  method  of  calculation  is  then  precisely  the 
same,  the  only  difference  being  that  the  horizontal  thrust  at  the  end  of  the  arch  becomes  a 
horizontal  pull  at  the  ends  of  the  cable,  and  therefore  members  which  were  in  compression 
are  now  in  tension,  and  vice  versa. 

Suspension  System. — A  common  construction  for  long  spans,  however,  is  that  shown 
in  the  figure.  Such  a  structure  we  may  call  a  ' '  com- 
posite "  system,  that  is,  it  consists  of  two  different  systems 
which  act  together.  The  figure  represents  the  most 
important  of  these,  known  as  the  "suspension  system." 
It  consists  of  a  flexible  chain  or  cable  which  is  stiffened 
under  the  action  of  partial  loads  by  a  truss.  The  truss  is 
slung  from  the  cable  by  suspenders,  and  may  be  of  any 
design,  either  double  or  single  intersection,  Pratt,  etc.  The  cable  carries  the  entire  dead 
weight,  that  is,  the  suspenders  are  screwed  up  until  the  ends  of  the  truss  just  bear  on  the 
abutments.  The  office  of  the  truss  is  thus  to  stiffen  the  cable  and  prevent  change  of  shape 
and  oscillation  due  to  partial  and  moving  loads.  It  also  acts  to  support  its  share  of  the 
moving  load.  There  are  usually  side  spans  at  each  end.  In  any  case  the  cable  passes  over 
rollers  on  top  of  the  towers,  and  is  carried  on  beyond  and  firmly  fastened  to  large  anchorages 
of  masonry. 

Defects  of  the  System.— The  principal  defect  of  this  system  is  its  lack  of  rigidity.  The 
cable  possesses  little  inherent  rigidity,  and  the  stiffness  is  due  almost  entirely,  therefore,  to  the 
truss. 

A  second  disadvantage  is  that  a  rise  of  temperature,  by  increasing  the  deflection,  throws 
considerable  load  on  the  truss.  To  obviate  this  objection,  the  truss  may  be  hinged  at  the 
centre  and  placed  on  rollers  at  the  ends. 

Advantages  of  the  System.^It  is  evident  from  the  preceding  that  the  system  is  best 
applied  to  long  spans.  The  cable,  then,  carries  the  dead  weight,  and  by  reason  of  its  own 
very  considerable  weight  in  such  case  resists  in  some  degree  the  deforming  action  of  partial 
loads.  The  truss  can  thus  be  very  light  compared  to  what  it  would  have  to  be  if  there  were 
no  cable. 

Stays  Unnecessary. — The  system  is  accordingly  in  practice  applied  only  to  very  long 
spans.  In  such  case,  with  cables  made  of  steel  wire  it  admits  of 
great  economy.  But,  owing  to  lack  of  rigidity,  additional  stiff- 
ness is  sought  to  be  obtained  by  the  introduction  of  stays  reaching 
from  the  top  of  the  tower  to  various  points  of  the  truss,  as  shown 
in  the  accompanying  figure.  The  use  of  these  is  not  to  be 
recommended. 


stresses  indeterminate. 


They  render   the  correct  determination   of  the 
A  load  at  any  point  may  be  carried  entirely  by  the  suspender  and 

609 


6io  ST/tTICS  OF  ELMSTIC  SOLIDS.  [CHAP   XI. 

stay  at  that  point,  or  by  the  suspender  and  truss,  or  by  the  stay  and  truss.  It  is  impossible 
to  tell  exactly  the  duty  performed  by  each;  and  even  if  it  were  not,  it  would  be  impossible  to 
so  adjust  the  several  systems  that  each  shall  take  its  proper  share.  If  such  adjustment  could 
be  made,  it  would  not  last.  Variations  of  stress,  set,  and  elongation  of  members,  shocks 
and  vibrations,  rise  and  fall  of  temperature,  would  constantly  disturb  such  adjustment. 

The  stays  are  also  superfluous.  The  truss  is  a  rigid  construction.  It  ought  to  render 
rigid  the  system  of  which  it  forms  a  part,  and  should  be  so  designed  as  to  perform  its  duty 
without  help.  If  such  superfluous  members  are  introduced,  they  can  then  be  considered  as 
an  extra  addition,  contributing  to  strength  and  stiffness.  But  the  truss  should  be  designed 
without  reference  to  their  action. 

Horizontal  Pull  of  Cable. — Let  the  span  or  chord  of  the  cable  AB  be  c,  and  the  rise  or 

A c  B  versed  sine  Cc  be  r.  Then  if  w  be  the  load  per  unit  of  horizontal,  we 

\~  jT7  have  for  uniformly  distributed  load,  taking  moments  about  B,  if  H  is  the 

H«~~^  &  -4  '*  horizontal  pull, 


-/fr  +  ^-o.     or    *=£  ........     (,) 

Equation  (i)  gives  the  horizontal  pull  of  the  cable  for  uniform  load,  which  is  evidently 
the  same  at  every  point. 

Shape  of  Cable.  —  If  we  take  moments  about  any  point  P  distant  x  from  the  centre,  we 
have,  \i  y  is  the  ordinate  for  origin  at  C, 

x*  ivx* 

=  0      or        = 


Inserting  the  value  of  //"from  (i), 

*  =  *£  ............    <» 

which  is  the  equation  of  a  parabola.      Hence  the  curve  of  the  cable  or  of  a  flexible  string 
uniformly  loaded  along  the  horizontal  is  a  parabola. 

Length  of  Suspenders.—  Let  /0  be  the  length  of  the  suspender  at  the  centre,  then  from 
(2)  we  have  for  the  length  of  a  suspender  at  a  distance  x  from  the  centre 


(3) 


Length  of  Segment  of  Cable.—  Let  n  be  the  number  of  segments  of  the  cable,  the  sus- 
penders being  equally  spaced,  so  that  -  is  the  distance  between  suspenders,  or  the  horizontal 
projection  of  a  segment.  Then  from  (2)  we  have  for  the  ordinate  of  the  nearest  end 


CHAP.  XL]  SUSPENSION  SYSTEM,  STRESS  IN  SEGMENT  OF  CABLE.  611 

and  for  the  ordinate  of  the  farther  end 


^2  = 


Hence  the  vertical  projection  of  a  segment  is 


The  length  se  of  a  segment  is  then    • 

(4) 


where  x  is  the  ordinate  from  the  centre  to  the  nearest  end  of  segment. 

Stress  in  Segment  of  Cable.— The  secant  of  the  angle  of  inclination  a  at  any  point  is 


ns, 

sec  a  =  — £ 
c 


Since  the  horizontal  pull  H  is  the  same  at  every  point,  the  stress  for  any  segment  is 


Stress  in  segment  =  H  sec  a  =  --  -, 

or,  from  (i),  for  uniform  load 

0  nwcsc 

Stress  in  segment  =  ............     (5) 

Deflection  of  Cable  Due  to  Temperature.  —  Let  e  be  the  coefficient  of  expansion  or  con- 
traction, that  is,  the  ratio  of  the  change  of  length  to  the  original  length  of  a  member  for  one 
degree  change  of  temperature.  Let  t  be  the  change  of  temperature  in  degrees,  A  the  total 

change  of  length  and  sc  the  length  of  a  cable  segment.      Then  —  is  the  ratio  of  the  change 
of  length  to  original  length  for  /  degrees.      For  one  degree-  we  have  then 


=  .,     or     t=*s, 

where    e   is  given  by  experiment,   and  et  is  an  abstract   number   or   numerical  value.      For 
values  of  e  see  page  582. 

Suppose  a  load  P  at  the  centre  of  the  cable  would  produce  the  same  change  of  length. 
Since  the  structure  is  rigid  so  that  the  shape  of  the  cable  does  not  change,  this  load  must  be 


612  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  XI. 

p 
distributed  by  the  truss  over  the  cable  as  a  uniform  load  of  —  per  foot  of  horizontal  projection. 

Inserting  this  for  w  in  (5),  we  have 

Stress  in  segment  =  —  —  -. 
or 

From  page  515  the  work  is  one  half  the  product  of  the  stress  and  change  of  length,  or 

nPets* 
Work  on  a  segment  =  —  g  —  . 

The  total  work  for  all  the  segments  is  then 

Work  = 


. 
l6r 


PA 
If  A  is  the  deflection  of  cable  at  centre,  the  work  is  also  —  .      Hence 


where  «  and  et  are  abstract  numbers  and  sc  and  r  are  lengths.  If  we  take  sc  and  r  in  feet  or 
inches,  equation  t'6)  will  give  A  in  feet  or  inches. 

Deflection  of  Truss  for  Uniform  Load.  —  Let  »0  be  the  stress  in  pounds  in  any  member 
of  the  truss  for  a  uniformly  distributed  load  of  one  pound  per  foot  of  length,  that  is,  7/0  is  the 
stress  per  pound-  per-foot  distributed  load.  Then  the  stress  in  pounds  for  a  uniform  load  of  u> 
pounds  per  foot  will  -be  given  by  wuv  The  corresponding  strain  A.  is  then  (page  477) 

A  = 

a 

where  s  is  the  length  of  truss  member,  a  its  area  of  cross-section  in  square  inches  and  E  its 
coefficient  of  elasticity  in  pounds  per  square  inch.  If  s  is  taken  in  feet  or  inches,  A  will  be 
given  in  feet  or  inches. 

Let  /  be  the  stress  in  pounds  in  the  member  due  to  one  pound  placed  at  the  centre  of 
the  truss.  That  is,  /  is  the  stress  in  pounds  per  pound  of  load  at  centre.  Then  the  work  on 
the  member  due  to  this  load  is 


/A.  0 

i  pound  X  —  =         p    X  i  pound. 


The  total  work  on  all  the  members  is  then 

Work  =  -SE          X   T  P°und- 


CHAP.  XL]  OLD    THEORY   OF  SUSPENSION  SYSTEM.  613 

But  if  A  is  the  deflection  at  the  centre  the  work  is  also  i  pound  X  -    •      Hence  we  have 


A         w  —^  ——n 

i  pound  X  -  =  ^Sr-  X  I  pound,      or     A  =  ^>~f-,   ....      (7) 

where  E  and  a  are  as  already  specified.      If  s  is  taken  in  feet  or  inches,  A  will  be  given  in 
feet  or  inches. 

Temperature  Load  for  Truss.  —  When  the  cable  expands  or  contracts  the  centre  falls  or 
rises  a  distance  dt  given  by  (6),  and  the  centre  of  the  truss  falls  or  rises  with  the  cable  a  dis- 
tance given  by  (7).  Let  wt  be  the  uniformly  distributed  load  in  pounds  per  foot  which  would 
cause  this  deflection.  Then,  equating  (6)  and  (7),  we  have 


Old  Theory  of  Suspension  System. — The  theory  of  the  suspension  system  heretofore  in 
use  is  due  to  Rankine,  and  is  based  upon  the  assumption  that  the  cable  carries  the  entire  load, 
dead  and  live,  the  office  of  the  truss  being  simply  to  distribute  FIG.  i. 

a  partial  loading  over  the  cable,  and  thus  prevent  change  of 

SllaPe-  A     A    A    A    A    A   A* A jjt^TJ^t 

MAXIMUM  SHEAR  IN  TRUSS — OLD  METHOD. — Let  the  Ay  T  TTTTTT  I — r  /     "'^B 
uniform  live  load  w  for  unit  of  length  extend  over  the  distance     jt~~~ 
s  from  the  right  end  (see  Fig.  i). 

Then  the  load  is  wz,  and,  since  by  assumption  the  cable  carries  all  this  load,  the  upward 

load  on  the  truss  due  to  the  cable  is  wz  or,  -^  for  unit  of  length. 

Let  RA  be  the  reaction  at  the  left  end  A  of  the  truss.  We  have,  taking  moments  about 
the  right  end  B, 

c        wz*                                         wz(c—z} 
-RAc-w*.-+—  =  0,     or     RA  =          —^— 

Since  this  is  negative,  the  truss  should  be  tied  down  at  the  ends.  If  the  load  «'  extends 
over  the  distance  z  from  the  left  end  (Fig.  2),  we  have 


FIG.  2. 


p-4— 1 

tmttftj          je::,^^.         .  .  .  .  w 


In  the  first  case  (Fig.   i),  when  the  load  comes  on  from  the  right,  we  have  for  the  shear 
at  any  point  distant  x  from  the  left  end : 

wz 
when  x  is  less  than  c  —  z  Shear  =  RA  -}-  ~~^~x'* 

when  x  is  greater  than  c  —  z        Shear  =  RA  +  -~x  —  w\x~—  (c  —  z~)~] ; 


614  STATICS  OF  ELASTIC  SOLIDS.  [CHA1-.  XI. 

or,  inserting  the  value  of  RA  from  (i), 

when  x  <  c  —  z  Shear  =  —  ~ \2x  —  (c  —  ~)    ; 

when  x  >  c  —  z  Shear  =  —    2x  —  (c  —  z)  j  —  iv\x  —  (c  —  *)]. 

From  the  last  of  these  equations  we  see  that  the  shear  is  a  positive  maximum  when  the 
last  term  is  zero  or  when  z  =  c  —  x.  That  is,  the  shear  for  the  unloaded  portion  is  a  positive 
maximum  at  the  head  of  the  load. 

From    the    first    of  these    equations    we    have    the    shear  a   negative    maximum    when 

z  = x.     That  is,  the  shear  for  the  unloaded  portion  is  a  negative  maximum  at  any  point 

when  the  distance  covered  by  the  load  is  equal  to  the  distance  of  the  point  from  the  centre. 

Inserting    these    values   of  z  = x 

unloaded  portion  distant  x  from  the  left  end : 


Inserting    these    values   of  z  = x  and   z  =  c  —  x,   we  have  for  any  point  of  the 


w(c  —  x\x 
maximum  positive  Shear    =  -\-  - — , 

unloaded  portion  -j  c          \2    j- (3) 

[  maximum  negative  Shear  =  — . 

In  the  second  case  (Fig.  2),  when  the  load  comes  on  from  the  left,  we  have  for  the  shear 
at  any  point  distant  x  from  the  left  end : 

when  x  <  z  Shear  =  RA  -j "  x  —  wx, 

or.  inserting  the  value  of  RA  from  (2), 

when  x  <  z  Shear  =  — \2x  4-  (c  —  zV\  —  wx. 

2Cl 

This  is  a  negative  maximum  for  z  =  x.      That  is,  the  shear  for  the  loaded  portion   is 
negative  maximum  at  the  head  of  the  load. 

It  is  a  positive  maximum  when  z  —  — (-  x.      That  is,  the  shear  for  the  loaded  portion  is 

a  positive  maximum  at  any  point  when  the  distance  between  the  point  and  the  end  of  the  load 
is  equal  to  the  half  span. 

Inserting  the  values  of  z  =  x  and  z  =  —  -{-  x,  we  have  for  any  point  of  the  loaded  portion 
distant  x  from  the  left  end : 


(4) 


loaded  portion  j  —£ 

w(c  — 


maximum  negative  Shear  = 


CHAP.  XI.]  OLD   THEORY  OF  SUSPENSION  SYSTEM.  615 

We  see  from  equations  (3)  and  (4)  that  we  have  for  the  maximum  shears  for 

I  I 

x  =  o  —c  -c 

4  2 

c,  ,  we  ywc  we 

Shear==±y  ±  i_  ±_ 

That  is,  the  maximum  shear  is  practically  constant  and  varies  but  litle  from  -5-. 

•o 

It  is  therefore  customary  by  the  old  method  to  design  every  brace  for  the  maximum 
shears  due  to  live  load. 

we 
Shear  =  ±  y (i) 

MAXIMUM  MOMENT  IN  TRUSS — OLD  METHOD. — For  the  moment  M  at  any  point  of 
the  unloaded  portion  (Fig.  i)  distant ' x  from  the  left  end,  if  w  is  the  uniform  live  load  for  unit 
of  length,  we  have 


A*  "         2C   ' 

or,  substituting  the  value  of  RA  from  (i), 

M  =  -^[_x  -  (c  -  z}} (5) 

For  any  point  of  the  loaded  portion  (Fig.  2)  we  have 

wzx*       wx* 
M  —  —  RAx -1 -, 

2C  2 

or,  substituting  the  value  of  RA  from  (2), 

In  (5)  M  —  o  for  x  =  c  —  z,  and  in  (6)  M=  o  for  x  =  z.  That  is,  the  moment  at  the 
head  of  the  load  is  zero.  Also,  if*  is  less  than  c  —  z  in  (5),  the  moment  is  positive,  and  if 
greater  than  c  -  z,  the  moment  is  negative.  The  head  of  the  load  is  then  a  point  of  inflection, 
and  the  loaded  and  unloaded  portions  may  be  considered  as  simple  trusses  uniformly  loaded. 
The  greatest  moment  for  each  portion  will  then  be  at  the  centre  of  each  portion.  Making, 

then,  *  —  —  -  in  (5)  and  *  =  -  in  (6),  we  have  for  the  moment  at  the  centre  of  each  portion 


wz(c  — 


,  x 

and  --•(<-*)• 


_ 


These  are  a  maximum,  respectively,  for  z  =  -c  and  z  =  - 


616  STATICS  OF  ELASTIC  SOLIDS.  [CHAP.  XI. 

Hence  the  maximum  positive  moment  is  at  the  middle  of  the  unloaded  portion  when  tlie 
load  extends  over  one  tliird  tlie  span,  and  tlie  maximum  negative  moment  is  at  the  middle  of 
the  load  when  //  eovers  two  thirds  the  span. 

We  have  then  for  the  maximum  positive  moment  at  any  point  of  the  unloaded  left  half 

span,  by  putting  z  =  -c  in  (5), 


(7) 


and  for  the  maximum  negative  moment  at  any  point  of  the  loaded  left  half  span,  by  putting 


=  -cin  (6), 


(8) 


From  equations  (7)  and  (8)  we  have  the  maximum  moments  for 

*  =  °      \c        r        ? 

we1  we*  we* 

M=°     ±T*      ±54      ±^ 

I  1VC* 

That  is,  the  maximum  moment  beyond  -c  is  practically  constant  and  varies  but  little  from  —  . 

It  is  therefore  customary  by  the  old  method  to  design  every  chord  panel  for  the  maximum 
moments  due  to  live  load 


Temperature  Load  for  Truss.  —  From  (I)  and  (II)  we  can  then  easily  find  the  area  a  of 
each  truss  member  due  to  live  load.  Thus  for  straight  truss  of  height  /r,  if  tr  is  the  working 
stress,  we  have  for  the  area  a  of  the  chords 

we* 

•=  54*?  ;•••-.•'  .........     (9) 

and  for  the  area  of  a  brace  which  makes  the  angle  #  with  the  vertical 


We  have  then  from  (8),  page  613,  the  temperature  load  per  unit  of  length, 


(in) 


Stress  in  Truss.  —  This  temperature  load  should  be  taken  into  account  together  with  the 
live  load  in  finding  the  maximum  truss  stresses.  We  have  then  to  add  to  the  shear  and 
moment  given  by  (I)  and  (III)  the  shear  and  moment  due  to  the  temperature  load  wt.  The 
actual  stresses  in  the  truss  members,  then,  are  greater  than  those  due  to  the  live  load  only, 
and  hence  the  areas  assumed  in  (9)  and  (10)  are  too  small  i:ml  the  corresponding  value  of  wt 


CHAP.  XL] 


SUSPENSION  SYSTEM.     CABLE  STRESS   AND  AREA. 


617 


given  by  (III)  is  too  small.  We  should  therefore  assume  wt  somewhat  larger  than  given  by 
(III),  and  then  find  the  stresses  for  this  assumed  wt  and  the  live  load.  The  corresponding 
areas  should,  when  inserted  in  (III),  give  us  pretty  closely  the  value  for  wt  we  assumed.  If 
not,  we  can  make  another  approximation. 

Cable  Stress  and  Area. — We  can  now  find  the  stress  and  area  in  any  cable  segment. 
Thus  let  the  dead  load  per  unit  of  length  be  w0 ,  the  live  load  w,  and  the  temperature  load 
wt,  assumed  as  above.  Then  the  total  unit  load  for  the  cable  is  (w  -\-  WQ  -}-  wt},  and  from 
(5)  we  have  for  the  stress  in  any  segment  whose  length  is  sc 


n(w  -f- 
Stress  in  segment  —  ~s  — 


or 


(IV) 


If  the  working  stress  for  cable  is  <rc ,  we  have  only  to  divide  by  <rc  to  have  the  area  of 
cross-section. 

If  the  cable  is  made  of  links,  (IV)  will  give  the  stress  for  any  link  according  to  the  value 
of  sc ,  and  dividing  by  <rc  we  have  the  area  of  cross-section  of  the  link.  If  the  links  are  of 
constant  area  of  cross-section,  or  if  we  have  a  wire  cable,  we  must  take  for  sc  the  length  of 
the  end  segment  for  which  sc  is  greatest. 

Suspender  Stress  and  Area. — Let  n  be  the  number  of   segments  of  the  cable ;    then 


-  is  the  distance  between  suspenders.      The  unit  load  is  (w  -\-  WQ  -\-  wt}\  hence 


Stress  on  suspender  =  (w  -\-  u>0  -f-  w^)—. 


(V) 


If  <rs  is  the  working  stress,  we  have  only  to  divide  by  <rt  to  have  the  area  of  cross-section 
of  the  suspender. 

Example.— OLD  METHOD.— In  order  to  abridge  the  work  of  computation  we  take  a  short  span  for  illus- 
tration. 

Uaia.—Let  the  span  c  =  54  feet ;  the  versine  of  cable  r  =  9  feet;  the  depth  of  truss  //  =  12  feet ;  the 
panel  length  9  feet,  so  that  the  number  of  segments  of  the  cable  is  «  =  6 ;  the  truss  of  steel  and  coefficient 
of  elasticity  for  truss  members  E=  30000000  pounds  per  square  inch;  working  stress  for  truss  members 
and  steel  suspenders  <r  =  <r,  =  10000  pounds  per  square  inch;  cable  of  steel  wire  working  stress  =  <rf  = 
30000  pounds  pei  square  inch  ;  live  load  w  =  2000  pounds  per  foot;  dead  load  wa  =  1000  pounds  per  foot; 
coefficient  of  expansion  e  =  0.00000686  ;  range  of  temperature  /  =  80°. 

Let  the  members  be  notated  as  in  the  following  figure.     The  angle  0  of  the  braces  witli  the  vertical  is 

then  such  that  sec  8  =  -. 


6i8 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAP.  XI. 


Calculation.—  We  have,  from  (I), 

we      2000  x  54 
Shear  =  ±  —  =  -  ^  —  —  =  13500  pounds. 

This,  then,  is  the  stress  for  every  post.     For  every  brace  the  stress  is 

Shear  x  sec  0  =  19089  pounds. 
From  (II),  the  moment  for  any  chord  panel  is 


54 


54 


Hence  the  stress  for  each  chord  panel  is 


108000 


=  9000. 


If  the  working  stress  <r  =  loooo,  we  have  then  for  each  post 

post  area  =  1.35  sq.  in., 
for  every  brace 

brace  area  =  1.9  sq.  in., 
for  every  chord  panel 

panel  area  =  0.9  sq.  in. 

Since  we  have  disregarded  the  stress  due  to  temperature,  let  us  take  post  area  =  1.5  sq.  in.,  brace  area 
=  2  sq.  in.,  panel  area  =  i  sq.  in. 

The  length  of  a  cable  segment  is,  from  (4),  page  611, 


»e  —  Y    "     T  8l 

where  x  is  the  ordinate  from  the  centre  to  nearest  end  of  segment. 
We  have  then  for  the  length  of  cable  segments  : 


a'b 


US 
4/81 


Hence  Ss^  =  556,  and  this  is  to  be  inserted  in  (III)  in  order  to  find  the  temperature  load  ivt. 
We  can  now  draw  up  the  following  table. 


Member. 

Area  a 
in  &. 

LcnBth 
s 
in  Feet. 

Stress  »„ 
in  Pounds. 

Stress  / 
in  Pounds. 

P«»* 
ti 

fu0s 

a.  4 

•  5 

12 

-22.5 

-0.5 

+   90 

+  135 

0*4, 

•5 

12 

-13-5 

-0.5 

f    54 

+    81 

«,//a 

-5 

12 

-     4-5 

—  0.5 

+    18 

+    24 

aAi 

IS 

•4-  28.125 

+  0.625 

+  131-836 

+  261.672 

a\Ai 

'  5 

+  16.875 

+  0.625 

+    79-  1°' 

+  158.202 

a*A> 
aa\ 

: 

13 
9 

+    5.6*5 
-16.875 

+  0.625 
-0-375 

+    26.367 
+    56.953 

f    52.734 
+    56.953 

aiat 

9 

—  27 

-0.75 

+  182.25 

+  182.25 

a**, 

9 

-30.375 

—  1.  125 

+  307-547 

+  307.547 

AAi 

9 

o 

o 

o 

o 

AiA* 
AiA* 

9 
9 

+  16.875 
+  27 

+  0.375 
+  0.75 

+    56.953 
+  182.25 

+  56.953 
+  182.25 

+  "85.257 

For  each  member  in  the  half  truss  we  give  in  the  table  the  area  a  in  square  inches  already  found,  the 
length  s  in  feet,  the  stress  «„  in  pounds  due  to  a  uniform  load  of  one  pound  per  foot  of  length  or  9  pounds 
at  each  lower  apex,  and  the  stress  p  in  pounds  due  to  one  pound  at  the  centre-line  apex.  In  the  last  column 

we  have  then  the  quantities  futs  and  <^L  for  each  member  of  the  half  span. 


CHAP.  XL]  SUSPENSION  SYSTEM.    EXAMPLE.  619 

The  table  gives  us  for  the  half  span  ^-^-  =  +  1185.257,  and  hence  for  the  whole  span 


Inserting  this  and  the  value  of  2s<?=  556  in  (III),  we  have 

30000000  x  6  x  0.00000686  x  80  x  556 
8  x  9  x  2370.5.4 

This,  as  we  have  seen,  must  be  too  small.     Let  us  then  assume 

•wt  =  600  pounds  per  foot  of  length. 

Stresses  in  Truss.  —  If  we  take  for  the  braces  the  live-load  shear  ±  -r-  =  ±  13500  pounds,  as  given  by  (I), 

for  the  chords  the  moment  ±  —  =  ±  108000  as  given  by  (II),  and  in  addition  the  shear  and  moment  due  to 
u't  =  600  pounds  per  foot  of  length,  we  obtain  the  following  stresses  in  the  truss  members  in  pounds: 
For  the  posts 

a  A  =  —  27000  aiAi  =  —  21600  ayA*  =  —  16200               a3A»  —  —  13500 
For  the  braces 

aAi  —  +  33750  aiAi  =  +  27000  a3A3  =  +  20250 

Aai   —  +  33750  Aia?  —  +  27000  A*a3  =  +  20250 
For  the  chords 

aai  —  —  19125  a^zy  =  ±  25200  a*a3  =  ±  27225 

AAi  =  —  i<)\'z$  AiA,=  ±  19125  A*A»  =  ±  25200 


Taking  the  working  stress  cr  =  10000  pounds  per  square  inch,  these  stresses  give  us  the  following  areas 
of  cross-section  in  square  inches: 

aA  =  2.7  c^Ai  =2.  16         a*A?  =  1.62         a,A3  =  1.35         aA^  =  3.37         atA3  =  2.7 

a*A3  —  2.  02  aa\  =  1.91  a^a?  =  2.52          aya3  =  2.72       AA\  =  1.91 

AiAt  —  1.91       A-iA*  =  2.52 

If  we  use  these  values  of  a  in  place  of  the  values  of  a  in  the  table  page  618,  we  obtain 

^=,,65.06. 

and  i-serting  this  in  (III)  we  obtain 

wt  =  655, 

whereas  we  assumed  ivt  at  only  600. 

Let  us  then  again  assume  wt  —  700.     We  have  then  the  following  stresses  : 
For  the  posts 

a  A  =  —  29250  aiAi  =  —  22950  a*A*  =  —  16650  a»A*  =  —  13500 

For  the  braces 

aAi  _  +  365g2  aiAt  =  +  28687  a*A3  —  +  20812 

Aai  =  +  36562  Aia»  =  +  27000  A?a3  =  +  20812 

For  the  chords 

aai  —  —  20812  aia?  =  ±  27900  a,a3  =  ±  30262 

AAi  =  —  20812  AiAi—  ±  20812  A,A»  =  ±  27900 

We  have  then   for  cr  =  10000  the  areas  of  cross-section  aA  =  2.92.  a^Ai  =  2.29,  a,At  =  i  66.  a*A*  = 
1.35,  aAi  =  3.66,  a^Ai  =  2.87,  a?A3  =  2.08,  aai  =  2.08,  a,aa  =  2.79,  aati3  —  3.02,  AA\  =  2.08,  AiA*  —  2.08, 

AyA3  —  2.79. 

Taking  these  values  of  a  in  place  of  the  values  of  a  in  the  table  page  618,  we  have 


620  STATICS  OF  ELASTIC  SOLIDS.  [CHAT.  XI. 

and  inserting  this  in  (III)  we  obtain  «/<  =  711.  Since  we  assumed  wt  =  700,  we  obtain  practically  the  same 
value  we  assumed. 

We  have  then  wt  =  700,  and  the  stresses  last  found  are  the  truss  stresses. 

Since  the  live-load  post  stresses  are  13500,  the  live-load  brace  stresses  16875,  a"d  the  live-load  chord 
stresses  9000  pounds,  we  see  that  the  temperature-load  stresses  are  in  this  case  much  greater  and  of  greater 
importance  than  the  live-load  stresses. 

The  truss  stresses  just  found  should  be  divided  by  2  for  two  trusses.  The  calculation  supposes  all  the 
load  carried  l>y  one  truss  and  one  cable. 

CABLE  STRESS  AND  AREA. — We  have  found  already  for  the  lengths  of  the  cable  segments 


A'a'  =  Vto6,  a'b'  =  Vyo,  V<?  =  yS2  feet. 

Hence,  from  (IV),  we  have  for  the  stresses 

A'a'  =  171495,  a'b'  =  158175,  tic1  =  149850  pounds. 

If  the  working  stress  is  30000  pounds  per  square  inch,  the  areas  of  cross-section  should  be 

A'a'  =  5.72,  a'tf  =  5.27,  b'<?  =  4.99  square  inches. 

If  the  cable  is  to  have  a  constant  area  of  cross-section  or  is  a  wire  cable,  the  area  should  be  then  5.72 
square  inches.     Stresses  and  areas  should  be  divided  by  2  for  two  cables,  etc. 
Suspender  Stress  and  Area. —  For  the  suspender  stress  we  have,  from  (V), 

Suspender  stress  =  (2000  +  1000  +  7oo)-5/  =  30300  pounds. 

If  the  working  stress  is  loooo  pounds  per  square  inch,  the  area  of  cross-section  should  be  3  square 
inches.  For  two  cables  we  have  then  1.5  square  inches  cross-section  for  eacli  suspender. 

If  the  suspenders  are  also  of  steel  wire,  so  that  the  working  stress  is  30000  pounds  per  square  inch,  the 
area  of  cross-sectiosi  would  be  I  square  inch,  or  for  two  cables  0.5  square  inch  for  each  suspender. 

New  Theory  of  Suspension  System. — The  old  method  which  we  have  just  illustrated  is 
based  upon  the  assumption  that  the  cable  carries  the  entire  load,  dead  and  five,  so  that  the 
office  of  the  truss  is  simply  to  prevent  change  of  shape.  Unless,  however,  the  truss  swings 
clear  of  the  supports  even  when  fully  loaded,  this  assumption  is  not  correct.  If  we  suppose 
that  the  truss  just  bears  on  the  supports  when  unloaded,  then  when  the  live  load  comes  on, 
the  truss  must  bear  a  portion  of  this  load  as  well  as  prevent  change  of  shape,  and  the  cable 
carries  then  the  dead  load  and  only  a  portion  of  the  live  load.  The  portion  carried  by  the 
cable  and  by  the  truss  must  then  be  determined  by  the  principle  of  least  work. 

Work  on  Suspenders. — Let  the  truss  just  bear  on  the  supports  when  unloaded,  and 
suppose  a  load  P  to  be  placed  at  any  apex.  Let  a  certain  fraction  of  P  represented  by  0  be 
carried  by  the  cable.  Then  if  there  is  no  change  of  shape,  the  load  <pP  must  act  on  the  cable 

as  a  uniformly  distributed   load,  and  the  load  on  a  suspender  is  .      If  /is  the  length  of  a 

.  2  pij 

suspender,   the  work  on  the  suspender  is    (page    515)  — ^ — .      The  work  on  all  the  sus- 
penders is  then 

Work  on  suspenders  = 


where  at  is  the  area  of  cross-section  and  Et  is  the  coefficient  of  elasticity  for  the  suspenders 
Work  on  Cable. — The  load  per  foot  of  length  on  the  cable  for  a  load  P  placed  on  the 

truss  is         .     We  have  then,  by  putting  —  —  in  place  of  u<  in  equation  (5),  page  61 1,  for  the 
stress  on  a  segment  of  the  cable  of  length  sc 


CHAP-  XI'J  SUSPENSION  SYSTEM.      WORK  ON   TRUSS.  621 

Hence  (page  515)  the  work  on  a  segment  is 


and  the  total  work  on  cable  is 

Work  on  cable  =       :  ,„  ^>  -^, ^ 


where  ae  is  the  area  of  cross-section  of  cable  segment,  and  Ee  is  the  coefficient  of  elasticity 
for  the  cable. 

Work  on  Truss.  —  The  truss  is  subjected  to  a  uniformly  distributed  upward  load  due  to 

0/> 

the  cable  of  —  pounds  per  foot  of  length,  and  also  supports  a  load  P  at  any  distance  z  from 
the  left  end. 

Let  »0  be  the  stress  in  pounds  in  any  member  for  a  uniformly  distributed  load  of  one 

fhP 

pound  per  foot  of  length.      Then  the  stress  for  —  —  pounds  per  foot  will  be 


Let  /  be  the  stress  in  pounds  in  any  member  due  to  one  pound  at  the  distance  z  from 
the  left  end.      Then  the  stress  due  to  P  will  be 

Pp. 

The  stress,  then,  in  any  member  can  be  written 

Stress  =PP-  <^=P(f-^}.       .     .     .     .....      (4) 

where  the  stresses  u0  and  /  are  to  be  inserted  with  their  proper  signs  (-[-)  for  tension  and  (—  ) 
for  compression. 

Now  the  work  of  the  member  is,  from  page  515, 

(Stress)2.? 

Work  =  =  -  F^—  , 
2Ea 

where  s  is  the  length,  a  the  area  of  cross-section  and  E  the  coefficient  of  elasticity.      Insert- 
ing the  value  of  the  stress  just  found,  we  have  for  the  work  of  all  the  members,  or 


Work  of  truss  =  P~^\p — )  —^-.     ...  (5) 

^r          c  i  2Ea 


Value  of  0.  —  We  can  now  find  the  value  of  0  or  that  fraction  of  the  load  P  carried  by 
the  cable. 

Thus,  adding  the  works  given  by  (i),  (3)  and  (5),  we  have 


Total  work  =  + 


sc*  /         0«0\2    s 

+  ^*2(/  ~  TV  25' 


The  value  of  0  is  that  which  makes  this  work  a  minimum. 

</(work) 
If  then  we  differentiate  with  reference  to  0  and  put  —  -TT  —    =  o,  we  have 


Ea 


622 


STATICS  Of-  ELASTIC  SOLIDS. 


[CHAP.  XI. 


J I cnce  we  have 


c^-Ea 


(VI) 


This  value  of  0  requires  that  the  values  of  the  cross-sections,  a,  af,  and  at  of  the  truss 
members,  cable,  and  suspenders  shall  be  known. 

New  Method.  —  We  therefore  assume  for  a  first  approximation  the  values  of  a,  ac  and  at 
as  already  found  by  the  old  method.  Then,  from  (VI),  we  can  find  the  value  of  0  for  each 
apex  live  load  P,  and  we  thus  know  for  each  apex  live  load  P  the  amount  <f>P  carried  by  the 
cable  which  acts  as  a  uniformly  distributed  upward  load  on  the  truss.  This  gives  an  upward 

<f>P 
apex  load  at  every  apex  of  --  . 

We  can  now  find  and  tabulate  the  stress  in  every  truss  member  due  to  each  apex  live 

4>P 

load  P  and  the  upward  apex  load  --  at  every  apex,  and  thus  obtain  the  maximum  live-load 

stresses.      To   these   must  be  added  the  stresses  due  to  temperature   load  wt   as   given  by 
equation  (III). 


The  cable  stress  is  then  given  from  (IV)  by  substituting 


for  w\ 


Stress  in  cable  segment  = 


. 

h 


Sr 


.     .      .     .     (VII) 


The  suspender  stress  is  in  the  same  way,  from  (V), 

Stress  on  suspender  = \-  IVQ  -j-  wt   -. 


(VIII) 


From  these  stresses  the  corresponding  areas  a,  ac  and  ar  can  be  determined,  and  if  not 
sufficiently  close  to  those  assumed,  another  approximation  can  be  made. 

Example. — NEW  METHOD. — We  take  the  same  example  as  before. 

Data.—c  =  54  ft.,  r  =  9  ft.,  h  =  12  ft.,  «  =  6,  E  =  Ec  =  E,  =  30000000  Ibs.  per  sq.  in.,  tre  =  30000, 
and  <rt  =  <r  =  10000  Ibs.  per  sq.  in.,  apex  live  load  P  =  18000  Ibs.,  dead  load  «/„  =  1000  Ibs.  per  ft.,  e  = 
0.00000686,  /  =  80°,  length  of  centre  suspender  /0  =  14  ft. 


CHAP.  XI.j  SUSPENSION  SYSTEM.    EXAMPLE  623 

Members  notated  as  in  the  figure.     The  angle  0  of  braces  with  vertical  such  that  sec  6  =  — . 

Calculation. — We  proceed  first  by  the  old  method  and  find,  as  already  shown,  the  areas  of  cross-section 
for  suspenders,  cable  and  truss,  and  the  temperature  load  «/*. 
We  have  already  found  by  the  old  method 

it/t  =  700        and        a,  =  3, 
and  for  cable  of  uniform  cross-section 


ac  =  5-72,         se  • 
If  the  cable  is  of  links,  we  have 

Atf  a'b'  Vtf 

ac  =  5.72  5.27  5 

sc  =  /To6  4/90  4/82 

We  have  then  for  uniform  cross-section  of  cable 


and  for  cable  of  links 


•Sc 

» —  =    IOOO. 

•ac 


We  have  from  equation  (3),  page  612,  the  length  of  a  suspender, 

'-«*.+£ 

Hence  we  obtain 

2/  =  14  +  2(15  +  18  +  23)  =  126. 

Inserting  these  values  in  (VI),  we  have,  since  Ec  =  E,  =  E,  for  cable  of  uniform  cross  section 


492-375  +     ^-" 


(6) 


For  cable  of  links  we  have  438  in  place  of  492-375- 

We  can  now  form  the  table  on  page  624. 

For  each  member  in  the  truss  (omitting  counters)  we  give  in  the  table  the  area  a  ,n  square  inches  as 
found  already  by  the  old  method,  the  length  ,  in  feet,  the  stress  „.  in  pounds  due  to  a  un.form  load  of  one 
pound  per  foot  of  length  or  9  pounds  at  each  lower  apex. 

We  can  determine  for  each  member  the  quantity^,  and  we  have 


^^-^  =  36210. 

^— -    a 

Equation  (6)  thus  becomes,  for  cable  of  uniform  section, 

^~^  .     (7) 


<}>  = 


492.375  +  670.55 


624 


STATICS  OF  ELASTIC  SOLIDS. 


[CHAI-.  XI. 


s 

a 

«o 

Ut's 

a 

A      i 

A 

A 

A 

A 

aA 

12 

2.93 

-22.5 

+  2080 

—  0.833 

-0.666 

-0.5 

-0.333 

-0.166 

OiAt 

13 

2.29 

—  13-5 

+    955 

+  0.166 

-  0.666 

-0.5 

-0.333 

-0.166 

Ot.'lt 

12 

.66 

-    4-5 

+    146 

+  0.166 

+  0.333 

-0.5 

-0.333 

-0.166 

at  A  i 

13 

•35 

o 

0 

o 

o 

o 

o 

0 

fl«//4 

12 

.66 

-    4-5 

+    146 

-0.166 

-0.333 

-0.5 

+0.333 

+  0.166 

atAt 

12 

.29 

-13-5 

+    955 

-0.166 

-0.333- 

-0.5 

-0.666 

+  0.166 

bB 

12 

.92 

-22.5 

+  2080 

-0.166 

—  0.333 

—  0.5 

-  0.666 

-  0.833 

aA, 

15 

.66 

+  28.125 

+  3241 

+  1.041 

+  0.833 

+  0.625 

+  0.416 

+0.208 

aiAt 

15 

.87 

+  16.875 

+  1488 

—  o.  208 

+  0.833 

+0.625 

+  0.416 

+  o.  208 

a,A* 

15 

.08 

+    5-625 

+    228 

—  o.  208 

—  0.416 

+  0.625 

+  0.416 

+  0.208 

a4At 

15 

.oS 

+   5-625 

+    228 

+  o.  208 

+  0.416 

+  0.625 

—  0.416 

—  0.208 

atA< 

15 

.87 

+  16.875 

+  1488 

+0.208 

+  0.416 

+  0.625 

+  0.833 

-0.208 

bAt 

15 

.66 

+  28.125 

+  3241 

+  0.208 

+  0.416 

+  0.625 

+  0.833 

+  1.041 

aai 

9 

.08 

-  16.875 

+  1232 

—  0.625 

-0.5 

-0.375 

—  0.25 

—  0.125 

Oirti 

9 

•79 

-27 

+  2352 

-0.5 

—  I.O 

-0.75 

-0.5 

-025 

at<i, 

9 

.02 

-  30.375 

+  2749 

-0.375 

-0.75 

-1.125 

-0.75 

-0.375 

0*04 

9 

.02 

-  30-375 

+  2749 

-0.375 

-0.75 

-1.  125 

-0.75 

-0.375 

0*at 

9 

•79 

-27 

+  2352 

-0.25 

-0.5 

-0.75 

—  i.o 

-0.5 

*l 

9 

.08 

-16.875 

+  1232 

—  0.125 

—  0.25 

-0.375 

-0.5 

-A625 

AA, 

9 

.08 

o 

0 

o 

0 

o 

o 

o 

A,A, 

9 

.08 

+  16.875 

+  1232 

+  0.625 

+  0.5 

+  0.375 

+  0.25 

+  0.125 

AtAt 

9 

•79 

+  27 

+  2352 

+  0.5 

+  1.0 

+  0.75 

+  0.5 

+  0.25 

AtA4 

9 

•79 

+  27 

+  2352 

+  0.25 

+  0.5 

+  0.75 

+  1.0 

+0.5 

A4At 

9 

.08 

+  16.875 

+  1232 

+  0.125 

+  0.25 

+  0.375 

+  0.5 

+  0.625 

AtB 

9 

.08 

0 

o 

o 

0 

0 

o 

0 

36210 

For  cable  of  links  we  have  438  in  place  of  492.375. 

We  also  give  in  the  table  the  stress  in  every  member  due  to  loads  p,,p*,pttp*,  PI  of  one  pound  placed 
at  each  lower  apex. 

We  can  now  form  the  following  table  giving  the  quantity  1—2-  for  each  member,  and  hence  ^>^— s£  fOr 
one  pound  at  A, ,  At ,  At ,  A4 ,  and  At. 


a 

a 

a 

a 

a 

aA 

+    77-05 

+  61.64 

+'  46.23 

+  30.82 

+    15-41 

a,Ai 

—     "-79 

+  47.16 

+    35-37 

+  23-58 

+     n-79 

a,At 

-      5-4 

-  10.8 

+    16.2 

+  10.8 

+      5-4 

atAt 

0 

o 

0 

0 

0 

a4A4 

+      5-4 

+  10.8 

+    16.2 

-  10.8 

5-4 

atAt 

+    11-79 

+  23-58 

+  35.37 

+  47.16 

-     "-79 

bB 

+    I5-4I 

+  30.82 

+  46.23 

+  61.64 

+    77-05 

aA, 

+  M9-87 

+  95-90 

h   71-92 

+  47-95 

+    23.97 

atAt 

-     18.34 

+  73.36 

+    55-02 

+  36.68 

-     18.34 

atAt 

-      8.44 

+  16.87 

+    25  3i 

+  16.87 

+      8.44 

a4At 

+      8.44 

-  16.87 

+    25.31 

-  16.87 

-       8.44 

atAt 

+    18.34 

+  36.68 

+    55-02 

+  73.36 

-     18.34 

bAt 

+    23.97 

+  47-95 

+    7I-92 

+  95.90 

+  119-87 

aa, 

+    45-65 

+  36-52 

+    27-39 

+  18.26 

+      9-13 

a,at 

I-    43-54 

+  86.08 

+    65.31 

+  43-54 

h    21.77 

a  tat 

+    33-94 

+  67.88 

+  101.82 

+  67.88 

h    33-94 

otat 

H    33-94 

+  67.88 

+  101.82 

+  67.88 

r    33-94 

a4at 

+    21.77 

+  43-54 

+    65.31 

+  86.08 

f-    43-54 

atb 

+      9-13 

+  18.26 

+    27.39 

+  36.52 

+    45-65 

AA, 

0 

o 

o 

o 

O 

A,At 

1-    45-65 

+  36.52 

-f  27.39 

+  18.26 

+    9-13 

AtAt 

+    43-54 

+  86.08 

+  65.31 

+  43-54 

+  21.77 

AtA4 

+    21.77 

+  43-54 

f-  65.31 

+  86.08 

i-    O-54 

A4At 

+      9-13 

+  18.26 

+  27.39 

+  36.52 

+    45-65 

AtB 

o 

0 

o 

O 

o 

\ 

+  544-36 

+  921-65     ,  +  1074.54    1    +  921.65 

+  544-36 

i                                                                       i 

SUSPENSION  SYSTEM.     EXAMPLE. 
We  have  then  from  (7)  for  the  apex  loads  A ,  A ,  7% ,  A  ,  /%  ,  for  cable  of  links, 


625 


/>, 

0  =  0.49 


0.83 


A 

0.96 


Pt 

0.83 


0.49 


We  have  P  =  ,8000  pounds  in  the  present  case  and  the  upward  load  on  truss  *£  =  3ooo0  pounds  at 
every  apex  of  the  loaded  chord.     We  also  have  the  temperature  load  «,,  =  700  pound's  per  foot 
We  can  then  draw  up  the  following  table  of  stresses  for  the  truss. 


aA 

atAt 

a?A* 

a3A3 

aA, 

aiAt 

a,A» 

P, 

P^ 
P^ 
P. 
P* 

-  "325 
-  5775 
—  1800 
o 

0 

0 

-  8265 
-  4680 
—  2265 
-   795 

-  5203 

0 

-  7560 
-  4755 
—  2265 

-  3735 
-  7245 
o 
—  7245 

—  3735 

+  14156 
+  7219 
+  2250 
-   281 
-   844 

-  6506 
+  10331 
+  5850 
+  2831 
+   994 

-  4668 
-  9056 
+  9450 
1-  5944 
+  2831 

*    \ 

+  19687 
-  19687 

+  11812 
—  11812 

+  3937 
-  3937 

—  15750 

-  15750 

-  9450 

-  3150 

Max.  Stresses.  ..  \ 

+  433" 
—  20812 

+  31818 
-  18318 

+  22162 
-17661 

-  34650 

—  31755 

-  29235 

—  25110 

aai 

aia* 

a?a3 

AA* 

AiA, 

A*A, 

Pi 

P* 

P\ 
P* 

-  8494 
-  4331 
-  1350 
o 

o 

—  8494 
—  10530 
-  4860 
-  1530 
-   90 

-  4590 
-  10530 
-  10530 
-  —  5096 
-  1789 

o 
o 
o 
—   168 
-   506 

+  4590 
+  4331 
+  1350 
-   168 
-   506 

+  1789 
+  5096 
+  4860 
+  1530 
+   90 

.,   { 

+  11812 
—  18900 

+  18900 
—  21262 

-f-  11812- 
—  18900 

+  18900 
-  21262 

—  11812 

—  11812 

i 

+  11812 
-  44404 

+  18900 
-  53797 

-f 

22083 
19574 

+  3226^ 

Max.  Stresses...] 

-  25987 

—  12486 

—  21262 

From  this  table  we  find  the  maximum  stresses  in  the  truss  members. 

We  have  now  from  (VII),  since  ivt  =  700,  tv0  =  1000,  2<f>  =  3.6,  for  the  stresses  in  the  cable  segments, 


A'a' 
I344I5 


123714 


118102 


The  suspender  stress  is,  from  (VIII),  26100  pounds. 

We  can  now  find  the  corresponding  areas  of  cross-section  of  the  members,  and  if  these  areas  differ  too 
much  from  those  assumed,  should  make  a  new  calculation  with  the  new  areas. 


626  STATICS  OF  ELASTIC  SOLIDS.  [CHAI-.  XI. 

COMPARISON  OF  THE  Two  METHODS. — In  the  present  case  we  have  then  the  following  results: 


Old  Method. 

New  Method. 

Old  Method. 

New  Method. 

oA 

—  29250 
—  22950 

-  34650 

-  3«755 

atat 

±  30262 

j  +  18900 
1  ~  53797 

atAt 

—  16050 

-  29230 

AA, 

—  2O8I2 

-  12486 

at  At 

-  13500 

-  25II5 

A  A 

±  20812 

j  +  22083 

aAl 

±  36562 
(  +  28687 

+  433" 
—  20812 

AtAt 

±  27900 

1  ~  19574 
j  +  32265 
(  —  21262 

a\A\ 
atAt 

J  -  27000 

'  -  18318 
+  22162 
—  17661 

A'a' 

/if 

+  I7I495 
+  I58I75 
+  149850 

+  I344I5 
+  123714 
+  118102 

aa, 

—  2O8I2 

-  25987 

Susp. 

+  30300 

+  26100 

aiat 

±  27900 

I  +II8I2 

I    444O4 

We  see  that  the  cable  and  suspender  stresses  are  much  less  by  the  new  method.     For  the  truss  members 
by  the  new  method  tke  direct  stresses  are  greater  and  the  counter-stresses  less. 


INDEX. 


ABUTTING  joints 487 

Acceleration,  angular 82 

axial  and  normal 83 

couple 144 

"       in  terms  of  linear 85 

"       moment  of 90 

rectangular  components  of     83 
resolution  and  composition 

of 83 

"       resultant 83,  159 

axial  and  normal 83,  160 

"  central m,  160 

directly  as  distance 125 

inversely  as   square   of  dis- 
tance  . 112 

of  centre  of  mass 299 

deflecting  and  deviating 80,  159 

of  gravity 100 

"        "        experimental   determina- 
tion of 340 

instantaneous  linear 75 

axis  of 151,  161 

linear,  resolution  and  composition 

of 76 

linear  and  angular  combined 150 

"  mean  linear 75 

"  moment  of 89,  90,  160 

of  moment  of  momentum 301 

radial  and  axial 80 

rectangular  components  of 77 

resultant 159 

analytic  determination  of      77 

spontaneous  axis  of 152 

"  tangential 158 

"          and  central 77 

uniform 100 

"  "       and  variable 78 

"  "       motion  on  an  inclined  plane   136 

"  "       motion  in  a  curve 103 

variable 109 

Adhesion 220 

Advantage,  mechanical 267 

Alphabet,  Greek i 

Analytic  determination  of  resultant  acceleration      77 
"                      "                "         "           velocity  ....     67 
'•                     "                *         "           angular    ve- 
locity      60 


Angle,  conical 7 

of  friction 263 

of  rupture  for  earth ' 468 

unit  of 6 

Angular  and  linear  acceleration  combined 150 

rate  of  change  of  speed 73 

velocity  combined. .' 144 

displacement 52 

resolution  and  composition 

of 1 58 

in  terms  of  linear  velocity 70 

speed  and  velocity 61 

speed,  rate  of  change  of 96 

velocity,  composition  and  resolution  of.     68 

"          moment  of 90 

rectangular  components  of. ...     69 

"          resultant 155 

"        analytic      determina- 
tion of 69 

"          uniform  and  variable 65 

Arch,  conditions  of  stability  of 599 

' '      dam 450 

"      framed,  fixed  at  ends 586 

"         hiriged  at  ends 579 

"         three  hinges 578 

"      metal 578 

"      pressure  curve  of 598 

"      reduced  surcharge  for 598 

"      solid,  fixed  at  ends 592 

hinged  at  ends 592 

stone 597 

"      straight . 607 

Area,  material 23 

"       moment  of 22 

Astronomical  unit  of  mass 206 

Attraction  of  homogeneous  shell  or  sphere 203 

Axes,  principal .  / 35 

Axial  acceleration   160 

"      moment 194 

"      and  normal  angular  acceleration 83 

"         "     radial  acceleration 80 

Axis  of  acceleration,  instantaneous 161 

"     instantaneous 145,  157,  371 

"     invariable 302,  378,  391 

"     moment  about 180 

"     neutral 493,  509 

"     of  rotation,  instantaneous 145,  157 

627 


($28 


INDEX. 


Axis  of  symmetry 23 

Axle  friction 227 

"          "        work  of 266 

BALANCE,  determination  of  mass  by 212 

Ballistic  pendulum 362 

Beams,  deflection  of 530 

"        fixed  horizontally  at  ends 522 

"        impact  of 358 

"        shearing  stress  in 553 

"       strength  of 499 

"       of  uniform  strength 503 

"        work  of  shearing  force  in 554 

Bending  and  compression  combined 528 

"  "     tension  "  528 

moment 494 

stress 493 

"         work  of 521 

Blackburn's  pendulum   135 

Butt-joint 487 

Brae  histoch  rone 142 

Brake,  friction 265 

CARTESIAN  co-ordinates 12 

Central  acceleration ill,  160 

directly  as  the  distance 125 

inversely  as  square   of  dis- 
tance    112 

"       and  tangential  acceleration 77 

impact 342,  344,  347 

Centre  of  gravity 22,  190,  207 

"  mass  20,  23,  189 

acceleration  of 299 

conservation  of 299 

motion  of 299 

"        "      "      resultant  force  at 193 

"      velocity  of 298 

"  oscillation 337 

"  parallel  forces 188,  414 

"         "  percussion 339 

Centrifugal  force 245 

Change  of  motion  of  a  point  of  a  rigid  body 157 

Coefficient  of  cohesion  for  earth 468 

"  elasticity 346,  476 

"         for  earth 464 

"  friction 221,  224,  263,  267 

for  shear 511,  555 

"  resilience 515 

"  rupture 499,  510 

Cohesion  of  earth 466 

Column,  the  ideal 559 

Columns,  factor  of  safety  for  long 569 

formulas  for  long 565 

Euler's  formula  for  long 563 

Gordon's       "       "        "   568 

Merriman's  "       "        "    568 

parabolic       "       "        "   566 

"         Rankine's     "       "        "   567 

the  straight-line  formula  for  long 565 

"         strength  of 559 

Combined  bending  and  tension 528 


,  CAOt 

Combined  bending  and  compression 528 

"         flexure  and  torsion 512 

tension  and  compression  or  shear 484 

Components  of  change  of  motion,  invariant  for.  iM 

"           "  motion,  invariant  for 157 

"           rectangular 55 

of  acceleration 77 

"  angular  acceleration.  83 

"         "         velocity 69 

"  velocity 67 

Composition  and  resolution  of  angular  accelera- 
tions   83 

"           and  resolution  of  angular  displace- 
ments   58 

and  resolution  of   angular  veloci 

ties 68 

and  resolution  of  forces 176 

"            "  linear    accelera- 
tions   76 

and  resolution  of  linear   displace- 
ments   54 

and  resolution  of  linear  velocities.  66 
"  moments. ...  88,  180 

"           of  harmonic  motions 129 

Compound  harmonic  motion 129 

"           pendulum 337 

Compression  and  bending  combined 528 

'     tension 484 

Compressive  stress  and  force 397 

Concurring  coplanar  forces 404 

forces 1 79 

"      resultant  for 179 

Cone  of  friction 222 

Conical  angle,  unit  of 7 

"        pendulum,  simple 244 

Conservation  of  areas 303 

"  centre  of  mass 299 

"  energy 276,  304 

"  momentum 300 

"              "  moment  of  momentum,  302,  320,  369, 

377,  390 

Consolidation  of  earth 350 

Constant  of  gravitation 205 

Constrained  motion  curved  path,  uniform  accel- 
eration   137 

"       in  a  circle,  uniform  accelera- 
tion   138 

"                "        of  a  point 136 

Co-ordinates,  Cartesian 12 

polar ii 

Cords  and  chains,  friction  of 231 

Cosines,  direction 13 

Couple,  angular  acceleration 144 

force 184 

••            "      resultant  of 185 

"      or  wrench 194 

Curvature,  unit  of 7 

Cycloid,  motion  in,  uniform  acceleration 138 

Cylinders,  strength  of 486 

D'ALEMBERT'S  principle, 242,  297 


INDEX. 


629 


Dams,  masonry 424 

design  of 434 

"          economic  section 440 

Deflecting  force 243 

'     at  earth's  surface 246 

"  and  deviating  acceleration 80,  159 

Deflection  of  beams 530 

"  framed  structures 516 

influence  of  shearing  force  on 554 

Degree 6 

Density 15 

Derived  unit 2 

Determination  of  mass  by  balance 212 

"  centre  of  mass 23 

Deviating  and  deflecting  acceleration 80,  159 

Deviation  of  falling  body 252 

Dimensions  of  a  derived  unit 2 

Direction  cosines 13 

Displacements,  angular,   resolution  and  compo- 
sition of 58 

line  representative  of 52 

linear  and  angular 52 

"       resolution    and    composi- 
tion of 54 

moment  of 89 

"  relative 55 

"         triangle  and  polygon  of     56 

Dynamic  components  of  motion 192 

"         equilibrium 210 

Dynamics i,  169 

Dyne 170 

EARTH,  angle  of  rupture  for 468 

"        coefficient  of  cohesion  for 468 

"  "         "  friction  for 466 

' '        cohesion  of 466 

"        consolidation 35° 

"        ellipticity  of 247 

"  .     mass  equilibrium  of 467 

"        pressure 454 

"        slope 468 

"        surface,  motion  on 251 

Eccentric  impact 3^3 

Effective  forces....' 242,  297 

"  "      fixed  axis,  moment  of 316 

"  "       rotation  and  translation,  moment 

of 389 

"             "       rotation  about  a  fixed  point,  mo- 
ment of 375 

"  "       translating  axis,  moment  of 367 

Efficiency 267 

Elastic  limit 475 

"       solids,  statics  of 473 

Elasticity 473 

"         coefficient  of 346.  47& 

"     for  shear 511.  555 

law  of 474 

"         modulus  of 347 

Elementary  mass  or  particle 20 

Ellipsoid  of  inertia 36 

h.lipticity  of  earth 247 


Energy,  change  of  potential., 


PAGE 
278 

"        conservation  of 276,  304 

"        kinetic t 271 

"  "      fixed  axis   320 

of  a  system 303 

rotation  about  a  fixed  point 378 

"       rotation  and  translation 391 

"      translating  axis 370 

"        law  of 275,304 

' '        potential 274,  304 

Epoch 128 

Equations,  homogeneous 3,  86 

Equilibrium  of  earth  mass 467 

"  "     forces 209 

"  "     material  system 310 

"     a  particle 277 

"     a  system  of  bodies 398 

limiting 221 

"          of  non-concurring  forces 211 

polygon 413 

stable 311 

"          static,  molar,  dynamic,  molecular  ..   210 

Equipotential  surface 287 

Erg 261 

Euler's  formula  for  long  columns 563 

"       geometric  equations 167 

External  forces - 297 

Eyebars  and  pins,  theory  of 506 


FACTOR  of  safety 

"        "       "       for  long  columns. 

Falling  body,  deviation  of 

Fixed  axis,  rotation 

' '      point,  rotation  about 

Flexure 

"        and  torsion  combined 


Force 

"  centrifugal 

"  couple 

"  "  resultant  of 185, 

"  deflecting 

"  "  at  earth  surface 

"  gravitation  unit  of 

"  impressed  and  effective 169, 

"  line  representative  of 

"  lines  of 

"  moment 

"  "  unit  of 

"  resultant  at  centre  of  mass 

"  and  stress 

"  system,  invariant  for 

"  tangential 

Forces,  centre  of  parallel 

"        equilibrium  of 209, 

"        impressed  and  effective 

"        non-concurring 

"  "  resultant  for 

"        revolution  and  composition  ot 

"        triangle  and  polygon  of 

"  work  of  non-concurring 

Formulas  for  long  columns 


481 
569 
252 
315 
375 
493 
512 
170 
245 
184 
194 
243 
246 
171 
242 
176 
287 
1 80 
iBi 
193 
473 
197 
255 
188 

211 
297 
IQI 

jSf> 


630 


1NDI-X. 


Framed  arch 578 

44  "  three  hinges 578 

"  hinged  at  ends 579 

4  "  fixed  at  ends 586 

"  structures 397,400 

"  "  deflection  of 516 

Friction,  angle  of 263 

41  axle 227 

"  brake 265 

44  coefficient  of 221,224,263,265,267 

"  "  for  earth 464 

44  cone  of 222 

"  kinds  of 220 

44  kinetic 263 

laws  of 222 

"  of  cords  and  chains 231 

"  "  masonry 424 

44  44  oblique  central  impact 353 

"  4<  pivots  224 

44  rolling 235 

*4  static 220 

"  wheels 230 

44  work  of  axle 266 

GEOMETRIC  equations,  Euler's 167 

Gordon's  formula  for  long  struts 568 

Graphical  statics 404 

Gravitation,  constant  of 205 

force  of 203 

"             law  of 203 

44             potential 286 

Gravity 14,  173 

"       acceleration  of 22 

"       centre  of 22,  190,  207 

41       experimental  determination  of  accelera- 
tion of 340 

"       specific 16 

Greek  alphabet I 

Gyration,  radius  of 32,  338 

HARMONIC  motion 125 

"      simple 126 

"               "      compound 129 

Hemp  ropes 235 

High  dam 434 

44      wall 428 

Hodograph 79 

Homogeneous  body A 15 

"  equations 3,  86 

shell  or  sphere,  attraction  of 203 

ICE-  and  wave-pressure 432 

Ideal  column 559 

Impact 342 

"      of  beams 358 

1 '      direct  central 342,  344,  347 

' '      eccentric 363 

imperfectly  elastic 347 

"      non-elastic 342 

"      oblique  central. . 352 


PAGE 

Impact  oscillating  body 361 

perfectly  elastic 344 

14      rotating  bodies 360 

"      and  strength 356 

Impressed  force 169 

"          and  effective  forces 242,  297 

Impulse 172,  257 

and  momentum 172,257 

Inclined  plane,  motion  on 136 

Indeterminate  stresses 407 

Inertia 169 

"      moment  of 31,  317 

"        *4  determination  of 38 

"         "  experimental  determination  of    340 

44        "  ellipsoid  of 36 

"  any  axis  in  general 34 

44        "  relative  to  an  axis 33 

"  a  point 34 

44  "        "  reduction  of 32 

"polar 33 

Instantaneous  angular  acceleration 82 

axis 371 

"    of  acceleration 151,  161 

14    of  rotation 145,156,392 

linear  acceleration 75 

and  angular  speed  and  ve- 
locity      63 

rate  of  change  of  speed 75 

Internal  and  external  forces 297 

"         stresses  in  a  beam 557 

Invariable  axis 378 

14    and  plane 302 

"    rotation  and  translation 391 

Invariant  for  components  of  motion 157 

"  change  of  motion. .    161 
"  4'    force  system 197 

JOINT,  abutting 487 

"  lap 487 

"  single-riveted 487 

44  stability  of  masonry 425 

Joule 261 

KEPLER'S  laws 120 

Kinematics I,  51 

44           of  a  point 91 

14           of  a  rigid  body 143 

Kinetic  energy 271 

"       fixed  axis 320 

14       rotation  and  translation 391 

41         about  a  fixed  point 378 

"       of  a  system 303 

"       translating  axis 370 

friction. 263 

Kinetics I,  242 

"       of  a  material  system 297 

LAP-JOINT 487 

Law  of  elasticity - 474 

"     "energy 275.304 

"     "  gravitation 203 

44     "  the  lever 183 


INDEX. 


63* 


Laws  of  friction 222 

Least  work,  principle  of 3H(  517 

Length,  unit  of 4^ 

Lever,  law  of ^3 

Limiting  equilibrium 221 

Line,  material , 20,  23 

"            "         moment  of 22 

"      representative  of  angular  acceleration 83 

"        displacement....  52 

"        velocity 66 

"                                   "  force 176 

'     moment 181 

"  linear  acceleration 76 

displacement 52 

"                                   "       "       velocity 66 

"                  "                 "  moment 83 

Lines  of  force , 287 

Linear  acceleration,  mean 75 

"                                   instantaneous...... 75 

resolution   and    composition 

of 76 

"        and  angular  acceleration  "          150 

"          "            "         rate  of  change  of  speed 73 

"         "           "        speed  and  velocity 61 

"         velocity  combined 144 

"       density 15 

"        displacement 52 

"  resolution  and  composition  of     54 

"       in  terms  of  angular  acceleration 84 

"        "       "       "         "        velocity 70 

"        velocity,  resolution  and  composition  of  . .  66 

Long  struts,  Euler's  formula 563 

"           "        factor  of  safety  for 569 

"           "        formulas  for 565 

•'           "        Gordon's  formula 568 

"           "        Merriman's     "        568 

"           '•        parabolic         "       566 

Rankine's       "       567 

"           "        straight-line"        5^5 

strength  of 559 

Low  dam 434 

"     wall 429 


MASONRY  dams 

"          joint,  stability  of 

"          weight  and  friction  of 

Mass,   acceleration  of  centre  of 


424 
425 
424 
299 


astronomical  unit  of 206 

centre  of 2O 

conservation  of  centre  of 299 

determination  of,  by  balance 212 

elementary ......     20 

independent  of  gravity 14,  173 

measurement  of r4.  *73 

moment  of 22 

motion  of *74 

"       "  centre  of 299 

notation  for *5 

properties  of  centre  of l89 

reduction  of 322 

and  space,  measurable  relations  of 14 


Mass,  specific  .................................  16 

"      standard  unit  of  .....................  4,  5,  14 

velocity  of  centre  of  .....................  298 

Material  line  .................................  20,  23 

"         particle  ....................  ..........  169 

"         surface  ......................    .......  20 

"         system,  equilibrium  of  ................  310 

"         kinetics  of  ..................  .  297 

Matter,  states  of  ...............................  i 

Mean  linear  acceleration  .......................  75 

"          "        and  angular  velocity  ...............  61 

Measurable  relations  of  mass  and  space  .........  14 

Measurement  ..................................  2 

of  mass  .......................    14.  173 

Measures,  table  of  .............................  8 

Mechanical  advantage  .........................  267 

Mechanics  .....................................  I 

Merriman's  formula  for  long  struts  .............  568 

Metal  arch  ....................................  578 

"      springs  .................................  537 

Middle-third  rule  ..............................  426 

Modulus  of  elasticity  ..........................  347 

Molar  equilibrium  ............................  210 

Molecular     "           .............................  210 

Moment  about  an  axis  .......................  88,  180 

axial  ..................................  194 

"        bending  ...................  ...........  494 

"         normal  ................................  194 

"        of  acceleration  ....................  89,  160 

"aline  ..............................  22 

"  angular  acceleration  ................  90 

"          "         "         velocity  ...................  90 

"area  .............................  22 

"         "  a  vector  quantity  ...................  88 

"          "  displacement  ..................    ....  89 

"          "  effective  forces,  fixed  axis  ..........  316 

"         "  "  "         rotation  about  fixed 

point  ............  375 

"          ««  "  "         rotation   and    trans- 

lation ............  389 

"          "           "              "     ,  translating  axis  .....  367 

"  force  ..............................  181 

"  inertia  ......................   31.34.3*7 

"          "         "     determination  of  .............  38 

••          "         •«                                   "  experimental  340 

"          "        "     polar  ........................  33 

"          "        "     reduction  of  .................  32 

"          ««         '•     relative  to  an  axis  ...........  33 

"          •'         "           "         "  a  point  ............  34 

"mass  ...............................  22 

"          "  momentum  ........................  3°° 

«<         ««          "           acceleration  of  ..........  301 

•«          "  "  conservation  of..  302,  320.  369 

'«         ««          "            rotation  about  fixed  point  377 

•«          «•           "            rotation  and  translation.  390 

«         ««          "            rotation,  fixed  axes  .....  318 

«          «           «•            translating  axis  .......    .  368 

"  velocity  .........................  89.  *55 

"          "volume  ............................  22 


resistng  ....... 

resolution  and  composition  of 


180 


63* 


/v/vr.v. 


Momentum 1 72,  255 

"  and  impulse 172,  257 

"          conservation  of 300 

"  ol  a  system 298 

"          rotation,  fixed  axis 318 

about  fixed  point 376 

translating  axis   368 

"  translation  and  rotation 390 

Motion,  change  of 157 

"         dynamic  components  of 192 

"         in  a  cycloid,  uniform  acceleration 136 

"         Newton's  first  law 169 

"  "          second  law 172 

third  law 174 

"         of  a  point  of  a  rigid  body 153 

"         of  centre  of  mass 174,  299 

of  particle  on  earth  surface 251 

on  an  inclined  plane 136 

NEUTRAL  axis 493.  509 

Newton's  laws  of  motion 169,  172,  174 

Non-concurring  forces 179,  191,  412 

"         equilibrium  of 211 

"         resultant 186 

"  "         work  of 215 

Normal  and  axial  angular  acceleration 83 

moment 194 

Notation  for  mass 15 

OSCILLATING  body,  impact  of 361 

Oscillation,  centre  of 337 

PARABOLA  formula  for  long  struts 566 

Parallel  forces,  centre  of 188 

"    graphic  construction. .  414 

Particle , 20 

equilibrium  of 277 

"         kinetics  of 242 

material 169 

moving  on  earth  surface —  251 

"         stable  equilibrium  of 277 

"        unstable         "           " 277 

Path  of  a  point 51 

Pendulum,  ballistic 362 

Blackburn's 135 

"           compound 337 

"           simple 336 

"       conical 244 

"                 "       time  of  vibration 138 

Per,  meaning  of 3 

Percussion,  centre  of 339 

Perpetual  motion 304 

Phase 128 

Physical  science i 

Pile-driving 351 

Pins  and  eye-bars,  theory  of 506 

Pipes,  strength  of 486 

Pivot-  or  swing-bridge 571 


Pivots,  friction  of. 


224 


Plane,  and  axis  of  symmetry 23 

"        invariable 302 


Planetary  motion 120 

Point,  position  of n 

"       of  reference n 

Polar  co-ordinates n 

"       moment  of  inertia 33 

Polygon,  equilibrium 413 

offerees 176 

"    linear  displacement 54 

"         "    relative           "              56 

Position  of  centre  of  mass 21 

Potential  energy 274 

change  of 278 

"         of  a  system 304 

"         gravitation 286 

Poundal 170 

Power 262 

41       transmission  of,  by  shafts 512 

Pressure  curve  for  arch 598 

"         on  fixed  axis 319 

Principal  a*es 35 

1     properties  of 35 

Principle  of  least  work 311,  517 

Properties  of  centre  of  mass 189 

RADIAL  and  axial  acceleration 80 

Radian 6 

"       square 7 

Radius  of  gyration 32,  338 

Rankine's  formula  for  long  struts 567 

Rate  of  change  of  speed 91 

"     ' "      angular 96 

"      linear  and  angular 73 

'     work 262 

Reaction,  influence  of  shearing  force  on 556 

Rectangular  components 55 

of  acceleration 77,83 

"    angular  acceleration  83 

"         "         velocity....  69 

"     velocity 67 

Reduction  of  mass -. 322 

"     moment  of  inertia 32 

Redundant  members 520 

Reference,  point  of u 

Relations,  measurable,  of  mass  and  space 14 

Relative  displacement 55 

."                                     triangle  and  polygon  of. .  56 

Resilience,  work  of 515 

Resisting  moment 496 

Resolution  and  composition  of  angular  accelera- 
tion    83 

Resolution  and  composition  of  angular  displace- 
ment   58 

Resolution  and  composition  of  angular  velocity..  68 

"     forces 176 

"     linear      accelera- 
tion    76 

Resolution   and  composition   of  linear  displace- 
ment    54 

Resolution  and  composition  of  linear  velocity. ..  66 

"  moments 88,  iSo 

Resultant  acceleration 159 


INDEX. 


633 


Resultant  acceleration,  analytic  determination  of. 

angular  acceleration 

analytic     determi- 
nation of 

velocity 

analytic     determina- 
tion of 

concurring  forces 

couple  or  wrench 

force  at  centre  of  mass 

for  non-concurring  forces 

moment 

44          of  a  force  couple 

velocity 

analytic  determination  of  .... 

"  work  of 

Ropes,  hemp 

"         rigidity  of 

"         wire 

Rigid  body,  change  of  motion  of  a  point  of 

kinematics  of 

motion  of  a  point  of 

Rigidity  of  ropes 

Rivet  table 

Riveting,  theory  of 

Rolling  contact,  stability  in 

"          friction •. 

Rotating  bodies,  impact  of 

momentum  of 

moment  of  momentum 

pressure  on  fixed  axis 

Rotation,  and  translation 153, 

fixed  axis 

fixed  point 154, 

instantaneous  axis  of 145,  157, 

invariable  axis  of 302,378, 

spontaneous  axis  of 

"         translating  axis 

"  "  "  momentum. .  . 


77 
159 

33 
155 

69 

179 
194 
193 
1  86 

89 
185 
154 

67 
215 

235 
234 
235 
157 
143 
153 
234 
490 
487 
312 
235 
360 


319 
389 
315 
375 
37i 
39i 
145 
367 
390 


SAFETY,  factor  of 481 

Science,   physical i 

Screw  spin 145 

"        wrench 192,197 


Set 


475 


Shear,  coefficient  of  elasticity  for 511,  555 

"         and  tension  combined 484 

Shearing  force  in  beams,  work  of 534 

influence  of,  upon  deflection 554 

"  "      "      reaction 556 

Shearing  stress 553 

"  "      and  force 397 

Shell  or  sphere,  attraction  of  homogeneous 203 

Simple  conical  pendulum 244 

"  •     harmonic  motion 126 

"       pendulum '. 336 

"          time  of  vibration 138 

Single-riveted  joint 487 

Solid  arch  fixed  at  ends 592 

44     hinged  at  ends 583 

Space  and  mass,  measurable  relations  of. .......     14 


PACK 

Specific  mass  and  gravity j6 

Speed,  and  velocity (Jl 

rate  of  change  of 73    gi 

S/ln : •••.'••..  -'i45 

Spontaneous  ax-s  of  acceleration 152 

"      "  rotation 145.392 

Springs,  metal 537 

Square  radian 7 

Stability,  conditions  of,  for  arch 599 

of  masonry  joint 425 

in  rolling  contact 3I2 

of  a  wall 426 

Stable  equilibrium 3II 

of  a  particle 277 

Standard  units ^ 

"     of  lengch 5 

41     "  mass 5 


Statement  of  a  quantity 2 

States  of  matter r 

Static  equilibrium 2ro 

friction 22O 

Statics 209 

of  elastic  solids 473 

graphical 412 

Stone  arch 597 

Straight  arch 607 

"       -line  formula  for  long  struts 565 

Strain... 


473 

'     due  to  weight 479 

Straining,  work  of 515 

Strength  and  impact 356 

"         of  beams 499 

41         "  long  struts 559 

"          "  pipes  and  cylinders 486 

"         ultimate 475 

Stress 174-  397 

"     and  force 473 

"     bending 493 

"     in  framed  structures 400 

44     internal,  in  a  beam 557 

44     shearing 553 

"     working 481 

Struts,  Euler's  formula  for 563 

"        factor  of  safety  for 569 

41       formulas  for  long 565 

41       Rankine's  formula  for 567 

"       strength  of  long 559 

Superfluous  members 402,  520 

Surface,  material 20 

Surcharge,  reduced  for  arch 598 

Surface  density 15 

Suspension  system 609 

•4<       new  theory 620 

"       o'd  theory 613 

44       stiffening  truss     612 

41  "       temperature  load 613 

Swing-bridge ...   571 

Symmetry,  plane  and  axis  of 23 


TABLE  of  measures. 


634 


INDEX. 


Tangent  acceleration 158 

Tangential  and  central  acceleration 77 

force 255 

Temperature  stress,  framed  arch  fixed  at  ends..   591 
"  "        framed  arch  hinged  at  ends  581 

••  "        solid  arch       fixed  at  ends..    585 

"  "        solid  arch       hinged  at  ends  595 

Tensile  stress  and  force 397 

Tension  and  bending  combined 528 

"  "     compression "         484 

"          "     shear  "         484 

Time,  standard  unit  of 4 

"      of  vibration,  simple  pendulum 138 

Torsion 5°9 

"       and  flexure  combined 512 

work  of 512 

Translating  axis,  conservation  of  moment  of  mo- 
mentum    369 

•'  "     kinetic  energy 370 

"  "     moments  of  effective  forces... .   367 

"  "     moment  of  momentum 368 

"     momentum 368 

Translation  and  rotation 153,  389 

Transmission  of  power  by  shafts 512 

Triangle  and  polygon  of  displacements 54.  56 

'•          "  "         "   forces 176 

Twisting  moment 510 

ULTIMATE  strength 475 

Uniform  acceleration 100 

motion  in  curve 103 

"                                       "       on  inclined  plane. ..  136 

"         strength,  beams  of 503 

and  variable  acceleration 78 

"           "            "        velocity 65 

Unit  of  angle 6 

"     "   conical  angle » 7 

|    "  "  density 16 

"  derived 2 

"  offeree 170 

"  "       "     moment 181 

"  "   length 4 

"  "   mass 4.14 

"  "       "    astronomical 206 

"  "   time 4 

"  "   work 261 


Unstable  equilibrium  of  a  particle 277 

VALUES  of  g ico 

Variable  and  uniform  acceleration 78,  109 

velocity 65 

Vector  quantity,  moment  of 88 

Velocity  along  axis  of  rotation 155 

"        angular,  composition  and  resolution  of     68 
"         rectangular  components  of. ...     69 

linear  and  angular 61 

linear  in  terms  of  angular 70 

"        moment  of. 89 

"        normal  to  axis  of  rotation 155 

"        of  centre  of  mass 298 

rectangular  components  of 67 

resolution  and  composition  of 66 

resultant 67,69 

Virtual  work 215 

Volume,  material. . .    23 

WALI 424 

"     design  of  high 429 

"low 429 

"     retaining 454 

"     stability  of 426 

Water-pressure 430 

Wave-        "         432 

Weight  of  a  body 170 

"       and  friction  of  masonry 424 

Weights  and  measures 8 

Wheels,  friction 230 

Wire  ropes 235 

Work 215,  260 

"     of  axle-friction 266 

"     "   bending 521 

'     principle  of  least 311,  517 

'     rate  of 262 

"     of  resultant 215 

"     "   resilience 516 

"     "   shearing  force  in  beams 554 

"     "   straining 515 

"     "   torsion -. 512 

"     unit  of 261 

"     virtual 217 

Working  stress  481 

Wrench 192 

"       resultant 194 

"       screw 197 


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